the contest problem book ix 2001-2007

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IXIThe Contest Problem Book IX

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2001-2007

Compiled and Edited by David Wells and J. Douglas Faires

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The Contest ProblemBook IX

American Mathematics Competitions(AMC 12)2001-2007

© 2008 byThe Mathematical Association of America (Incorporated)

Library of Congress Catalog Card Number 2008920907

ISBN 978-0-88385-826-4

Printed in the United States of America

Current Printing (last digit):10987654321

The Contest ProblemBook IX

American Mathematics Competitions(AMC 12)2001-2007

Compiled and Edited byDavid Wells & J. Douglas Faires

Published and distributed byThe Mathematical Association of America

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MAA PROBLEM BOOKS SERIES

Problem Books is a series of the Mathematical Association of America consist-ing of collections of problems and solutions from annual mathematical com-petitions; compilations of problems (including unsolved problems) specific toparticular branches of mathematics; books on the art and practice of problemsolving, etc.

Council on PublicationsPaul Zorn, Chair

Richard A. Gillman EditorRoger Nelsen

Mark SaulTatiana Shubin

The Contest Problem Book VII: American Mathematics Competitions, 1995-2000 Con-tests, compiled and augmented by Harold B. Reiter

The Contest Problem Book VIII: American Mathematics Competitions (AMC 10), 2000-2007 Contests, compiled and edited by J. Douglas Faires & David Wells

The Contest Problem Book IX: American Mathematics Competitions (AMC 12), 2001-2007 Contests, compiled and edited by David Wells & J. Douglas Faires

A Friendly Mathematics Competition: 35 Years of Teamwork in Indiana, edited by RickGillman

The Inquisitive Problem Solver, Paul Vaderlind, Richard K. Guy, and Loren C. Larson

International Mathematical Olympiads 1986-1999, Marcin E. Kuczma

Mathematical Olympiads 1998-1999: Problems and Solutions From Around the World,edited by Titu Andreescu and Zuming Feng

Mathematical Olympiads 1999-2000: Problems and Solutions From Around the World,edited by Titu Andreescu and Zuming Feng

Mathematical Olympiads 2000-2001: Problems and Solutions From Around the World,edited by Titu Andreescu, Zuming Feng, and George Lee, Jr.

The William Lowell Putnam Mathematical Competition Problems and Solutions: 1938-1964, A. M. Gleason, R. E. Greenwood, L. M. Kelly

The William Lowell Putnam Mathematical Competition Problems and Solutions: 1965-1984, Gerald L. Alexanderson, Leonard F. Klosinski, and Loren C. Larson

The William Lowell Putnam Mathematical Competition 1985-2000: Problems, Solu-tions, and Commentary, Kiran S. Kedlaya, Bjorn Poonen, Ravi Vakil

USA and International Mathematical Olympiads 2000, edited by Titu Andreescu andZuming Feng




USA and International Mathematical Olympiads 2004, edited by Titu Andreescu, Zum-ing Feng, and Po-Shen Loh

MAA Service CenterP. 0. Box 91112

Washington, DC 20090-11121-800-331-1622 fax: 1-301-206-9789

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Contents

Preface ............................................................ ixA Brief History of the Contests . . . . . . . . . . . . . . . . . . . ix

Construction of the Contests . . . . . . . . . . . . . . . . . . . . . xii

Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . xv

Problems .......................................................... 1

2001 AMC 12 Problems . . . . . . . . . . . . . . . . . . . . . . . 1

2002 AMC 12A Problems . . . . . . . . . . . . . . . . . . . . . . 8

2002 AMC 12B Problems . . . . . . . . . . . . . . . . . . . . . . 13

2003 AMC 12A Problems . . . . . . . . . . . . . . . . . . . . . . 17

2003 AMC 12B Problems . . . . . . . . . . . . . . . . . . . . . . 22

2004 AMC 12A Problems . . . . . . . . . . . . . . . . . . . . . . 28

2004 AMC 12B Problems . . . . . . . . . . . . . . . . . . . . . . 33

2005 AMC 12A Problems . . . . . . . . . . . . . . . . . . . . . . 38

2005 AMC 12B Problems . . . . . . . . . . . . . . . . . . . . . . 43

2006 AMC 12A Problems . . . . . . . . . . . . . . . . . . . . . . 47

2006 AMC 12B Problems . . . . . . . . . . . . . . . . . . . . . . 53

2007 AMC 12A Problems . . . . . . . . . . . . . . . . . . . . . . 57

2007 AMC 12B Problems . . . . . . . . . . . . . . . . . . . . . . 61

Solutions .......................................................... 652001 AMC 12 Solutions . . . . . . . . . . . . . . . . . . . . . . . 662002 AMC 12A Solutions . . . . . . . . . . . . . . . . . . . . . . 78

2002 AMC 12B Solutions . . . . . . . . . . . . . . . . . . . . . . 86

2003 AMC 12A Solutions . . . . . . . . . . . . . . . . . . . . . . 95

2003 AMC 12B Solutions . . . . . . . . . . . . . . . . . . . . . . 106

2004 AMC 12A Solutions . . . . . . . . . . . . . . . . . . . . . . 117

2004 AMC 12B Solutions . . . . . . . . . . . . . . . . . . . . . . 126

2005 AMC 12A Solutions . . . . . . . . . . . . . . . . . . . . . . 135

2005 AMC 12B Solutions . . . . . . . . . . . . . . . . . . . . . . 145

vii

VIII

2006 AMC 12A Solutions . . . . . . . . . . . . . . . . . . . . . . 1572006 AMC 12B Solutions . . . . . . . . . . . . . . . . . . . . . . 169

2007 AMC 12A Solutions . . . . . . . . . . . . . . . . . . . . . . 1802007 AMC 12B Solutions . . . . . . . . . . . . . . . . . . . . . . 189

Index of Problems ................................................. 199

About the Editors ................................................. 213

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Preface

A Brief History of the Contests

Overview

The contests from which the problems in this book were drawn are direct de-scendants of one created in 1950 by the Metropolitan New York Section of theMathematical Association of America (MAA). Originally given only to stu-dents in the New York City area, the contests were first offered nationally in1957 under the sponsorship of the MAA and the Society of Actuaries. The pri-mary objective of the contests was, and still is, to promote the study of mathe-matics by providing high school students with a positive experience in creativeproblem solving. A secondary objective of identifying mathematically talentedstudents was introduced in 1972, when about 100 top scorers were invited toparticipate in the USA Mathematical Olympiad (USAMO), an extremely chal-lenging proof-oriented contest.

The 1950 contest was entitled simply, "Mathematical Contest." The word"Annual" was added to the name in 1953. The solution booklet from 1959 istitled, "Annual H.S. Mathematics Examination." Between 1959 and 1982 thetitles of the contest and solution booklets used either "Contest" or "Examina-tion," the latter appearing more frequently in the later years. The words "HighSchool" or the initials H.S. were sometimes, but not always, included in thetitle. In 1983 the contest officially became the American High School Mathe-matics Examination (AHSME). In the same year a third contest, the AmericanInvitational Mathematics Examination (AIME), was introduced as a steppingstone between the AHSME and the USAMO. Computational in nature, butrequiring more creativity than the AHSME, the AIME was deemed a morereliable vehicle for identifying suitable USAMO participants. Two years latera contest for middle school students was created. The American Junior HighSchool Mathematics Examination (AJHSME) was modeled after the AHSME.Shortly thereafter, the entire program of contests became known as the Amer-ican Mathematics Competitions (AMC).

For reasons that will be mentioned later, it was decided to create a newcontest in 2000, limited to students in grades 10 and below. At that time the

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x The Contest Problem Book IX

AHSME and AJHSME were renamed the AMC 12 and AMC 8, respectively,and the new contest was called the AMC 10.

There was some discussion within the Committee on American Mathemat-ics Competitions (CAMC) about whether top scorers on the AMC 10 shouldbe invited to take the AIME. The argument on one side was that 10th-gradersbright enough to qualify for the AIME should be taking the AMC 12. The op-posing argument was that a 10th-grader whose previously undiscovered talentis first revealed by a high AMC 10 score should not be denied the opportunityto take the AIME. It was finally decided that AIME invitations would go tothose students who obtained a score of 120 out of 150, or else scored in the top1%, on the AMC 10. The corresponding criteria for the AMC 12 are a score of100 or placement in the top 5%.

Since 2002 two versions of both the AMC 10 and the AMC 12 have beencreated every year and offered about two weeks apart. This has been done fortwo reasons. The immediate impetus for change was a conflict between the2002 contest date and state holidays in Illinois and Louisiana, which wouldhave precluded participation by students in both states. The new arrangementalso addressed a growing concern among CAMC members about contest se-curity. In the past, contests were administered for official credit at any timeduring a window of a few days. With the advent of contest-focused web sites,it was recognized that contest answers would become public knowledge withinhours. Setting a specific date for each form greatly reduces the probability thata student will learn the answers before taking the contest.

The AMC contests have attracted many new sponsors since the early years.In addition to the MAA and the Society of Actuaries, the list of sponsoringorganizations grew to include the high school and two-year college honorarymathematics society Mu Alpha Theta, the National Council of Teachers ofMathematics, and the Casualty Actuarial Society by 1971. There are currentlymore than 21 Contributors and Sponsors of the contests, which demonstratesthe wide range of interest in the program.

Format and Scoring

Like their common ancestor, the AHSME, the AMC 10 and AMC 12 aremultiple-choice contests with five answer choices for each question. The timeallotted for work was initially 80 minutes, increasing to 90 minutes in 1978and decreasing to 75 minutes in 2000. Although the highest possible score onthe contests has always been 150, there have been numerous changes in thenumber of questions and the scoring formulas used over the years. The firstcontests had 50 questions, divided into three sections with increasing levels ofdifficulty. The number of questions in Parts I, II, and III, was 15 (2 points each),20 (3 points each), and 15 (4 points each), respectively. In 1958 a 20/20/10

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division was used with a 2/3/5 scoring. In 1960 the number of questions wasreduced to 40, with a 20/10/10 division and a 3/4/5 scoring. In 1968 the num-ber of questions was further reduced to 35, with a 10/10/10/5 division and a3/4/5/6 scoring. Beginning in 1956, in an effort to discourage random guessingwhile still encouraging educated guessing, a penalty for incorrect responseswas imposed. A student's score was calculated as PR - 0.2Ptt. (changed toPR - 0.25Ptt. in 1958), where PR and Ptt. represent the total point values ofquestions with correct and incorrect responses.

Beginning in 1974 the contest was no longer partitioned into sections. Thecontest consisted of 30 5-point questions with the scoring formula 5R - W,where R and W represent the number of questions with correct and incorrectresponses. This formula was changed to 30 + 4R - W in 1977. The latterformula is neutral with respect to guessing and precludes the possibility ofa negative score, a potentially devastating result for an unlucky student! Re-newed concern about random guessing prompted another scoring change in1986, this time to the formula 5R + 2B, where B is the number of responsesleft blank. The number of questions was reduced once more in 2000, result-ing in a contest of 25 questions each worth 6 points with the scoring formula6R + 2B, altered to 6R + 2.5B in 2002. The last change, instituted in part toreduce the probability of ties among award recipients, appears to have discour-aged students from attempting the more difficult problems. For that reason, thescoring formula changed again in 2007, to 6R + 1.5B.

Content of the AMC 10 and AMC 12

The scope of the AMC 12 contest is restricted to topics covered in a stan-dard high school curriculum. This seemingly simple guideline leaves signifi-cant room for debate. For example, does it exclude a problem whose solutionrequires Pick's Theorem or the formula for the area of an ellipse? The listof topics considered acceptable for the contest has also changed with time.Probability did not make its first appearance on the contest until 1970, andtrigonometry not until 1972. Although many AMC 12 contestants have stud-ied calculus, the CAMC does not consider it to be a standard high school topicat this time.

During the early years of the AHSME, many problems were closely relatedto routine exercises that students were likely to have done in their classes. Thatis less the case today, for two reasons. First, in the process of reducing the num-ber of questions on the contest from the original 50 to the present 25, it was de-sired to maintain a large number of problems whose solutions require creativeinsight. Consequently, the number of routine problems on each contest hasbeen greatly reduced. Second, beginning in 1994 students were allowed to usecalculators on the AHSME. Calculators, including some models with built-in

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computer algebra systems, have been permitted on the AMC 8, AMC 10 andAMC 12, although not on the AIME or USAMO. Thus, for example, a prob-lem that asks students only to solve an algebraic equation is not appropriate fora calculator-permitted AMC 12. Many good problems involving trigonometryand logarithms are also rendered trivial by calculators. The problems includedin this volume were all on calculator-permitted contests, but beginning in 2008calculators will not be permitted on either the AMC 10 or the AMC 12 con-tests.

By the late 1990s it was clear that many students in grades 9 and 10 wereunable to find many approachable problems on the AHSME. This situation leddirectly to the creation of the AMC 10. The scope of that contest is restrictedto topics that are typically studied in grades 10 and below. The majority ofthe problems on the AMC 10 contests involve topics from algebra, geometry,probability, and elementary counting and number theory. Specifically excludedfrom this contest are problems that involve trigonometry, logarithms, complexnumbers, and the concept of functions and properties of their graphs. Problemsthat cover more advanced topics in geometry are also excluded. For example,those that require Heron's formula for finding the area of a triangle, extensionsof the Central Angle Theorem, and problems requiring the Power of a PointTheorem are excluded.

Construction of the Contests

The Process

The period between the end of January and mid-February of each year seesthe administration of four AMC contests. The two forms of the AMC 12 con-test, referred to as the 12A and 12B, are given about two weeks apart. The twoforms must be comparable in terms of difficulty and topical balance, yet notso similar as to give an advantage to students taking the later 12B. The AMC10A and lOB contests are each given on the same day as their AMC 12 coun-terparts. This arrangement allows the same question to appear on both contestsgiven on the same day, and in fact there is usually an overlap of 40-50% be-tween corresponding forms of the AMC 10 and AMC 12. Because of thoseconsiderations, the content of each contest is inextricably linked to the contentof the other three, so that all the AMC 10 and 12 contests must be constructedtogether.

For the contests given in year n, the process begins early in year n - 2,when a call goes out to a panel of about 50 problem posers. By early summera packet of 150 to 200 problems is sent out to a panel of packet reviewers,many of whom are also problem posers. Each reviewer submits a ballot of fa-vorite problems from the packet and a set of comments, including corrections,

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alternate solutions, and suggestions for improved wording. A tentative draft ofeach contest is written in the fall and sent out to a panel of draft reviewers forediting. In January of year n - 1, a second draft of the contests is sent to mem-bers of the AMC 10/12 Subcommittee, who meet to discuss it in February. Itis common for many of the problems and solutions to be significantly rewrit-ten at the meeting. In addition, the Subcommittee often reorders problems,shifts them from one contest to another, or replaces them entirely. Followingthe meeting, another draft of each contest is written. Over the next few monthsthat draft undergoes additional reviews, and the contest galleys are printed atthe AMC office and sent out for proofreading. The contests are printed duringthe summer of year n - 1. By that time the problem packet for year n + 1 is

being reviewed, and the process continues.Performance statistics over the last several years indicate that we have been

reasonably successful in judging the relative difficulty of the A and B formsof each contest. Average scores on the two forms have usually been within afew points of each other. We have not always been as successful in judging therelative difficulty of individual problems. For example, Problem 20 on the 2005AMC 12B had a higher rate of correct responses (42.8%) than did Problem 6(31.3%).

In spite of our best efforts, ambiguities and errors occasionally creep intothe final product. Fortunately, we have yet to have a difficulty on the AMC 10contests, but on the 2002 12A contest, the distracter "N < 1" was inadver-tently written as "N > 1", resulting in two correct answer choices.

The Problems

Of the problems that are submitted in each year's problem packet, fewer thanhalf survive the reviewing and editing process. Each problem that ultimatelyappears on one of the contests must meet several criteria.

It must be original. The easier problems are often somewhat similar totextbook exercises. However, if a problem or a nearly identical one hasbeen used in another contest or published elsewhere, it must be rejected.

The best students should be able to make a reasonable attempt to solveall 25 problems within the 75-minute time frame, so problem statementsshould be brief, and questions should be simple to state. The higher-numbered problems should be difficult by virtue of the insight requiredfor the solution, not the length of the computations.

It must be possible to state the problem unambiguously.

The correct answer must not be easily guessable.

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Whenever possible, the wrong answer choices, or distracters, should reflectanswers that result from common errors in thinking.

If a problem has a solution that uses calculus, it must have an equally easysolution that does not.

Calculators should not provide students with a significant advantage insolving any problem.

The distracters should trap students who think carelessly, but not those whosimply read carelessly. For example, the answer obtained by a student whomisreads "x ? 0" as "x < 0" should usually not be included as a distracter. Anumber of top students missed Problem 3 on the 2004 AMC 1OA (Problem Ion 2004 AMC 12A) contest because they read the problem hastily, expressedtheir answers in dollars instead of cents, and found the answer they obtainedas one of the choices.

It is often difficult to decide what sort of calculator usage constitutes a"significant advantage". A problem that asks only for the value of

20072 - 2006.2008

is clearly not suitable, but if the value must be calculated as one step in the so-lution of a more difficult problem, the advantage provided by a calculator maybe judged to be insignificant. Solutions by programming raise unique issues.For example, if a problem requires the 20071h term of a sequence, has a stu-dent gained a significant advantage by programming a calculator to generatesuccessive terms of the sequence? Here the answer depends on the nature ofthe problem. In some cases it has been decided that the creativity required towrite the program is comparable to that required to solve the problem withouta calculator.

Conventions in Stating Problems

Contest problems should be stated precisely, but not at the expense of clarity.Thus, for example, the choice of a point in an interval according to a uniformprobability distribution is referred to simply as a choice made "at random", and"divisor" is usually substituted for the more precise "positive integer divisor".It is usually understood that triangles cannot have an area of zero, althoughthe phrase "triangle with positive area" has occasionally appeared in problemsinvolving the selection of three points from a lattice of points.

In stating geometry problems, the AMC 10 and AMC 12 contests bridgea stylistic gap between the AMC 8, where problems often consist of only adiagram and a question, and the AIME, in which students must usually con-struct their own diagrams from a prose description. Problem statements on the

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AMC 12 are nearly always diagram-independent, although diagrams are of-ten provided in order to save time when the construction of the diagram isnot deemed to be a significant component of the solution. By contrast, thephrase "as shown" appears more frequently in AMC 10 problem statementsthat would otherwise become lengthy.

Problems in a Physical Context

Problems can often be made more interesting by putting them into a physicalcontext, but attempts to do so occasionally backfire. The problem packet for the2003 contests contained a submission that read, "Jay wants to cut his heatingcosts. He takes four wooden cubes 3 feet on each side, and places each of thefour at one of the top corners of his 15 foot by 10 foot by 8 foot office. By whatpercentage is the airspace in the room reduced?" After a reviewer calculatedthe probable weight of each cube, the wood was changed to styrofoam, but asecond reviewer then pointed out that the resulting dimensions of the officewould make it useless as a workspace. In its third incarnation, the submissionappeared as Problem 3 on the 2003 12A contest.

Problems arising from real-life situations can be especially interesting, butit is usually difficult to describe the problem-inspiring situation concisely. Theidea for Problem 3 on the 2006 AMC 12B came from a radio broadcast of aPitt-Nebraska football game, in which an announcer said, "After a first half inwhich 34 points were scored, it's been a scoreless third quarter, so now we goto the fourth quarter with the Panthers still trailing by 14." After a moment'sthought, the problem poser realized that he had enough information to deter-mine the score. However, it was decided that the unneeded information aboutquarters might be confusing to students unfamiliar with American football.

AcknowledgmentsThis book consists of a collection of problems submitted by a large group ofpeople. Quite often a submitted problem will be edited so that it might actuallybe unrecognizable to the original author, but it is the germ of the idea for thefinal problem that is so valuable for those creating the final contests. We havea complete record of the source of the problems since 2003, and are able todetermine the originator of most of those on the earlier contests, but we apol-ogize in advance for any names that have been inadvertently omitted. Thosewho have submitted the problems that make these contests possible include:

Bernardo Abrego, Titu Andreescu, Scott Annin, Sam Baethge, CraigBailey, John Benson, Arapad Benyi, Tom Butts, Barbara Clanton, GregCrow, Steven Davis, Steve Dunbar, Tim Eckert, Jacek Fabrykowski,

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xvi The Contest Problem Book IX

Zuming Feng, Sylvia Fernandez, Micah Fogel, Sr. Josanne Furey, DickGibbs, Darren Glass, David Hankin, John Haverhals, B. Henry, JohnJensen, Chris Jeuell, Elgin Johnston, Paul Karafiol, Dan Kennedy, JoeKennedy, Thomas Koshy, Bonnie Leitch, Kurt Ludwick, Glenn Mart,John Merzel, Thomas Mildorf, Roland Minton, Roger Patterson, PetePedersen, Krassimir Penev, Tom Reardon, Harold Reiter, Cecil Rousseau,Vince Schielack, Kelly Schultz, Peter Shiue, Charlyn Shepherd, DanStronger, Zoran Sunik, Dan Ullman, Pat Vennebush, Don Vestal, WilliamWardlaw, Steve Weissburg, LeRoy Wenstrom, and Ron Yannone.

We would like to add a special thanks to Steve Dunbar and the staff atAMC Headquarters for making life easier for us in numerous ways. Finally, wegreatly appreciate the work of Carrie Davis, a student assistant at YoungstownState University, for doing much of the editing work on the manuscript.

Dave Wells Doug FairesPennsylvania State University Youngstown State Universityat New Kensington [email protected]@psu.edu

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Problems

2001 AMC 12 ProblemsThe solutions for the 2001 AMC 12 begin on page 66.

1. The sum of two numbers is S. Suppose 3 is added to each number and theneach of the resulting numbers is doubled. What is the sum of the final twonumbers?

(A) 2S + 3 (B) 3S + 2 (C) 3S + 6 (D) 2S + 6(E) 2S + 12

2. Let P(n) and S(n) denote the product and the sum, respectively, of thedigits of the integer n. For example, P(23) = 6 and S(23) = 5. SupposeN is a two-digit number such that N = P(N) + S(N). What is the unitsdigit of N?

(A) 2 (B) 3 (C) 6 (D) 8 (E) 9

3. The state income tax where Kristin lives is levied at the rate of p% of thefirst $28000 of annual income plus (p + 2)% of any amount above $28000.Kristin noticed that the state income tax she paid amounted to (p + 0.25)%of her annual income. What was her annual income?

(A) $28000 (B) $32000 (C) $35000 (D) $42000

(E) $56000

4. The mean of three numbers is 10 more than the least of the numbers and 15less than the greatest. The median of the three numbers is 5. What is theirsum?

(A) 5 (B) 20 (C) 25 (D) 30 (E) 36

5. What is the product of all positive odd integers less than 10000?10000! 10000! 9999!

(A)(5000!)2 (B) 25000 (C) 25000

10000! 5000!(D) 25000.5000! (E) 25000

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6. A telephone number has the form ABC-DEF-GHIJ, where each letter rep-resents a different digit. The digits in each part of the number are in de-creasing order; that is, A > B > C, D > E > F, and G > H > I > J.Furthermore, D, E, and F are consecutive even digits; G, H, I, and J areconsecutive odd digits; and A + B + C = 9. What is A?

(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

7. A charity sells 140 benefit tickets for a total of $2001. Some tickets sellfor full price (a whole dollar amount), and the rest sell for half price. Howmuch money is raised by the full-price tickets?

(A) $782 (B) $986 (C) $1158 (D) $1219 (E) $1449

8. Which of the cones below can be formed from a 252° sector of a circle ofradius 10 by aligning the two straight sides?

(C)

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2001 AMC 12 Problems 3

9. Let f be a function satisfying f (xy) = f (x)/y for all positive real num-bers x and y. If f (500) = 3, what is the value of f (600)?

5 18(A) 1 (B) 2 (C) (D) 3 (E) -

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10. The plane is tiled by congruent squares and congruent pentagons as indi-cated. Which of the following is closest to the percent of the plane that isenclosed by the pentagons?

(A) 50 (B) 52 (C) 54 (D) 56 (E) 58

1 2 3 4 5 6 7

11. A box contains exactly five chips, three red and two white. Chips are ran-domly removed one at a time without replacement until all the red chipsare drawn or all the white chips are drawn. What is the probability that thelast chip drawn is white?

3 21 (D)

3 7(A) (B) (C) (E)

10 5 2 5 10

12. How many positive integers not exceeding 2001 are multiples of 3 or 4 butnot 5?

(A) 768 (B) 801 (C) 934 (D) 1067 (E) 1167

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13. The parabola with equation y = ax2 + bx + c and vertex (h, k) is reflect-ed about the line y = k. This results in the parabola with equation y =dx2 + ex + f. Which of the following equals a + b + c + d + e + f?

(A) 2b (B) 2c (C) 2a + 2b (D) 2h (E) 2k

14. Given the nine-sided regular polygon Al A2A3A4A5A6A7A8A9, how manydistinct equilateral triangles in the plane of the polygon have at least twovertices in the set {A1, A2, ... A9)?

(A) 30 (B) 36 (C) 63 (D) 66 (E) 72

15. An insect lives on the surface of a regular tetrahedron with edges of length1. It wishes to travel on the surface of the tetrahedron from the midpointof one edge to the midpoint of the opposite edge. What is the length of theshortest such trip? (Note: Two edges of a tetrahedron are opposite if theyhave no common endpoint.)

(B) 1 (C)3

(D)2

(E) 2

16. A spider has one sock and one shoe for each of its eight legs. In how manydifferent orders can the spider put on its socks and shoes, assuming that,on each leg, the sock must be put on before the shoe?

(A) 8! (B) 28 8! (C) (8!)2 (D) 2 (E) 16!

17. A point P is selected at random from the interior of the pentagon withvertices A = (0, 2), B = (4, 0), C = (2n + 1, 0), D = (2,-r + 1, 4), andE = (0, 4). What is the probability that LAPB is obtuse?

(A)1 (B) 1 (C) 5 (D) 3 (E) 1

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18. A circle centered at A with a radius of 1 and a circle centered at B with aradius of 4 are externally tangent. A third circle is tangent to the first twoand to one of their common external tangents as shown. What is the radiusof the third circle?

(A) 1 (B)2

(C) 5 (D)4

(E) 13 5 12 9 2

19. The polynomial P(x) = x3 +ax2 +bx +c has the property that the meanof its zeros, the product of its zeros, and the sum of its coefficients are allequal. If the y-intercept of the graph of y = P(x) is 2, what is b?

(A) -11 (B) -10 (C) -9 (D) 1 (E) 5

20. Points A = (3, 9), B = (1, 1), C = (5, 3), and D = (a, b) lie in the firstquadrant and are the vertices of quadrilateral ABCD. The quadrilateralformed by joining the midpoints of AB, BC, CD, and DA is a square.What is the sum of the coordinates of point D?

(A) 7 (B) 9 (C) 10 (D) 12 (E) 16

21. Four positive integers a, b, c, and d have a product of 8! and satisfy

ab + a + b = 524,be + b + c = 146, and

cd+c+d=104.

6 Problems

What is a - d?

(A) 4 (B) 6 (C) 8 (D) 10 (E) 12

22. In rectangle ABCD, points F and G lie on AB so that AF = FG = GBand E is the midpoint of DC. Also, AC intersects EF at H and EG at J.The area of rectangle ABCD is 70. What is the area of A EHJ?

5 35(B)(A)

2 12

D

A

(C) 3 (D)7

E

35(E)

8

C

n,

GFB

23. A polynomial of degree four with leading coefficient I and integer coeffi-cients has two real zeros, both of which are integers. Which of the follow-ing can also be a zero of the polynomial?

(A)

(E)

l+i2

11(B) 12 (C)

2

+ i (D) 1 + 2

l+i 13

2

24. In triangle ABC, LABC = 45°. Point D is on BC so that 2. BD = CDand /DAB = 15°. What is /ACB?

(A) 54° (B) 60° (C) 72° (D) 75° (E) 90°

OS'


25. Consider sequences of positive real numbers of the form x, 2000, y.....in which every term after the first is 1 less than the product of its twoimmediate neighbors. For how many different values of x does the term2001 appear somewhere in the sequence?

(A) 1 (B) 2 (C) 3 (D) 4 (E) more than 4

CA

D

--r

8 Problems

2002 AMC 12A ProblemsThe solutions for the 2002 AMC 12A begin on page 78.

1. What is the sum of all the roots of (2x + 3) (x - 4) + (2x + 3)(x - 6) = 0?

(A)2

(B) 4 (C) 5 (D) 7 (E) 13

2. Cindy was asked by her teacher to subtract 3 from a certain number andthen divide the result by 9. Instead, she subtracted 9 and then divided theresult by 3, giving an answer of 43. What would her answer have been hadshe worked the problem correctly?

(A) 15 (B) 34 (C) 43 (D) 51 (E) 138

3. According to the standard convention for exponentiation,

22(22)

22 = 2(2

= 216 = 65, 536.

If the order in which the exponentiations are performed is changed, howmany other values are possible?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

4. What is the degree measure of an angle whose complement is 25% of itssupplement?

(A) 48 (B) 60 (C) 75 (D) 120 (E) 150

5. Each of the small circles in the figure has radius one. The innermost circleis tangent to the six circles that surround it, and each of those circles istangent to the large circle and to its small-circle neighbors. What is thearea of the shaded region?

(A) r (B) 1.57r (C) 2 r (D) 37r (E) 3.5ir

6. For how many positive integers m does there exist at least one positiveinteger n such that m n < m + n?

(A) 4 (B) 6 (C) 9 (D) 12 (E) infinitely many

001

't7

2002 AMC 12A Problems 9

7. An arc of 45° on circle A has the same length as an arc of 30° on circle B.What is the ratio of the area of circle A to the area of circle B?

(A)4 2

(B)

?(C) 5 (D) 3 (E) 9

9 3 6 2 4

8. Betsy designed a flag using blue triangles ( ), small white squares (0),and a red center square as shown. Let B be the total area of the bluetriangles, W the total area of the white squares, and R the area of the redsquare. Which of the following is correct?

(A) B = W (B) W = R (C) B = R (D) 3B = 2R(E) 2R = W

9. Jamal wants to store 30 computer files on floppy disks, each of which hasa capacity of 1.44 megabytes (mb). Three of his files require 0.8 mb ofmemory each, 12 more require 0.7 mb each, and the remaining 15 require0.4 mb each. No file can be split between floppy disks. What is the minimalnumber of floppy disks that will hold all the files?

(A) 12 (B) 13 (C) 14 (D) 15 (E) 16

10. Sarah pours four ounces of coffee into an eight-ounce cup and four ouncesof cream into a second cup of the same size. She then transfers half thecoffee from the first cup to the second and, after stirring thoroughly, trans-fers half the liquid in the second cup back to the first. What fraction of theliquid in the first cup is now cream?

(A) 1 (B) 1 (C) 3 (D) ? (E)4 3 8 5 2

11. Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. everymorning. When he averages 40 miles per hour, he arrives at his workplacethree minutes late. When he averages 60 miles per hour, he arrives threeminutes early. At what average speed, in miles per hour, should Mr. Birddrive to arrive at his workplace precisely on time?

(A) 45 (B) 48 (C) 50 (D) 55 (E) 58

12. Both roots of the quadratic equation x2 - 63x + k = 0 are prime numbers.How many values of k are possible?

(A) 0 (B) 1 (C) 2 (D) 4 (E) more than four

1NN

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10 Problems

13. Two different positive numbers a and b each differ from their reciprocalsby 1. What is a + b?

(A) 1 (B) 2 (C) (D) 66- (E) 3

14. For all positive integers n, let f (n) = 1092002 n2. Let

N = f(11)+f(13)+f(14).

Which of the following relations is true?

(A) N<1 (B) N=1 (C) 1<N<2 (D) N = 2(E) N>2

15. The mean, median, unique mode, and range of a collection of eight integersare all equal to 8. What is the largest integer that can be an element of thiscollection?

(A) 11 (B) 12 (C) 13 (D) 14 (E) 15

16. Tina randomly selects two distinct numbers from the set { 1, 2, 3, 4, 5}, andSergio randomly selects a number from the set { 1 , 2, ... , 10}. What is theprobability that Sergio's number is larger than the sum of the two numberschosen by Tina?

(A) (B) (C) (D) (E)5 20 2 20 25

17. Several sets of prime numbers, such as {7, 83, 421, 659}, use each of thenine nonzero digits exactly once. What is the smallest possible sum such aset of primes could have?

(A) 193 (B) 207 (C) 225 (D) 252 (E) 477

18. Let C1 and C2 be circles defined by

and

(x-10)2+y2=36

(x+15)2+y2= 81,

respectively. What is the length of the shortest line segment PQ that istangent to Cl at P and to C2 at Q?

(A) 15 (B) 18 (C) 20 (D) 21 (E) 24

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19. The graph of the function f is shown below. How many solutions does theequation f (f (x)) = 6 have?

(A) 2 (B) 4 (C) 5 (D) 6 (E) 7

20. Suppose that a and b are digits, not both 9 and not both 0, and the repeatingdecimal 0.ab is expressed as a fraction in lowest terms. How many differentdenominators are possible?

(A) 3 (B) 4 (C) 5 (D) 8 (E) 9

2 1 . Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6. .... For n > 2, thenth term of the sequence is the units digit of the sum of the two previousterms. Let S denote the sum of the first n terms of this sequence. What isthe smallest value of n for which S > 10, 000?

(A) 1992 (B) 1999 (C) 2001 (D) 2002 (E) 2004

22. Triangle ABC is a right triangle with LACB as its right angle, rn/ABC =60°, and AB = 10. Let P be randomly chosen inside L ABC, and extendBP to meet AC at D. What is the probability that BD > 5,/2-?

(A) 2- (B) 1 (C) 3 - (D) 1 (E) 5- Vs2 3 3 2 5

G1.

12 Problems

23. In triangle ABC, side AC and the perpendicular bisector of BC meet inpoint D, and BD bisects LABC. If AD = 9 and DC = 7, what is thearea of triangle ABD?

(A) 14 (B) 21 (C) 28 (D) 14,/5- (E) 28v24. What is the number of ordered pairs of real numbers (a, b) such that

(a + bi)2002 = a - bi?

(A) 1001 (B) 1002 (C) 2001 (D) 2002 (E) 2004

25. The nonzero coefficients of a polynomial P with real coefficients are allreplaced by their mean to form a polynomial Q. Which of the followingcould be a graph of y = P (x) and y = Q (x) over the interval -4 < x <4?

(C) (D)

(E)

+-0

,y'

,.p

2002 AMC 12B Problems 13

2002 AMC 12B ProblemsThe solutions for the 2002 AMC 12B begin on page 86.

1. The arithmetic mean of the nine numbers in the set {9,99,999,9999, ... ,999999999} is a 9-digit number M, all of whose digits are distinct. Whichof the following digits does not appear in the number M?

(A) 0 (B) 2 (C) 4 (D) 6 (E) 8

2. What is the value of

(3x - 2)(4x + 1) - (3x - 2)4x + 1

when x = 4?

(A) 0 (B) 1 (C) 10 (D) 11 (E) 12

3. For how many positive integers n is n2 - 3n + 2 a prime number?

(A) none (B) one (C) two (D) more than two, but finitely many

(E) infinitely many

4. Let n be a positive integer such that + 3 + + n is an integer. Whichof the following statements is not true?

(A) 2 divides n (B) 3 divides n (C) 6 divides n (D) 7 divides n

(E) n > 84

5. Let v, w, x, y, and z be the degree measures of the five angles of a pen-tagon. Suppose v < w < x < y < z and v, w, x, y, and z form anarithmetic sequence. What is x?

(A) 72 (B) 84 (C) 90 (D) 108 (E) 120

6. Suppose that a and b are nonzero real numbers, and that the equationx2 + ax + b = 0 has solutions a and b. What is the pair (a, b)?

(A) (-2, 1) (B) (-1, 2) (C) (1, -2) (D) (2, -1) (E) (4, 4)

7. The product of three consecutive positive integers is 8 times their sum.What is the sum of their squares?

(A) 50 (B) 77 (C) 110 (D) 149 (E) 194

8. Suppose July of year N has five Mondays. Which of the following mustoccur five times in August of year N? (Note: Both months have 31 days.)

(A) Monday (B) Tuesday (C) Wednesday (D) Thursday(E) Friday

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Ir.

14 Problems

9. Let a, b, c, and d be positive real numbers such that a, b, c, d form an in-creasing arithmetic sequence and a, b, d form a geometric sequence. Whatis a/d?

(A) 12 (B) 6 (C) 4 (D) 3 (E) 2

10. How many different integers can be expressed as the sum of three distinctmembers of the set 11, 4, 7, 10, 13, 16, 19}?

(A) 13 (B) 16 (C) 24 (D) 30 (E) 35

11. The positive integers A, B, A - B, and A + B are all prime numbers. Thesum of these four primes is which of the following?

(A) even (B) divisible by 3 (C) divisible by 5 (D) divisible by 7

(E) prime

12. For how many integers n is n/(20 - n) the square of an integer?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 10

13. The sum of 18 consecutive positive integers is a perfect square. What is thesmallest possible value of this sum?

(A) 169 (B) 225 (C) 289 (D) 361 (E) 441

14. Four distinct circles are drawn in a plane. What is the maximum number ofpoints where at least two of the circles intersect?

(A) 8 (B) 9 (C) 10 (D) 12 (E) 16

15. How many four-digit numbers N have the property that the three-digitnumber obtained by removing the leftmost digit is one ninth of N?

(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

16. Juan rolls a fair regular octahedral die marked with the numbers 1 through8. Then Amal rolls a fair six-sided die. What is the probability that theproduct of the two rolls is a multiple of 3?

7(A) 1 (B) 1 (C) 1 (D) ? (E)

2

12 3 2 12 3

17. Andy's lawn has twice as much area as Beth's lawn and three times asmuch area as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth'smower and one third as fast as Andy's mower. If they all start to mow theirlawns at the same time, who will finish first?

(A) Andy (B) Beth (C) Carlos (D) Andy and Carlos tie for first.

(E) All three tie.

I!'


18. A point P is randomly selected from the rectangular region with vertices(0, 0), (2, 0), (2, 1), and (0, 1). What is the probability that P is closer tothe origin than it is to the point (3, 1)?

l 2 3 4(A) 2 (B)

3(C)

4(D) 5 (E) 1

19. Let a, b, and c be positive real numbers such that a(b + c) = 152,b(c + a) = 162, and c(a + b) = 170. What is abc?

(A) 672 (B) 688 (C) 704 (D) 720 (E) 750

20. Let AXOY be a right triangle with mXXOY = 90°. Let M and N bethe midpoints of legs OX and 0 Y, respectively. Given that XN = 19 andYM = 22, what is X Y?

(A) 24 (B) 26 (C) 28 (D) 30 (E) 32

21. For all positive integers n less than 2002, let

11, if n is divisible by 13 and 14;

_ 13, if n is divisible by 14 and 11;an

14, if n is divisible by 11 and 13;

0, otherwise.

2001

What is Y, an?n=1

(A) 448 (B) 486 (C) 1560 (D) 2001 (E) 2002

22. For all integers n greater than 1, define an = (loge 2002)-1. Let b =a2+a3+a4+a5 andc =aio+all +a12+a13+a14. Whatisb-c?

(A) -2 (B) -1 (C)1

2002 (D)

1

1001 (E)

1

2

23. In AABC, we have AB = 1 and AC = 2. Side BC and the median fromA to BC have the same length. What is BC?

1 2(A) (B) I + (C) (D) 2 (E)2

24. A convex quadrilateral ABCD with area 2002 contains a point P in itsinterior such that PA = 24, PB = 32, PC = 28, and PD = 45. What isthe perimeter of ABCD?

(A) 4,/2002 (B) 2,/8465 (C) 2 (48 + 2002)

(D) 2,/8633 (E) 4 (36+ 113)

16 Problems

25. Let f (x) = x2 + 6x + 1, and let R denote the set of points (x, y) in thecoordinate plane such that

f(x) + f(Y) -< 0 and f(x) - f(Y) -0.

Which of the following is closest to the area of R?

(A) 21 (B) 22 (C) 23 (D) 24 (E) 25



1. What is the difference between the sum of the first 2003 even countingnumbers and the sum of the first 2003 odd counting numbers?

(A) 0 (B) 1 (C) 2 (D) 2003 (E) 4006

2. Members of the Rockham Soccer League buy socks and T-shirts. Sockscost $4 per pair, and each T-shirt costs $5 more than a pair of socks. Eachmember needs one pair of socks and a shirt for home games and anotherpair of socks and a shirt for away games. If the total cost is $2366, howmany members are in the League?

(A) 77 (B) 91 (C) 143 (D) 182 (E) 286

3. A solid box is 15 cm by 10 cm by 8 cm. A new solid is formed by removinga cube 3 cm on a side from each comer of this box. What percent of theoriginal volume is removed?

(A) 4.5 (B) 9 (C) 12 (D) 18 (E) 24

4. It takes Mary 30 minutes to walk uphill 1 km from her home to school, butit takes her only 10 minutes to walk from school to home along the sameroute. What is her average speed, in km/hr, for the round trip?

(A) 3 (B) 3.125 (C) 3.5 (D) 4 (E) 4.5

5. The sum of the two 5-digit numbers AMC 10 and AMC12 is 123422.What is A + M + C?

(A) 10 (B) 11 (C) 12 (D) 13 (E) 14

6. Define xC7y to be Ix - yj for all real numbers x and y. Which of thefollowing statements is not true?

(A) xc7y = ycx for all x and y

(B) 2(xc7y) = (2x)O(2y) for all x and y

(C) x00 = x for all x (D) xOx = 0 for all x (E) xc7y > 0 if x 54 y

7. How many non-congruent triangles with perimeter 7 have integer sidelengths?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

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BC

D

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18 Problems

8. What is the probability that a randomly drawn positive factor of 60 is lessthan 7?

(A) 1 (B) - (C) - (D) - (E) 110 6 4 3 2

9. A set S of points in the xy-plane is symmetric about the origin, both co-ordinate axes, and the line y = x. If (2, 3) is in S, what is the smallestpossible number of points in S?

(A) 1 (B) 2 (C) 4 (D) 8 (E) 16

10. Al, Bert, and Carl are the winners of a school drawing for a pile of Hal-loween candy, which they are to divide in a ratio of 3 : 2 : 1, respec-tively. Due to some confusion they come at different times to claim theirprizes, and each assumes he is the first to arrive. If each takes what hebelieves to be his correct share of candy, what fraction of the candy goesunclaimed?

(A) 18 (B) 6 (C) 9 (D) 18 (E) 12

11. A square and an equilateral triangle have the same perimeter. Let A be thearea of the circle circumscribed about the square and B be the area of thecircle circumscribed about the triangle. What is A/B?

9 3 27(A)

16(B)

4(C)

32(D)

3 g(E) 1

12. Sally has five red cards numbered 1 through 5 and four blue cards num-bered 3 through 6. She stacks the cards so that the colors alternate and sothat the number on each red card divides evenly into the number on eachneighboring blue card. What is the sum of the numbers on the middle threecards?

(A) 8 (B) 9 (C) 10 (D) 11 (E) 12

13. The polygon enclosed by the solid lines in the figure consists of 4 congruentsquares joined edge-to-edge. One more congruent square is attached to anedge at one of the nine positions indicated. How many of the nine resultingpolygons can be folded to form a cube with one face missing?


- - - - I - - - - - - - - - - - -

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6

14. Points K, L, M, and N lie in the plane of the square ABCD so that AKB,BLC, CMD, and DNA are equilateral triangles. The area of ABCD is 16.What is the area of KLMN?

M

(A) 32 (B) 16 + 16/ (C) 48 (D) 32 + 16/ (E) 64

15. A semicircle of diameter 1 sits at the top of a semicircle of diameter 2, asshown. The shaded area inside the smaller semicircle and outside the largersemicircle is called a tune. What is the area of this lune?

(i.'CS

C/1

[--

20 Problems

2

1 1 1 1

(A)6 4

(B)4 12

7r (C)4 24

7r (D)4 + 24 7r

(E) + 1 7r

4 12

16. A point P is chosen at random in the interior of equilateral triangle ABC.What is the probability that L A BP has a greater area than each of L ACPand ABCP?

(A) 1 (B) 1 (C) 1 (D) 1 (E) 26 4 3 2 3

17. Square ABCD has sides of length 4, and M is the midpoint of CD. Acircle with radius 2 and center M intersects a circle with radius 4 andcenter A at points P and D. What is the distance from P to AD?

16 13 7(A)

3(B)

5 (C) 4 (D) 2

D M C

18. Let n be a 5-digit number, and let q and r be the quotient and remainder,respectively, when n is divided by 100. For how many values of n is q + rdivisible by 11?

(A) 8180 (B) 8181 (C) 8182 (D) 9000 (E) 9090

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19. A parabola with equation y = ax 2 + bx + c is reflected about thex-axis. The parabola and its reflection are translated horizontally fiveunits in opposite directions to become the graphs of y = f (x) and y =g(x), respectively. Which of the following describes the graph of y =(f + g)(x)?(A) a parabola tangent to the x-axis

(B) a parabola not tangent to the x-axis (C) a horizontal line

(D) a non-horizontal line (E) the graph of a cubic function

20. How many 15-letter arrangements of 5 A's, 5 B's, and 5 C's have no A's inthe first 5 letters, no B's in the next 5 letters, and no C's in the last 5 letters?

I(B) 35.25 (C) 215 (D)(5 )3

(E) 315(A)

(,),k=50

21. The graph of the polynomial

P(x) = x5 + ax4 + bx3 + cx2 + dx + e

has five distinct x-intercepts, one of which is at (0, 0). Which of the fol-lowing coefficients cannot be zero?

(A) a (B) b (C) c (D) d (E) e

22. Objects A and B move simultaneously in the coordinate plane via a se-quence of steps, each of length one. Object A starts at (0, 0) and each ofits steps is either right or up, both equally likely. Object B starts at (5, 7)and each of its steps is either left or down, both equally likely. Which ofthe following is closest to the probability that the objects meet?

(A) 0.10 (B) 0.15 (C) 0.20 (D) 0.25 (E) 0.30

23. How many perfect squares are divisors of the product 1!.2!.3! 9! ?

(A) 504 (B) 672 (C) 864 (D) 936 (E) 1008

24. If a > b > 1, what is the largest possible value of log,, (alb) + logo (b/a)?

(A) -2 (B) 0 (C) 2 (D) 3 (E) 4

25. Let f (x) = axe + bx. For how many real values of a is there at leastone positive value of b for which the domain of f and the range of f arethe same set?


22 Problems


1. Which of the following is the same as

2-4+6-8+10-12+143-6+9- 12+ 15- 18+21

2 2 14(A) -1 (B) - (C) 3 (D) 1 (E)

3 3

2. Al gets the disease algebritis and must take one green pill and one pink pilleach day for two weeks. A green pill costs $1 more than a pink pill, andAl's pills cost a total of $546 for the two weeks. How much does one greenpill cost?

(A) $7 (B) $14 (C) $19 (D) $20 (E) $39

3. Rose fills each of the rectangular regions of her rectangular flower bed witha different type of flower. The lengths, in feet, of the rectangular regions inher flower bed are as shown in the figure. She plants one flower per squarefoot in each region. Asters cost $1 each, begonias $1.50 each, cannas $2each, dahlias $2.50 each, and Easter lilies $3 each. What is the least possi-ble cost, in dollars, for her garden?

4 7

3

5

3

1

6 5

(A) 108 (B) 115 (C) 132 (D) 144 (E) 156

4. Moe uses a mower to cut his rectangular 90-foot by 150-foot lawn. Theswath he cuts is 28 inches wide, but he overlaps each cut by 4 inches tomake sure that no grass is missed. He walks at the rate of 5000 feet per

r..

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.-.

co,

Hei

ght

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O


hour while pushing the mower. Which of the following is closest to thenumber of hours it will take Moe to mow his lawn?

(A) 0.75 (B) 0.8 (C) 1.35 (D) 1.5 (E) 3

5. Many television screens are rectangles that are measured by the length oftheir diagonals. The ratio of the horizontal length to the height in a standardtelevision screen is 4 : 3. The horizontal length of a "27-inch" televisionscreen is closest, in inches, to which of the following?

Length

(A) 20 (B) 20.5 (C) 21 (D) 21.5 (E) 22

6. The second and fourth terms of a geometric sequence are 2 and 6. Whichof the following is a possible first term?

(A) - (B) - 2 (C) - (D) (E) 33 3

7. Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, allnickels, dimes, and quarters, whose total value is $8.35. It does not neces-sarily contain coins of all three types. What is the difference between thelargest and smallest number of dimes that could be in the bank?

(A) 0 (B) 13 (C) 37 (D) 64 (E) 83

8. Let 4(x) denote the sum of the digits of the positive integer x. For exam-ple, 4(8) = 8 and 4(123) = 1 + 2 + 3 = 6. For how many two-digitvalues of x is 4(4(x)) = 3?

(A) 3 (B) 4 (C) 6 (D) 9 (E) 10

9. Let f be a linear function for which f (6) - f (2) = 12. What is f (12) -.f (2)?

(A) 12 (B) 18 (C) 24 (D) 30 (E) 36

'C7

24 Problems

10. Several figures can be made by attaching two equilateral triangles to theregular pentagon ABCDE in two of the five positions shown. How manynon-congruent figures can be constructed in this way?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

11. Cassandra sets her watch to the correct time at noon. At the actual time of1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assumingthat her watch loses time at a constant rate, what will be the actual timewhen her watch first reads 10:00 PM?

(A) 10:22 PM and 24 seconds (B) 10:24 PM (C) 10:25 PM

(D) 10:27 PM (E) 10:30 PM

12. What is the largest integer that is a divisor of

(n + 1)(n + 3)(n + 5)(n + 7)(n + 9)for all positive even integers n?

(A) 3 (B) 5 (C) 11 (D) 15 (E) 165

13. An ice cream cone consists of a sphere of vanilla ice cream and a right cir-cular cone that has the same diameter as the sphere. If the ice cream melts,it will exactly fill the cone. Assume that the melted ice cream occupies75% of the volume of the frozen ice cream. What is the ratio of the cone'sheight to its radius?

(A) 2 : 1 (B) 3 : 1 (C) 4 : 1 (D) 16 : 3 (E) 6 : 1

14. In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CDso that DF = 1 and GC = 2. Lines AF and BG intersect at E. What isthe area of AAEB?

E"'


21

(A) 10 (B) 2(C) 12 (D)

25(E) 15

15. A regular octagon ABCDEFGH has an area of one square unit. What isthe area of the rectangle ABEF?

A B

H//_ C

G

F E

D

(A) 1 - (B) (C) v - 1 (D) 1 (E) 1 +2 4 2 4

16. Three semicircles of radius 1 are constructed on diameter AB of a semi-circle of radius 2. The centers of the small semicircles divide AB into fourline segments of equal length, as shown. What is the area of the shadedregion that lies within the large semicircle but outside the smaller semi-circles?

nil

E

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.-.

26 Problems

(A) it - (B) it - (C)

(E) 6 it - 2 22

(D)7r

2

/3

17. If log(xy3) = 1 and log(x2y) = 1, what is log(xy)?1(A) -2

(B) 0 (C) (D) 52

(E) 1

18. Let x and y be positive integers such that 7x5 = 11y13. The minimumpossible value of x has a prime factorization a`bd . What is a + b + c +d?

(A) 30 (B) 31 (C) 32 (D) 33 (E) 34

19. Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whichthe first term is not 1. A permutation is chosen randomly from S. Theprobability that the second term is 2, in lowest terms, is a/b. What is a +b?

(A) 5 (B) 6 (C) 11 (D) 16 (E) 19

20. Part of the graph of f (x) = ax3 + bx2 + cx + d is shown. What is b?

(A) -4 (B) -2 (C) 0 (D) 2 (E) 4

21. An object moves 8 cm in a straight line from A to B, turns at an anglea, measured in radians and chosen at random from the interval (0, 7r), andmoves 5 cm in a straight line to C. What is the probability that AC < 7?

(A) (B) (C) (D) I (E) I6 54 3 2

22. Let ABCD be a rhombus with AC = 16 and BD = 30. Let N be a pointon AB, and let P and Q be the feet of the perpendiculars from N to AC

...


and BD, respectively. Which of the following is closest to the minimumpossible value of PQ?

DL vC

(A) 6.5 (B) 6.75 (C) 7 (D) 7.25 (E) 7.5

23. Which of the following is closest to the number of x-intercepts on the graphof y = sin(l/x) in the interval (0.000 1, 0.001)?

(A) 2900 (B) 3000 (C) 3100 (D) 3200 (E) 3300

24. Positive integers a, b, and c are chosen so that a < b < c, and the systemof equations

2x+y=2003 and y=Ix-al+Ix-bl+Ix-clhas exactly one solution. What is the minimum value of c?

(A) 668 (B) 669 (C) 1002 (D) 2003 (E) 2004

25. Three points are chosen randomly and independently on a circle. What isthe probability that all three pairwise distances between the points are lessthan the radius of the circle?

(A) 36 (B) 24 (C) 18 (D) 12 (E) 9

.

,cam

.n.

C'7

28 Problems


1. Alicia earns $20 per hour, of which 1.45% is deducted to pay local taxes.How many cents per hour of Alicia's wages are used to pay local taxes?

(A) 0.0029 (B) 0.029 (C) 0.29 (D) 2.9 (E) 29

2. On the AMC 12, each correct answer is worth 6 points, each incorrectanswer is worth 0 points, and each problem left unanswered is worth 2.5points. If Charlyn leaves 8 of the 25 problems unanswered, how many ofthe remaining problems must she answer correctly in order to score at least100?

(A) 11 (B) 13 (C) 14 (D) 16 (E) 17

3. For how many ordered pairs of positive integers (x, y) is x + 2y = 100?

(A) 33 (B) 49 (C) 50 (D) 99 (E) 100

4. Bertha has 6 daughters and no sons. Some of her daughters have 6 daugh-ters, and the rest have none. Bertha has a total of 30 daughters and grand-daughters, and no great-granddaughters. How many of Bertha's daughtersand granddaughters have no daughters?

(A) 22 (B) 23 (C) 24 (D) 25 (E) 26

5. The graph of a line y = mx + b is shown. Which of the following is true?

y 2

1 22,x

(A) mb < -1 (B) -1 < mb < 0 (C) mb = 0(D) 0< m b< 1 (E) m b> 1

6. Let U = 2 2004200s V = 2004200s W = 2003 20042004 X =

2.20042004 Y = 20042004 and Z = 20042003. Which of the following islargest?

(A) U- V (B) V- W (C) W- X (D) X- Y(E) Y - Z

;10

C17


7. A game is played with tokens according to the following rule. In eachround, the player with the most tokens gives one token to each of the otherplayers and also places one token into a discard pile. The game ends whensome player runs out of tokens. Players A, B, and C start with 15, 14, and13 tokens, respectively. How many rounds will there be in the game?

(A) 36 (B) 37 (C) 38 (D) 39 (E) 40

8. In the Figure, /EAB and /ABC are right angles, AB = 4, BC = 6,AE = 8, and AC and BE intersect at D. What is the difference betweenthe areas of LADE and ABDC?

E

(A) 2 (B) 4 (C) 5 (D) 8 (E) 9

9. A company sells peanut butter in cylindrical jars. Marketing research sug-gests that using wider jars will increase sales. If the diameter of the jars isincreased by 25% without altering the volume, by what percent must theheight be decreased?

(A) 10 (B) 25 (C) 36 (D) 50 (E) 60

10. The sum of 49 consecutive integers is 75. What is their median?

(A) 7 (B) 72 (C) 73 (D) 74 (E) 75

11. The average value of all the pennies, nickels, dimes, and quarters in Paula'spurse is 20 cents. If she had one more quarter, the average value would be21 cents. How many dimes does she have in her purse?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

12. Let A = (0, 9) and B = (0, 12). Points A' and B' are on the line y = x,and AA' and BB' intersect at C = (2, 8). What is the length of A'B'?

(A) 2 (B) 2 2 (C) 3 (D) 2 + 12 (E) 3'13. Let S be the set of points (a, b) in the coordinate plane, where each of a

and b may be -1, 0, or 1. How many distinct lines pass through at leasttwo members of S?

(A) 8 (B) 20 (C) 24 (D) 27 (E) 36

30 Problems

14. A sequence of three real numbers forms an arithmetic progression with afirst term of 9. If 2 is added to the second term and 20 is added to the thirdterm, the three resulting numbers form a geometric progression. What isthe smallest possible value for the third term of the geometric progression?

(A) 1 (B) 4 (C) 36 (D) 49 (E) 81

15. Brenda and Sally run in opposite directions on a circular track, starting atdiametrically opposite points. They first meet after Brenda has run 100 me-ters. They next meet after Sally has run 150 meters past their first meetingpoint. Each girl runs at a constant speed. What is the length of the track inmeters?

(A) 250 (B) 300 (C) 350 (D) 400 (E) 500

16. The set of all real numbers x for which

1092004009200300920020092001 x)))

is defined is {x I x > c}. What is the value of c?

(A) 0 (B) 20012002 (C) 20022003 (D) 20032004

(E)200120021003

17. Let f be a function with the following properties:

(i) f(1) = 1, and

(ii) f (2n) = n f (n) for any positive integer n.

What is the value of f (2100)?

(A) I (B) 299 (C) 2100 (D) 24950 (E) 29999

18. Square ABCD has side length 2. A semicircle with diameter AB is con-structed inside the square, and the tangent to the semicircle from C inter-sects side AD at E. What is the length of CE?

B

(A) 2 2 (B) 15 (C) (D)2

(E) 5 -

'-ra-?

QC

)fl.

--N

r°+

IVY


19. Circles A, B, and C are externally tangent to each other and internallytangent to circle D. Circles B and C are congruent. Circle A has radius 1and passes through the center of circle D. What is the radius of circle B?

(A) 2 (B) (C) 7(D)

8 (E) 1 +3 2 8 9 3

20. Select numbers a and b between 0 and 1 independently and at random,and let c be their sum. Let A, B, and C be the results when a, b, and c,respectively, are rounded to the nearest integer. What is the probability thatA + B = C?

(A) 1 (B) 1 (C) 1 (D)2

(E) 34 3 2

21. If Fn°_o cos2, 0 = 5, what is the value of cos 20?

3 4

(A) 12

(B) - (C)113

(D)3

(E)4

5 5 5 5 5

22. Three mutually tangent spheres of radius 1 rest on a horizontal plane. Asphere of radius 2 rests on them. What is the distance from the plane to thetop of the larger sphere?

(A) 3 + 30 (B) 3 + (C) 3 +123 (D) 52

2 3 4 9

(E) 3 + 2/'2-

23. A polynomial

P(x) = c2004x2004 + c2003x2003 + + clx + co

has real coefficients with C2004 # 0 and 2004 distinct complex zeros zkak + bki, 1 < k < 2004 with ak and bk real, al = b, = 0, and

2004 2004

T ak = E bk.k=1 k=1

.-.

+--

32 Problems

Which of the following quantities can be a nonzero number?2004

(A) co (B) C2003 (C) b2b3...b2004 (D) Y, ak

(E)

2004

Ck

k=1

k=1

24. A plane contains points A and B with AB = 1. Let S be the union of alldisks of radius 1 in the plane that cover AB. What is the area of S?

(A) 27r + (B)83

(C) 3yr - 2 (D)

(E) 47r - 2'25. For each integer n > 4, let a denote the base-n number 0.133,,. The

product a4a5 ... a99 can be expressed as m/n!, where m and n are positiveintegers and n is as small as possible. What is the value of m?

(A) 98 (B) 101 (C) 132 (D) 798 (E) 962

ooh

...11

0

00.

00.

ran

C17



1. At each basketball practice last week, Jenny made twice as many freethrows as she made at the previous practice. At her fifth practice she made48 free throws. How many free throws did she make at the first practice?

(A) 3 (B) 6 (C) 9 (D) 12 (E) 15

2. In the expression c ab - d, the values of a, b, c, and d are 0, 1, 2, and 3,although not necessarily in that order. What is the maximum possible valueof the result?

(A) 5 (B) 6 (C) 8 (D) 9 (E) 10

3. If x and y are positive integers for which 2X3' = 1296, what is the valueofx+y?(A) 8 (B) 9 (C) 10 (D) 11 (E) 12

4. An integer x, with 10 < x < 99, is to be chosen. If all choices are equallylikely, what is the probability that at least one digit of x is a 7?

(A)1

(B) 1 (C) 19 (D) 2 (E) 1

9 5 90 9 3

5. On a trip from the United States to Canada, Isabella took d U.S. dollars. Atthe border she exchanged them all, receiving 10 Canadian dollars for every7 U.S. dollars. After spending 60 Canadian dollars, she had d Canadiandollars left. What is the sum of the digits of d?

(A) 5 (B) 6 (C) 7 (D) 8 (E) 9

6. Minneapolis-St. Paul International Airport is 8 miles southwest of down-town St. Paul and 10 miles southeast of downtown Minneapolis. Which ofthe following is closest to the number of miles between downtown St. Pauland downtown Minneapolis?

(A) 13 (B) 14 (C) 15 (D) 16 (E) 17

7. A square has sides of length 10, and a circle centered at one of its verticeshas radius 10. What is the area of the union of the regions enclosed by thesquare and the circle?

(A) 200 + 257r (B) 100 + 757r (C) 75 + 100 r

(D) 100 + 1007r (E) 100 + 125,-r

dab

..a

34 Problems

8. A grocer makes a display of cans in which the top row has one can andeach lower row has two more cans than the row above it. If the displaycontains 100 cans, how many rows does it contain?

(A) 5 (B) 8 (C) 9 (D) 10 (E) 11

9. The point (-3,2) is rotated 90° clockwise around the origin to point B.Point B is then reflected in the line y = x to point C. What are the coor-dinates of C?

(A) (-3, -2) (B) (-2, -3) (C) (2, -3) (D) (2, 3)(E) (3, 2)

10. An annulus is the region between two concentric circles. The concentriccircles in the figure have radii b and c, with b > c. Let OX be a radius ofthe larger circle, let XZ be tangent to the smaller circle at Z, and let OYbe the radius of the larger circle that contains Z. Let a = X Z, d = YZ,and e = X Y. What is the area of the annulus?

(A) nag (B) ,rb2 (C) ,rc2 (D) 7rd 2 (E) ire2

11. All the students in an algebra class took a 100-point test. Five studentsscored 100, each student scored at least 60, and the mean score was 76.What is the smallest possible number of students in the class?

(A) 10 (B) 11 (C) 12 (D) 13 (E) 14

12. In the sequence 2001, 2002, 2003, ... ,each term after the third is found bysubtracting the previous term from the sum of the two terms that precedethat term. For example, the fourth term is 2001 + 2002 - 2003 = 2000.What is the 2004`h term in this sequence?

(A) -2004 (B) -2 (C) 0 (D) 4003 (E) 6007

13. If f (x) = ax + b and f (x) = bx + a with a and b real, what is thevalue of a + b?

(A) -2 (B) -1 (C) 0 (D) 1 (E) 2

'_"


14. In AABC, AB = 13, AC = 5 and BC = 12. Points M and N lie on ACand BC, respectively, with CM = CN = 4. Points J and K are on ABso that MJ and NK are perpendicular to A B. What is the area of pentagonCMJKN?

A

C

81(A) 15 (B) 5

J

N B

205 240(C)(E) 20(D)

1312

15. The two digits in Jack's age are the same as the digits in Bill's age, but inreverse order. In five years Jack will be twice as old as Bill will be then.What is the difference in their current ages?

(A) 9 (B) 18 (C) 27 (D) 36 (E) 45

16. A function f is defined by f (z) = i z, where i = and z is thecomplex conjugate of z. How many values of z satisfy both 1zI = 5 and.f (z) = z?(A) 0 (B) 1 (C) 2 (D) 4 (E) 8

17. For some real numbers a and b, the equation

8x3 + 4ax2 + 2bx + a = 0

has three distinct positive roots. If the sum of the base-2 logarithms of theroots is 5, what is the value of a?

(A) -256 (B) -64 (C) -8 (D) 64 (E) 256

18. Points A and B are on the parabola y = 4x2 + 7x - 1, and the origin isthe midpoint of AB. What is the length of AB?

(A) 2/ (B) 5 + 2 (C) 5 + / (D) 7 (E) 519. A truncated cone has horizontal bases with radii 18 and 2. A sphere is

tangent to the top, bottom, and lateral surface of the truncated cone. Whatis the radius of the sphere?

(A) 6 (B) 4v (C) 9 (D) 10 (E) 6/3

20. Each face of a cube is painted either red or blue, each with probability 1/2.The color of each face is determined independently. What is the probability

Cr=

y

..-

0.l

°-°

36 Problems

that the painted cube can be placed on a horizontal surface so that the fourvertical faces are all the same color?(A) l (B) 5 (C) 3 (D) 7 (E) 1

4 16 8 16 2

21. The graph of 2x2 + xy + 3y2 - l lx - 20y + 40 = 0 is an ellipse in thefirst quadrant of the xy-plane. Let a and b be the maximum and minimumvalues of y/x over all points (x, y) on the ellipse. What is the value ofa + b?

(A) 3 (B) 10 (C)2

(D) 2 (E) 2 14

22. The square

50

e

C

d f2

is a multiplicative magic square. That is, the product of the numbers ineach row, column, and diagonal is the same. If all the entries are positiveintegers, what is the sum of the possible values of g?

(A) 10 (B) 25 (C) 35 (D) 62 (E) 136

23. The polynomial x3 - 2004x2 + mx + n has integer coefficients and threedistinct positive zeros. Exactly one of these is an integer, and it is the sumof the other two. How many values of n are possible?

(A) 250,000 (B) 250,250 (C) 250,500 (D) 250,750(E) 251,000

24. In AABC, AB = BC, and BD is an altitude. Point E is on the extensionof AC such that BE = 10. The values of tan LCBE, tan Z DBE, andtan LABE form a geometric progression, and the values of cot ZDBE,cot LCBE, cot LDBC form an arithmetic progression. What is the area ofAABC?

A D C E

(A) 16 (B) 50 (C) 10/3 (D) 8/5 (E) 18

V'1


25. Given that 22004 is a 604-digit number whose first digit is 1, how manyelements of the set S = {20 21 22 22003} have a first digit of 4?

(A) 194 (B) 195 (C) 196 (D) 197 (E) 198

38 Problems


1. Two is 10% of x and 20% of y. What is x - y?

(A) 1 (B) 2 (C) 5 (D) 10 (E) 20

2. The equations 2x + 7 = 3 and bx - 10 = -2 have the same solution forx. What is the value of b?

(A) -8 (B) -4 (C) -2 (D) 4 (E) 8

3. A rectangle with a diagonal of length x is twice as long as it is wide. Whatis the area of the rectangle?

(A) 1x2 (B) 2x2 (C) 1x2 (D) x2 (E) 3x2

4. A store normally sells windows at $100 each. This week the store is offer-ing one free window for each purchase of four. Dave needs seven windowsand Doug needs eight windows. How many dollars will they save if theypurchase the windows together rather than separately?

(A) 100 (B) 200 (C) 300 (D) 400 (E) 500

5. The average (mean) of 20 numbers is 30, and the average of 30 other num-bers is 20. What is the average of all 50 numbers?

(A) 23 (B) 24 (C) 25 (D) 26 (E) 27

6. Josh and Mike live 13 miles apart. Yesterday Josh started to ride his bicycletoward Mike's house. A little later Mike started to ride his bicycle towardJosh's house. When they met, Josh had ridden for twice the length of timeas Mike and at four-fifths of Mike's rate. How many miles had Mike riddenwhen they met?

(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

7. In the figure, the length of side AB of square ABCD is 50, E is betweenB and H, and BE = 1. What is the the area of the inner square EFGH ?

A g

I.:

°-)


(A) 25 (B) 32 (C) 36 (D) 40 (E) 42

8. Let A, M, and C be digits with

(100A+1OM+C)(A+M+C)=2005.

What is A?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

9. There are two values of a for which the equation 4x2 + ax + 8x + 9 = 0has only one solution for x. What is the sum of those values of a?

(A) -16 (B) -8 (C) 0 (D) 8 (E) 20

10. A wooden cube n units on a side is painted red on all six faces and then cutinto n3 unit cubes. Exactly one-fourth of the total number of faces of theunit cubes are red. What is n?

(A) 3 (B) 4 (C) 5 (D) 6 (E) 7

11. How many three-digit numbers satisfy the property that the middle digit isthe average of the first and the last digits?

(A) 41 (B) 42 (C) 43 (D) 44 (E) 45

12. A line passes through A = (1, 1) and B = (100, 1000). How many otherpoints with integer coordinates are on the line and strictly betweenA and B?

(A) 0 (B) 2 (C) 3 (D) 8 (E) 9

13. In the five-sided star shown, the letters A, B, C, D and E are replacedby the numbers 3, 5, 6, 7 and 9, although not necessarily in that order.The sums of the numbers at the ends of the line segments AB, BC, CD,DE and EA form an arithmetic sequence, although not necessarily in thatorder. What is the middle term of the arithmetic sequence?

(A) 9 (B) 10 (C) 11 (D) 12 (E) 13

40 Problems

14. On a standard die one of the dots is removed at random with each dotequally likely to be chosen. The die is then rolled. What is the probabilitythat the top face has an odd number of dots?

(A) 5 (B) 10 (C) 1 (D) 11 (E) 6

11 21 2 21 11

15. Let AB be a diameter of a circle and C be a point on AB with 2 AC =BC. Let D and E be points on the circle such that DCIAB and DE isa second diameter. What is the ratio of the area of ADCE to the area ofAABD?

(A) 64 (D)3 2

2(E)

3

16. Three circles of radius s are drawn in the first quadrant of the xy-plane.The first circle is tangent to both axes, the second is tangent to the firstcircle and the x-axis, and the third is tangent to the first circle and the y-axis. A circle of radius r > s is tangent to both axes and to the second andthird circles. What is r/s?

(A) 5

(B)

(B) 6

(C)

(C) 8 (D) 9 (E) 10

17. A unit cube is cut twice to form three triangular prisms, two of which arecongruent, as shown in Figure 1. The cube is then cut in the same manner

NIA

-

nQ,

«T.

-IN


along the dashed lines shown in Figure 2. This creates nine pieces. What isthe volume of the piece that contains vertex W?

Figure 1 Figure 2

1 1 1

(A) (B) (C)12 9 8

1 1

(D) 6 (E)4

18. Call a number "prime-looking" if it is composite but not divisible by 2, 3,or 5. The three smallest prime-looking numbers are 49, 77, and 91. Thereare 168 prime numbers less than 1000. How many prime-looking numbersare there less than 1000?

(A) 100 (B) 102 (C) 104 (D) 106 (E) 108

19. A faulty car odometer proceeds from digit 3 to digit 5, always skippingthe digit 4, regardless of position. For example, after traveling one milethe odometer changed from 000039 to 000050. If the odometer now reads002005, how many miles has the car actually traveled?

(A) 1404 (B) 1462 (C) 1604 (D) 1605 (E) 1804

20. For each x in [0, 1], define

f(X) = 2x, if o <X < z2-2x, if z <x < 1.

Let f 12] (x) = f (f (x)), and f1'+11(x) = f 1"1(f (x)) for each integern > 2. For how many values of x in [0, 1] is f[2005](X) = 1/2?

(A) 0 (B) 2005 (C) 4010 (D) 20052 (E) 22005

21. How many ordered triples of integers (a, b, c), with a > 2, b > 1, andc > 0, satisfy both log,, b = c2005 and a + b + c = 2005?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

22. A rectangular box P is inscribed in a sphere of radius r. The surface areaof P is 384, and the sum of the lengths of its 12 edges is 112. What is r?

(A) 8 (B) 10 (C) 12 (D) 14 (E) 16

CA

D

42 Problems

23. Two distinct numbers a and b are chosen randomly from the set{2, 22, 23'...,2 25}. What is the probability that logo b is an integer?

2 31(A) (B)

25 300

13 7 1

(C) (D) (E)100 50 2

24. Let P(x) = (x - 1)(x - 2)(x - 3). For how many polynomials Q(x)does there exist a polynomial R(x) of degree 3 such that P (Q(x)) _P(x) R(x)?(A) 19 (B) 22 (C) 24 (D) 27 (E) 32

25. Let S be the set of all points with coordinates (x, y, z), where x, y, and zare each chosen from the set {0, 1, 2}. How many equilateral triangles haveall their vertices in S?

(A) 72 (B) 76 (C) 80 (D) 84 (E) 88

.fl

.fl

.fl

ten

..e

/'1



1. A scout troop buys 1000 candy bars at a price of five for $2. They sell allthe candy bars at a price of two for $1. What was their profit, in dollars?

(A) 100 (B) 200 (C) 300 (D) 400 (E) 500

2. A positive number x has the property that x% of x is 4. What is x?

(A) 2 (B) 4 (C) 10 (D) 20 (E) 40

3. Brianna is using part of the money she earned on her weekend job to buyseveral equally-priced CDs. She used one fifth of her money to buy onethird of the CDs. What fraction of her money will she have left after shebuys all the CDs?

(A) I (B) 1 (C) 2 (D) 2 (E) 45 3 5 3 5

4. At the beginning of the school year, Lisa's goal was to earn an A on at least80% of her 50 quizzes for the year. She earned an A on 22 of the first 30quizzes. If she is to achieve her goal, on at most how many of the remainingquizzes can she earn a grade lower than an A?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

5. An 8-foot by 10-foot floor is tiled with square tiles of size 1 foot by 1 foot.Each tile has a pattern consisting of four white quarter circles of radius 1/2foot centered at each corner of the tile. The remaining portion of the tile isshaded. How many square feet of the floor are shaded?

(A) 80 - 207r (B) 60 - 107r (C) 80 - 107r (D) 60 + 107r(E) 80 + 10ir

6. In DABC, we have AC = BC = 7 and AB = 2. Suppose that D is apoint on line A B such that B lies between A and D and CD = 8. What isBD?

(A) 3 (B) 2J (C) 4 (D) 5 (E) 4/

7. What is the area enclosed by the graph of 13x I + 14y I = 12?

(A) 6 (B) 12 (C) 16 (D) 24 (E) 25

0.C

.On

ac'

C'7

C.,

44 Problems

8. For how many values of a is it true that the line y = x + a passes throughthe vertex of the parabola y = x2 + a2?


9. On a certain math exam, 10% of the students got 70 points, 25% got 80points, 20% got 85 points, 15% got 90 points, and the rest got 95 points.What is the difference between the mean and the median score on thisexam?

(A) 0 (B) 1 (C) 2 (D) 4 (E) 5

10. The first term of a sequence is 2005. Each succeeding term is the sum ofthe cubes of the digits of the previous term. What is the 20051h term of thesequence?

(A) 29 (B) 55 (C) 85 (D) 133 (E) 250

11. An envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties. Twobills are drawn at random without replacement. What is the probability thattheir sum is $20 or more?

1

2 3 2(A) (B) 5 (C) - (D) 2 (E) 3

4

12. The quadratic equation x2 + mx + n = 0 has roots that are twice those ofx2 + px + m = 0, and none of m, n and p is zero. What is the value ofn/p?(A) 1 (B) 2 (C) 4 (D) 8 (E) 16

13. Suppose that 4X1 = 5, 5X2 = 6, 6x3 = 7, ... , 127X124 = 128. What isx1x2...x124?

5 7(A) 2 (B)

2(C) 3 (D)

2(E) 4

14. A circle having center (0, k), with k > 6, is tangent to the lines y = x,y = -x and y = 6. What is the radius of this circle?

(A) 6'h - 6 (B) 6 (C) 6' (D) 12 (E) 6 + 6,/'2-

15. The sum of four two-digit numbers is 221. None of the eight digits is 0and no two of them are the same. Which of the following is not includedamong the eight digits?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

16. Eight spheres of radius 1, one per octant, are each tangent to the coordinateplanes. What is the radius of the smallest sphere, centered at the origin, thatcontains these eight spheres?

(A) (B) (C) 1 + (D) 1 + (E) 3

`ID

.n.

+u+

...d-0

fl.

L].

'-.


17. How many distinct four-tuples (a, b, c, d) of rational numbers are therewith

a loglo 2 + b loglo 3 + c loglo 5 + d loglo 7 = 2005?


18. Let A(2, 2) and B(7, 7) be points in the plane. Define R as the region inthe first quadrant consisting of those points C such that AABC is an acutetriangle. What is the closest integer to the area of the region R?

(A) 25 (B) 39 (C) 51 (D) 60 (E) 80

19. Let x and y be two-digit integers such that y is obtained by reversing thedigits of x. The integers x and y satisfy x2 - y2 = m2 for some positiveinteger m. What is x + y + m?

(A) 88 (B) 112 (C) 116 (D) 144 (E) 154

20. Let a, b, c, d, e, f, g and h be distinct elements in the set

{-7, -5, -3, -2, 2, 4, 6, 13}.

What is the minimum possible value of

(a+b+c+ d)2+(e+f +g+h)2?

(A) 30 (B) 32 (C) 34 (D) 40 (E) 50

21. A positive integer n has 60 divisors and 7n has 80 divisors. What is thegreatest integer k such that 7k divides n?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

22. A sequence of complex numbers z0, zl, Z2.... is defined by the rule

IZnZn+1 =

Zn

where zn is the complex conjugate of zn and i2 = -1. Suppose that Izo l _1 and Z2005 = 1. How many possible values are there for zo?

(A) 1 (B) 2 (C) 4 (D) 2005 (E) 22005

23. Let S be the set of ordered triples (x, y, z) of real numbers for which

loglo(x + y) = z and log10(x2+y2)=z+1.

There are real numbers a and b such that for all ordered triples (x, y, z) inS we have x3 + y3 = a 103z + b 102z. What is the value of a + b?

(A) 2 (B)29

(C) 15 (D)29

(E) 24

46 Problems

24. All three vertices of an equilateral triangle are on the parabola y = x2,and one of its sides has a slope of 2. The x-coordinates of the three verticeshave a sum of m/n, where m and n are relatively prime positive integers.What is the value of m + n?

(A) 14 (B) 15 (C) 16 (D) 17 (E) 18

25. Six ants simultaneously stand on the six vertices of a regular octahedron,with each ant at a different vertex. Simultaneously and independently, eachant moves from its vertex to one of the four adjacent vertices, each withequal probability. What is the probability that no two ants arrive at thesame vertex?

5 21 11 23 3(A)

256(B)

1024 (C) 512(D)

1024(E)

128

C/1



1. Sandwiches at Joe's Fast Food cost $3 each and sodas cost $2 each. Howmany dollars will it cost to purchase 5 sandwiches and 8 sodas?

(A) 31 (B) 32 (C) 33 (D) 34 (E) 35

2. Define x®y=x3-y. Whatish®(h®h)?(A) -h (B) 0 (C) h (D) 2h (E) h3

3. The ratio of Mary's age to Alice's age is 3: 5. Alice is 30 years old. Howold is Mary?

(A) 15 (B) 18 (C) 20 (D) 24 (E) 50

4. A digital watch displays hours and minutes with AM and PM. What is thelargest possible sum of the digits in the display?

(A) 17 (B) 19 (C) 21 (D) 22 (E) 23

5. Doug and Dave shared a pizza with 8 equally-sized slices. Doug wanteda plain pizza, but Dave wanted anchovies on half of the pizza. The costof a plain pizza was $8, and there was an additional cost of $2 for puttinganchovies on one half. Dave ate all the slices of anchovy pizza and oneplain slice. Doug ate the remainder. Each then paid for what he had eaten.How many more dollars did Dave pay than Doug?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

6. The 8 x 18 rectangle ABCD is cut into two congruent hexagons, as shown,in such a way that the two hexagons can be repositioned without overlap toform a square. What is y?

A B

(A) 6 (B) 7 (C) 8 (D) 9 (E) 10

7. Mary is 20% older than Sally, and Sally is 40% younger than Danielle. Thesum of their ages is 23.2 years. How old will Mary be on her next birthday?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11

,-.

CO

D

CO

D

o-0

..O

,.O

48 Problems

8. How many sets of two or more consecutive positive integers have a sum of15?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

9. Oscar buys 13 pencils and 3 erasers for $1.00. A pencil costs more than aneraser, and both items cost a whole number of cents. What is the total cost,in cents, of one pencil and one eraser?

(A) 10 (B) 12 (C) 15 (D) 18 (E) 20

10. For how many real values of x is 120 - an integer?

(A) 3 (B) 6 (C) 9 (D) 10 (E) 11

11. Which of the following describes the graph of the equation (x + y)2 =x2+y2?(A) the empty set (B) one point (C) two lines (D) a circle(E) the entire plane

12. A number of linked rings, each 1 cm thick, are hanging on a peg. The topring has an outside diameter of 20 cm. The outside diameter of each of theother rings is 1 cm less than that of the ring above it. The bottom ring hasan outside diameter of 3 cm. What is the distance, in cm, from the top ofthe top ring to the bottom of the bottom ring?

(A) 171 (B) 173 (C) 182 (D) 188 (E) 210

13. The vertices of a 3-4-5 right triangle are the centers of three mutuallyexternally tangent circles, as shown. What is the sum of the areas of thesecircles?

C17

°o.

KID

-a7

KIN

KIM

KID


25mr(A) 12ir (B)

2

27,-r(C) 137r (D)

2(E) 147r

14. Two farmers agree that pigs are worth $300 and that goats are worth $210.When one farmer owes the other money, he pays the debt in pigs or goats,with "change" received in the form of goats or pigs as necessary. (For ex-ample, a $390 debt could be paid with two pigs, with one goat receivedin change.) What is the amount of the smallest positive debt that can beresolved in this way?

(A) $5 (B) $10 (C) $30 (D) $90 (E) $210

15. Suppose cosx = 0 and cos(x + z) = 1/2. What is the smallest possiblepositive value of z?

7r7r 7r 57r 77r

(A) 6 (B) 3 (C)2

(D) 6 (E) 6

16. Circles with centers A and B have radii 3 and 8, respectively. A commoninternal tangent intersects the circles at C and D, respectively. Lines ABand CD intersect at E, and A E = 5. What is CD ?

(A) 13 (B) 34 (C) 221 (D) 25555

(E) 3

17. Square ABCD has side length s, a circle centered at E has radius r, andr and s are both rational. The circle passes through D, and D lies on BE.Point F lies on the circle, on the same side of BE as A. Segment A F istangent to the circle, and AF = 9 + What is r/s ?

-°=

50 Problems

(A) 1 (B) 5 (C)3

(D) 5 (E) 92 9 5 3 5

18. The function f has the property that for each real number x in its domain,1 /x is also in its domain and

f(x)+f(1)=x.X

What is the largest set of real numbers that can be in the domain of f ?

(A) {xIx 0} (B) {xIx<0} (C) {xIx>0}(D) {x I x 0 -1 and x 0 0 and x 0 1} (E) {-1, 1}

19. Circles with centers (2, 4) and (14, 9) have radii 4 and 9, respectively. Theequation of a common external tangent to the circles can be written in theform y = mx + b with m > 0. What is b?

(A) 908 (B) 909 (C) 130 (D) 911 (E) 912

119 119 17 119 119

20. A bug starts at one vertex of a cube and moves along the edges of the cubeaccording to the following rule. At each vertex the bug will choose to travelalong one of the three edges emanating from that vertex. Each edge has

,--.

...

s-.


equal probability of being chosen, and all choices are independent. What isthe probability that after seven moves the bug will have visited every vertexexactly once?

(A)2187

(B)729 (C) 243

(D)81

(E)243

21. Let

Si = {(x, Y) Ilogio(1 + x2 + y2) < 1 + 1og10(x + Y)}

and

S2 = {(x,y)Ilog10(2 + x2 + y2) < 2 + 1og10(x + y)}.

What is the ratio of the area of S2 to the area of Si?

(A) 98 (B) 99 (C) 100 (D) l01 (E) 102

22. A circle of radius r is concentric with and outside a regular hexagon of sidelength 2. The probability that three entire sides of the hexagon are visiblefrom a randomly chosen point on the circle is 1/2. What is r?

(A)2/2- + 2V (B)3/-3- + 52 (C)2' + (D)3' + 46-

(E) 6J -23. Given a finite sequence S = (a , , a2, ... , of n real numbers, let A(S)

be the sequence

a, + a2 a2 + a32 2

an )2 /f

of n - I real numbers. Define A' (S) = A(S) and, for each integer m,2 < m < n - 1, define At(S) = A(Am-'(S)). Suppose x > 0, and letS = (1, x, X2'. . . , x100). If A100(S) = 1/250), then what is x?

(A) 1 - 2 (B) /2 - 1 (C)2

(D) 2 - /2 (E)

24. The expression

(x + y + z)2006 + (x - y - z)2006

2

is simplified by expanding it and combining like terms. How many termsare in the simplified expression?

(A) 6018 (B) 671,676 (C) 1,007,514 (D) 1,008,016(E) 2,015,028

52 Problems

25. How many non-empty subsets S of 11, 2, 3, ... , 15} have the followingtwo properties?

(1) No two consecutive integers belong to S.

(2) If S contains k elements, then S contains no number less than k.

(A) 277 (B) 311 (C) 376 (D) 377 (E) 405

OP-

.

try



1. What is (-1)1 + (-1)2 + + (-1)2006?

(A) -2006 (B) -1 (C) 0 (D) 1 (E) 2006

2. For real numbers x and y, define x it y = (x + y)(x - y). What is3*(4*5)?(A) -72 (B) -27 (C) -24 (D) 24 (E) 72

3. A football game was played between two teams, the Cougars and the Pan-thers. The two teams scored a total of 34 points, and the Cougars won by amargin of 14 points. How many points did the Panthers score?

(A) 10 (B) 14 (C) 17 (D) 20 (E) 24

4. Mary is about to pay for five items at the grocery store. The prices ofthe items are $7.99, $4.99, $2.99, $1.99, and $0.99. Mary will pay witha twenty-dollar bill. Which of the following is closest to the percentage ofthe $20.00 that she will receive in change?

(A) 5 (B) 10 (C) 15 (D) 20 (E) 25

5. John is walking east at a speed of 3 miles per hour, while Bob is alsowalking east, but at a speed of 5 miles per hour. If Bob is now 1 mile westof John, how many minutes will it take for Bob to catch up to John?

(A) 30 (B) 50 (C) 60 (D) 90 (E) 120

6. Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400grams of water to make lemonade. There are 25 calories in 100 grams oflemon juice and 386 calories in 100 grams of sugar. Water contains nocalories. How many calories are in 200 grams of her lemonade?

(A) 129 (B) 137 (C) 174 (D) 223 (E) 411

7. Mr. and Mrs. Lopez have two children. When they get into their familycar, two people sit in the front, and the other two sit in the back. EitherMr. Lopez or Mrs. Lopez must sit in the driver's seat. How many seatingarrangements are possible?

(A) 4 (B) 12 (C) 16 (D) 24 (E) 48

8. The lines x = 4y + a and y = 4x + b intersect at the point (1, 2). Whatisa+b?

3 9(A) 0 (B)

4(C) I (D) 2 (E)

4

54 Problems

9. How many even three-digit integers have the property that their digits, readleft to right, are in strictly increasing order?

(A) 21 (B) 34 (C) 51 (D) 72 (E) 150

10. In a triangle with integer side lengths, one side is three times as long asa second side, and the length of the third side is 15. What is the greatestpossible perimeter of the triangle?

(A) 43 (B) 44 (C) 45 (D) 46 (E) 47

It. Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joedrank 2 ounces of his coffee and then added 2 ounces of cream. JoAnnadded 2 ounces of cream, stirred the coffee well, and then drank 2 ounces.What is the resulting ratio of the amount of cream in Joe's coffee to that inJoAnn's coffee?(A) 6 (B) 13 (C) I (D) 14 (E) 7

7 14 13 6

12. The parabola y = ax 2 +bx +c has vertex (p, p) and y-intercept (0, -p),where p 0. What is b?

(A) -p (B) 0 (C) 2 (D) 4 (E) p

13. Rhombus ABCD is similar to rhombus BFDE. The area of rhombusABCD is 24, and LBAD = 60°. What is the area of rhombus BFDE ?

D C

B

(A) 6 (B) 4/3 (C) 8 (D) 9 (E) 6V

14. Elmo makes N sandwiches for a fundraiser. For each sandwich he uses Bglobs of peanut butter at 4¢ per glob and J blobs of jam at 5¢ per blob.The cost of the peanut butter and jam to make all the sandwiches is $2.53.Assume that B, J, and N are positive integers with N > 1. What is thecost of the jam Elmo uses to make the sandwiches?

(A) $1.05 (B) $1.25 (C) $1.45 (D) $1.65 (E) $1.85

15. Circles with centers 0 and P have radii 2 and 4, respectively, and areexternally tangent. Points A and B are on the circle centered at 0, andpoints C and D are on the circle centered at P, such that AD and BC

ci,

i11

C17


are common external tangents to the circles. What is the area of hexagonAOBCPD?

(A) 18./ (B) 24s (C) 36 (D) 24,/ (E) 32,/2-

16. Regular hexagon ABCDEF has vertices A and C at (0, 0) and (7, 1), re-spectively. What is its area?

(A) 20./ (B) 22V (C) 25,f3- (D) 27 / (E) 50

17. For a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4,5, and 6 on each die are in the ratio 1 : 2: 3: 4: 5 : 6. What is the probabilityof rolling a total of 7 on the two dice?

(A) 63 (B) 8 (C) 63 (D) 6 (E) 7

18. An object in the plane moves from one lattice point to another. At eachstep, the object may move one unit to the right, one unit to the left, one unitup, or one unit down. If the object starts at the origin and takes a ten-steppath, how many different points could be the final point?

(A) 120 (B) 121 (C) 221 (D) 230 (E) 231

19. Mr. Jones has eight children of different ages. On a family trip his oldestchild, who is 9, spots a license plate with a 4-digit number in which each oftwo digits appears two times. "Look, daddy!" she exclaims. "That numberis evenly divisible by the age of each of us kids!" "That's right," repliesMr. Jones, "and the last two digits just happen to be my age." Which of thefollowing is not the age of one of Mr. Jones's children?

(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

20. Let x be chosen at random from the interval (0, 1). What is the probabilitythat

[log10 4x] - [loglo x] = 0?

Here [x] denotes the greatest integer that is less than or equal to x.

(A) g (B)20

(C) 6 (D) 5 (E) 4

0..m

.N,

KIN

V'1

56 Problems

21. Rectangle A B CD has area 2006. An ellipse with area 20067r passes throughA and C and has foci at B and D. What is the perimeter of the rectangle?(The area of an ellipse is 7rab, where 2a and 2b are the lengths of its axes.)

16 2006 1003(A)

7r

(B) 4 (C) 8 1003 (D) 6,/2006

(E)32 1003

7r

22. Suppose a, b, and c are positive integers with a + b + c = 2006, anda!b!c! = m 10n, where m and n are integers and m is not divisible by 10.What is the smallest possible value of n ?

(A) 489 (B) 492 (C) 495 (D) 498 (E) 501

23. Isosceles AABC has a right angle at C. Point P is inside AABC, suchthat PA = 11, PB = 7, and PC = 6. Legs AC and BC have lengths = a + b,12-, where a and b are positive integers. What is a + b ?

(A) 85 (B) 91 (C) 108 (D) 121 (E) 127

24. Let S be the set of all points (x, y) in the coordinate plane such that 0x < i and 0 < y < i . What is the area of the subset of S for which

sin2 x - sin x sin y + sine y < 3 ?4

(A)7F

2

9

7r2 7re 37r2(B)

8(C) 6 (D) 16 (E)

27r2

9

25. A sequence al, a2, ... of non-negative integers is defined by the rulean+2 = Ian+i - anI for n > 1. If al = 999, a2 < 999, and a2006 = 1,how many different values of a2 are possible?

(A) 165 (B) 324 (C) 495 (D) 499 (E) 660

=-'

,L'

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..oti.



1. One ticket to a show costs $20 at full price. Susan buys 4 tickets using acoupon that gives her a 25% discount. Pam buys 5 tickets using a couponthat gives her a 30% discount. How many more dollars does Pam pay thanSusan?

(A) 2 (B) 5 (C) 10 (D) 15 (E) 20

2. An aquarium has a rectangular base that measures 100 cm by 40 cm andhas a height of 50 cm. It is filled with water to a height of 40 cm. A brickwith a rectangular base that measures 40 cm by 20 cm and a height of 10cm is placed in the aquarium. By how many centimeters does the waterrise?

(A) 0.5 (B) 1 (C) 1.5 (D) 2 (E) 2.5

3. The larger of two consecutive odd integers is three times the smaller. Whatis their sum?

(A) 4 (B) 8 (C) 12 (D) 16 (E) 20

4. Kate rode her bicycle for 30 minutes at a speed of 16 mph, then walkedfor 90 minutes at a speed of 4 mph. What was her overall average speed inmiles per hour?

(A) 7 (B) 9 (C) 10 (D) 12 (E) 14

5. Last year Mr. John Q. Public received an inheritance. He paid 20% in fed-eral taxes on the inheritance, and paid 10% of what he had left in statetaxes. He paid a total of $10,500 for both taxes. How many dollars was theinheritance?

(A) 30,000 (B) 32,500 (C) 35,000 (D) 37,500 (E) 40,000

6. Triangles ABC and ADC are isosceles with AB = BC and AD = DC.Point D is inside AABC, /ABC = 40°, and LADC = 1400. What isthe degree measure of /BAD ?

(A) 20 (B) 30 (C) 40 (D) 50 (E) 60

7. Let a, h, c, d, and e be five consecutive terms in an arithmetic sequence,and suppose that a + b + c + d + e = 30. Which of the following can befound?

(A) a (B) b (C) c (D) d (E) e

C"7

001

..p

.....J

}C,

.N-,

0.^

+v.

100..O

.-..-.

.-.

58 Problems

8. A star-polygon is drawn on a clock face by drawing a chord from eachnumber to the fifth number counted clockwise from that number. That is,chords are drawn from 12 to 5, from 5 to 10, from 10 to 3, and so on,ending back at 12. What is the degree measure of the angle at each vertexin the star-polygon?

(A) 20 (B) 24 (C) 30 (D) 36 (E) 60

9. Yan is somewhere between his home and the stadium. To get to the stadiumhe can walk directly to the stadium, or else he can walk home and then ridehis bicycle to the stadium. He rides 7 times as fast as he walks, and bothchoices require the same amount of time. What is the ratio of Yan's distancefrom his home to his distance from the stadium?

(A) 2 (B) 3 (C) 4 (D) 5 (E) 63 4 5 6 7

10. A triangle with side lengths in the ratio 3: 4: 5 is inscribed in a circle ofradius 3. What is the area of the triangle?

(A) 8.64 (B) 12 (C) 5mr (D) 17.28 (E) 18

11. A finite sequence of three-digit integers has the property that the tens andunits digits of each term are, respectively, the hundreds and tens digits ofthe next term, and the tens and units digits of the last term are, respectively,the hundreds and tens digits of the first term. For example, such a sequencemight begin with terms 247, 475, and 756 and end with the term 824. LetS be the sum of all the terms in the sequence. What is the largest primenumber that always divides S?

(A) 3 (B) 7 (C) 13 (D) 37 (E) 43

12. Integers a, b, c, and d, not necessarily distinct, are chosen independentlyand at random from 0 to 2007, inclusive. What is the probability that ad -be is even?

(A) (B) 6 (C)2

(D) 6 (E)8 8

13. A piece of cheese is located at (12, 10) in a coordinate plane. A mouse isat (4, -2) and is running up the line y = -5x + 18. At the point (a, b) themouse starts getting farther from the cheese rather than closer to it. Whatisa+b?(A) 6 (B) 10 (C) 14 (D) 18 (E) 22

14. Let a, b, c, d, and e be distinct integers such that

(6 - a)(6 - b)(6 - c)(6 - d)(6 - e) = 45.

(CD

..O


What is a+b+c+d+e?(A) 5 (B) 17 (C) 25 (D) 27 (E) 30

15. The set {3, 6, 9, 10} is augmented by a fifth element n, not equal to any ofthe other four. The median of the resulting set is equal to its mean. What isthe sum of all possible values of n ?

(A) 7 (B) 9 (C) 19 (D) 24 (E) 26

16. How many three-digit numbers are composed of three distinct digits suchthat one digit is the average of the other two?

(A) 96 (B) 104 (C) 112 (D) 120 (E) 256

17. Suppose that sin a + sin bcos(a - b) ?

(A)3/_l (B) 1

3 2 3

18. The polynomial J'(x) = x4 + ax3 + bx2 + cx + d has real coefficients,and f (2i) = f (2 + i) = 0. What is a + b + c + d ?

(A) 0 (B) 1 (C) 4 (D) 9 (E) 16

19. Triangles ABC and ADE have areas 2007 and 7002, respectively, withB = (0, 0), C = (223, 0), D = (680, 380), and E = (689, 389). What isthe sum of all possible x-coordinates of A ?

(A) 282 (B) 300 (C) 600 (D) 900 (E) 1200

20. Corners are sliced off a unit cube so that the six faces each become regularoctagons. What is the total volume of the removed tetrahedra?

5,/2-7 10-7,/2- 3-2,/2- 8J-11(A) 3 (B) 3 (C) 3 (D) 3

(E)6-4/

3

21. The sum of the zeros, the product of the zeros, and the sum of the coef-ficients of the function f (x) = ax 2 + bx + c are equal. Their commonvalue must also be which of the following?

(A) the coefficient of x2 (B) the coefficient of x

(C) the y-intercept of the graph of y = f (x)

(D) one of the x-intercepts of the graph of y = f (x)

(E) the mean of the x-intercepts of the graph of y = f (x)

= V51-3 and cos a + cos b = 1. What is

(C) 1 (D)

?(E) 1

It)

'.D

C17

'"'

60 Problems

22. For each positive integer n, let S(n) denote the sum of the digits of n. Forhow many values of n is n + S(n) + S(S(n)) = 2007?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

23. Square ABCD has area 36, and AB is parallel to the x-axis. Vertices A,B, and C are on the graphs of y = log,, x, y = 2 log, x, and y = 3 log,, x,respectively. What is a?

(A) (B) 13 (C) (D) (E) 6

24. For each integer n > 1, let F(n) be the number of solutions of the equationsin x = sin nx on the interval [0, n]. What is yzoo. F(n)?

(A) 2,014,524 (B) 2,015,028 (C) 2,015,033 (D) 2,016,532

(E) 2,017,033

25. Call a set of integers spacy if it contains no more than one out of any threeconsecutive integers. How many subsets of { 1, 2, 3, ... , 12}, including theempty set, are spacy?

(A) 121 (B) 123 (C) 125 (D) 127 (E) 129

,CO

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1. Isabella's house has 3 bedrooms. Each bedroom is 12 feet long, 10 feetwide, and 8 feet high. Isabella must paint the walls of all the bedrooms.Doorways and windows, which will not be painted, occupy 60 square feetin each bedroom. How many square feet of walls must be painted?

(A) 678 (B) 768 (C) 786 (D) 867 (E) 876

2. A college student drove his compact car 120 miles home for the weekendand averaged 30 miles per gallon. On the return trip the student drove hisparents' SUV and averaged only 20 miles per gallon. What was the averagegas mileage, in miles per gallon, for the round trip?

(A) 22 (B) 24 (C) 25 (D) 26 (E) 28

3. The point 0 is the center of the circle circumscribed about AABC, withLBOC = 120° and LAOB = 140°, as shown. What is the degree mea-sure of /ABC ?

(A) 35 (B) 40 (C) 45 (D) 50 (E) 60

4. At Frank's Fruit Market, 3 bananas cost as much as 2 apples, and 6 applescost as much as 4 oranges. How many oranges cost as much as 18 bananas?

(A) 6 (B) 8 (C) 9 (D) 12 (E) 18

5. The 2007 AMC 12 contests will be scored by awarding 6 points for eachcorrect response, 0 points for each incorrect response, and 1.5 points foreach problem left unanswered. After looking over the 25 problems, Sarahhas decided to attempt the first 22 and leave the last 3 unanswered. Howmany of the first 22 problems must she solve correctly in order to score atleast 100 points?

(A) 13 (B) 14 (C) 15 (D) 16 (E) 17

_`n

=03

62 Problems

6. Triangle ABC has side lengths AB = 5, BC = 6, and AC = 7. Twobugs start simultaneously from A and crawl along the sides of the trianglein opposite directions at the same speed. They meet at point D. What isBD?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

7. All sides of the convex pentagon ABCDE are of equal length, and LA =LB = 90°. What is the degree measure of LE ?

(A) 90 (B) 108 (C) 120 (D) 144 (E) 150

8. Tom's age is T years, which is also the sum of the ages of his three children.His age N years ago was twice the sum of their ages then. What is T/N?

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6

9. A function f has the property that f (3x - 1) = x2 + x + 1 for all realnumbers x. What is f (5) ?

(A) 7 (B) 13 (C) 31 (D) 111 (E) 211

10. Some boys and girls are having a car wash to raise money for a class tripto China. Initially 40% of the group are girls. Shortly thereafter two girlsleave and two boys arrive, and then 30% of the group are girls. How manygirls were initially in the group?

(A) 4 (B) 6 (C) 8 (D) 10 (E) 12

11. The angles of quadrilateral ABCD satisfy LA = 2LB = 3LC = 4LD.What is the degree measure of LA, rounded to the nearest whole number?

(A) 125 (B) 144 (C) 153 (D) 173 (E) 180

12. A teacher gave a test to a class in which 10% of the students are juniorsand 90% are seniors. The average score on the test was 84. The juniors allreceived the same score, and the average score of the seniors was 83. Whatscore did each of the juniors receive on the test?

(A) 85 (B) 88 (C) 93 (D) 94 (E) 98

13. A traffic light runs repeatedly through the following cycle: green for 30seconds, then yellow for 3 seconds, and then red for 30 seconds. Leahpicks a random three-second time interval to watch the light. What is theprobability that the color changes while she is watching?

(A) 63 (B) 21 (C) 10 (D) 7 (E) 3

...


14. Point P is inside equilateral AABC. Points Q, R, and S are the feet ofthe perpendiculars from P to AB, BC, and CA, respectively. Given thatPQ = 1, PR = 2, and PS = 3, what is AB ?

(A) 4 (B) 3J (C) 6 (D) 4/3 (E) 9

15. The geometric series a + a r + a r2 + has a sum of 7, and the termsinvolving odd powers of r have a sum of 3. What is a + r ?(A) 4 (B) 12 (C) 3 (D) 7 (E) 5

3 7 2 3 2

16. Each face of a regular tetrahedron is painted either red, white, or blue. Twocolorings are considered indistinguishable if two congruent tetrahedra withthose colorings can be rotated so that their appearances are identical. Howmany distinguishable colorings are possible?

(A) 15 (B) 18 (C) 27 (D) 54 (E) 81

17. If a is a nonzero integer and b is a positive number such that ab2 = log10 b,what is the median of the set {0, 1, a, b, 1 /b} ?

(A) 0 (B) 1 (C) a (D) b (E)b

18. Let a, b, and c be digits with a 0. The three-digit integer abc lies onethird of the way from the square of a positive integer to the square of thenext larger integer. The integer acb lies two thirds of the way between thesame two squares. What is a + b + c ?

(A) 10 (B) 13 (C) 16 (D) 18 (E) 21

19. Rhombus ABCD, with side length 6, is rolled to form a cylinder of volume6 by taping AB to DC. What is sin (LA BC) ?

(A) -(B) 1 (C)

7r

(D) -a (E)9 2 6 4 2

20. The parallelogram bounded by the lines y = ax + c, y = ax + d, y =bx + c, and y = bx + d has area 18. The parallelogram bounded by thelines y = ax + c, y = ax - d, y = bx + c, and y = bx - d has area 72.Given that a, b, c, and d are positive integers, what is the smallest possiblevalue ofa+b+c+d?(A) 13 (B) 14 (C) 15 (D) 16 (E) 17

21. The first 2007 positive integers are each written in base 3. How many ofthese base-3 representations are palindromes? (A palindrome is a numberthat reads the same forward and backward.)

(A) 100 (B) 101 (C) 102 (D) 103 (E) 104

64 Problems

22. Two particles move along the edges of equilateral AABC in the direction

A -+ B-*C-*A,

starting simultaneously and moving at the same speed. One starts at A, andthe other starts at the midpoint of BC. The midpoint of the line segmentjoining the two particles traces out a path that encloses a region R. What isthe ratio of the area of R to the area of L ABC?

(A) 16 (B) 12 (C) 9 (D) 6 (E) 4

23. How many non-congruent right triangles with positive integer leg lengthshave areas that are numerically equal to 3 times their perimeters?

(A) 6 (B) 7 (C) 8 (D) 10 (E) 12

24. How many pairs of positive integers (a, b) are there such that gcd(a, b) _1 and

a 14b

b 9a

is an integer?


25. Points A, B, C, D, and E are located in 3-dimensional space with AB =BC = CD = DE = EA = 2 and LABC = LCDE = LDEA = 90°.The plane of AABC is parallel to DE. What is the area of L BDE?

(A) (B) (C) 2 (D) (E)

<`c

ON

O

--'

V"1

I'D.-r

SolutionsProblem Difficulty for 2001 12

A box indicates the correct answer choice.

Problem % A % B % C % D % E % Omitted

1 3.41 0.61 2.23 21.82 66.19 5.72

2 6.08 3.89 3.04 2.10 48.23 36.64

3 2.60 29.00 20.72 2.60 1.18 43.89

4 1.80 7.46 4.49 45.25 5.67 35.31

5 3.51 8.52 13.90 14.33 2.61 57.11

6 15.91 7.52 5.95 2.33 53.18 15.09

7 9.89 3.71 4.80 6.34 4.44 70.82

8 10.03 2.98 32.75 3.66 6.47 44.09

9 1.15 2.43 11.75 3.01 19.06 62.57

10 15.30 2.25 3.67 56.25 5.17 17.34

11 12.52 31.85 7.74 10.28 2.58 35.02

12 13.30 14.17 13.34 4.65 5.11 49.42

13 2.32 5.03 4.92 3.74 12.78 71.20

14 2.69 10.28 4.41 2.28 5.78 74.55

15 6.32 22.91 6.18 2.49 5.99 56.09

16 9.71 9.17 24.76 7.99 6.17 42.19

17 19.33 3.56 7.98 5.21 3.44 60.47

18 7.40 4.62 3.43 4.88 11.06 68.59

19 1.46 2.26 2.42 3.86 1.69 88.29

20 5.19 3.75 10.99 8.83 10.54 60.69

21 2.12 2.76 3.40 6.90 2.17 82.65

22 1.63 6.14 2.28 3.89 7.88 78.17

23 2.95 4.44 2.55 2.14 1.73 86.17

24 4.20 14.14 5.28 10.88 20.00 45.49

25 6.57 3.60 3.90 1.93 4.65 79.34

65

Aim G

..

66 Solutions

2001 AMC 12 SolutionsThe problems for the 2001 AMC 12 begin on page 1.

1. ANSWER (E) Suppose the two numbers are a and b. Then the desiredsum is

2(a + 3) + 2(b + 3) = 2(a + b) + 12 = 2S + 12.

2. ANSWER (E) Suppose N = 10a + b. Then 10a + b = ab + (a + b). Itfollows that 9a = ab, which implies that b = 9, since a # 0. Note thatthe numbers 19, 29, 39, ... , 99 all meet the required condition.

3. ANSWER (B) Because Kristin's tax was more than p% of her income,she earned more than $28,000. If her annual income is x dollars, then

100 .28000 + pl 02 . (x - 28000) = p10025

Multiplying both sides by 100 and expanding yields

28000p + px + 2x - 28000p - 56000 = px + 0.25x.

So 1.75x = 2 x = 56000 and x = 32000.

4. ANSWER (D) Since the median is 5, we can write the three numbers asx, 5, and y, where

1 13(x+5+y)=x+10 and 3(x+5+y)+15=y.

If we add these equations, we get

23(x+5+y)+15=x+y+ 10,

and solving for x + y gives x + y = 25. Hence the sum of the numbers isx+5+y= 30.

OR

The difference between the smallest number and the largest is 10 + 15 =25, so let the three numbers be x, 5, and x + 25. The mean of the three

inn

+

2001 AMC 12 Solutions 67

numbers is 10 more than x, so

3(x+5+(x+25))=x+10

Solving for x gives x = 0, so the three numbers are 0, 5, and 25, and theirsum is 30.

OR

Let m be the mean of the three numbers. Then the least of the numbers ism - 10 and the greatest is m + 15. The middle number of the three is themedian, 5. So

3((m-10)+5+(m+15))=m

and m = 10. Hence the sum of the three numbers is 3(10) = 30.

5. ANSWER (D) Note that

1 .3 9999--1 10000!

= _25000.1 .2...5000

10000!= 25000 .5000!

6. ANSWER (E) The last four digits (GHIJ) are either 9753 or 7531, andthe remaining odd digit (either 1 or 9) is A, B, or C. Since A + B + C= 9, the odd digit among A, B, and C must be 1. Thus the sum of thetwo even digits in ABC is 8. The three digits in DEF are 864, 642, or 420,leaving the pairs 2 and 0, 8 and 0, or 8 and 6, respectively, as the twoeven digits in ABC. Of those, only the pair 8 and 0 has sum 8, so ABC is810, and the required first digit is 8. The only such telephone number is810-642-9753.

7. ANSWER (A) Let n be the number of full-price tickets and p be the priceof each in dollars. Then

np+(140-n) 2 = 2001, so p(n + 140) = 4002.

Thus n + 140 must be a factor of 4002 = 2.3.23.29. Since 0 < n < 140,we have 140 < n + 140 < 280, and the only factor of 4002 that is in the

S-=

ago

V'1

68 Solutions

required range for n + 140 is 174 = 2 3 29. Therefore n + 140 = 174,so n = 34 and p = 23. The money raised by the full-price tickets is34. 23 = 782 dollars.

8. ANSWER (C) The slant height of the cone is 10, the radius of the sec-tor. The circumference of the base of the cone is the same as the lengthof the sector's arc. This is 252/360 = 7/10 of the circumference of thecircle from which the sector is cut. The base circumference of the cone is(7/10)(20n) = 14,-r, so its radius is 7.

9. ANSWER (C) Note that

f(600) = f (500. 6) = f(5 0) =63

5- 5.

5

For all positive x,

f(x) = f(1 x) = f(1)x

so xf(x) is the constant f(1). Therefore

600f (600) = 500f (500) = 500(3) = 1500,

so f (600) = 1500 = 5600 2'

NOTE. f (x) = 15x00 is the unique function satisfying the given condi-tions.

10. ANSWER (D) The pattern shown at left is repeated in the plane. In fact,nine repetitions of it are shown in the statement of the problem. Note thatthe three-by-three square shown is made up of four unit squares and fourpentagons, and that four of the nine squares in the three-by-three square arenot in the four pentagons that make up the three-by-three square. Thereforethe percentage of the plane that is enclosed by pentagons is

4 5 51--=-=55-%.9 9 9

;1?

t-4

C1.

L1.

^..

...

,-.


1 _ 1 3 S 6 7 % 9

11. ANSWER (D) Think of continuing the drawing until all five chips areremoved from the box. There are ten possible orderings of the colors:RRRWW, RRWRW, RWRRW, WRRRW, RRWWR, RWRWR, WRRWR,RWWRR, WRWRR, and WWRRR. The six orderings that end in R rep-resent drawings that would have ended when the second white chip wasdrawn.

OR

Imagine drawing until only one chip remains. If the remaining chip is red,then that draw would have ended when the second white chip was removed.The last chip will be red with probability 3/5.

12. ANSWER (B) For integers not exceeding 2001, there are L2001/3J = 667multiples of 3 and L2001/4] = 500 multiples of 4. The total of 1167 countsthe L2001/12] = 166 multiples of 12 twice, so there are 1167 - 166 =1001 multiples of 3 or 4. From these we exclude the L2001/15] = 133multiples of 15 and the L2001/20] = 100 multiples of 20, since these aremultiples of 5. However, this excludes the L2001/60J = 33 multiples of60 twice, so we must re-include these. The number of integers satisfyingthe conditions is 1001 - 133 - 100 + 33 = 801.

13. ANSWER (E) The equation of the first parabola can be written in the form

y = a(x - h)2 + k = axe - 2axh + ah2 + k,

and the equation for the second (having the same shape and vertex, butopening in the opposite direction) can be written in the form

y = -a(x - h)2 + k = -axe + 2axh - ah2 + k.

LL

.

'-O

j.,

..rfl.

70 Solutions

Hence

a+b+c+d+e+f= a + (-2ah) + (ah2 + k) + (-a) + (2ah) + (-ah2 + k) = 2k.

OR

The reflection of a point (x, y) about the line y = k is (x, 2k - y). Thusthe equation of the reflected parabola is

2k - y = ax 2 + bx + c, or equivalently, y = 2k - (ax2 + bx + c).

Hencea+b+c+d+e+f =2k.

NOTE: The case y = x2 + 2 demonstrates that none of the other ans-wer choices is true in every case.

14. ANSWER (D) Each of the (9) = 36 pairs of vertices determines twoequilateral triangles, for a total of 72 triangles. However, the three trianglesAI A4A7, A2A5A8, and A3A6A9 are each counted 3 times, resulting in anovercount of 6. Thus there are 66 distinct equilateral triangles.

15. ANSWER (B) Unfold the tetrahedron onto a plane. The two opposite-edge midpoints become the midpoints of opposite sides of a rhombus withsides of length 1, so are now one unit apart. Folding the triangle back to atetrahedron does not change the distance and it remains minimal.

16. ANSWER (D) Number the spider's legs from 1 through 8, and let ak andbk denote the sock and shoe that will go on leg k. A possible

0.'


arrangement of the socks and shoes is a permutation of the sixteen sym-bols a1, bl,... , as, b8, in which ak precedes bk for 1 < k < 8. Thereare 16! permutations of the sixteen symbols, and a 1 precedes b1 in exactlyhalf of these, or 16!/2 permutations. Similarly, a2 precedes b2 in exactlyhalf of those, or 16!/22 permutations. Continuing, we can conclude that akprecedes hk for 1 < k < 8 in exactly 16!/28 permutations.

17. ANSWER (C) Since ZAPB = 90° if and only if P lies on the semicirclewith center (2, 1) and radius the angle is obtuse if and only if thepoint P lies inside this semicircle. The semicircle lies entirely inside thepentagon, since the distance, 3, from (2, 1) to DE is greater than the radiusof the circle. Thus the probability that the angle is obtuse is the ratio of thearea of the semicircle to the area of the pentagon.

Let 0 = (0, 0). Then the area of the pentagon is

[ABCDE] = [OCDE] - [OAB] = 4 .(27r + 1) - 2 (2 - 4) = Sir,

and the area of the semicircle is

2n(.f) =zn.

The probability is

Zn _ 5

Sir 16

18. ANSWER (D) Let C be the intersection of the horizontal line through Aand the vertical line through B. In right triangle ABC, we have BC = 3and AB = 5, so AC = 4. Let x be the radius of the third circle, and

.-j

CIA

0..,L

'

72 Solutions

let D be the center. Let E and F be the points of intersection of the hori-zontal line through D with the vertical lines through B and A, respectively.

In ABED we have BD = 4 + x and BE = 4 - x, so

DE2 = (4 + x)2 - (4 - x)2 = 16x,

and DE = In LADF we have AD = 1+xandAF= 1-x,so

FD2 = (1 + x)2 - (1 - x)2 = 4x,

and FD = 2,17. Hence

4=AC = FD+DE=2vfx- +4/=6J,

so = 2, and x = 4.

19. ANSWER (A) The sum and product of the zeros of P(x) are -a and -c,respectively. Therefore

a-3 = -c= 1+a+b+c.

Since c = P(O) is the y-intercept of y = P(x), it follows that c = 2.Thus a =6andb=-11.

20. ANSWER (C) Let the midpoints of sides AB, BC, CD and DA be M, N,P, and Q, respectively. Then M = (2,5) and N = (3, 2). Since MNhas slope -3, the slope of MQ must be 1/3, and MQ = MN = 10.

An equation for the line containing MQ is thus y - 5 = (x - 2)/3, or


y = (x + 13)/3. So Q has coordinates of the form (a, (a + 13)/3). SinceMQ = 10, we have

((a - 2)2 +a +

313

5)2 = 10

(a-2)2+(a32) =10

9(a-2)2=10

(a - 2)2 = 9

a-2=+3

Since Q is in the first quadrant, a = 5 and Q = (5, 6). Since Q is themidpoint of AD and A = (3, 9), we have D = (7, 3), and the sum of thecoordinates of D is 10.

OR

Use translation by vectors. As before, M = (2, 5) and N = (3, 2), soNM = (-1, 3). The vector MN must have the same length as MN andbe perpendicular to it. Because Q is in the first quadrant, MN = (3, 1).Thus Q = (5, 6). As before, D = (7, 3), and the answer is 10.

OR

Each pair of opposite sides of the square is parallel to a diagonal of A B CD,so the diagonals of A BCD are perpendicular. Similarly, each pair of oppo-site sides of the square has length half that of a diagonal, so the diagonalsof ABCD are congruent. Since the slope of AC is -3 and AC is perpen-dicular to BD, we have

b-1 1 so a-1=3(b-1).a-1 3'

Since AC = BD, and (5 - 3)2 + (3 - 9)2 = 40, we have

40=(a-1)2+(b-1)2=9(b-1)2+(b-1)2=10(b-1)2.

Sob is positive, b = 3 and a = 1 + 3(b - 1) = 7, and the answer is 10.

.°,, `".J

_`

74 Solutions


(a+1)(b+1)=ab+a+b+1 =524+1 =525=3.52.7,

and

(b+l)(c+1)=be+b+c+1=146+1=147=3.72.

Since (a + 1)(b + 1) is a multiple of 25 and (b+ 1)(c + 1) is not a multipleof 5, it follows that a + 1 must be a multiple of 25. Since a + 1 divides525, a is one of 24, 74, 174, or 524. Among these only 24 is a divisor of 8!,so a = 24. This implies that b + 1 = 21, and b = 20. From this it followsthat c + 1 = 7 and c = 6. Finally, (c + 1) (d + 1) = 105 = 3 5 7, sod + 1 = 15 and d = 14. Therefore a - d = 24 - 14 = 10.

22. ANSWER (C) The area of triangle EFG is (1/6)(70) = 35/3. Tri-angles AFH and CEH are similar, so 3/2 = EC/AF = EH/HFand EH/EF = 3/5. Triangles AGJ and CEJ are similar, so 3/4 =EC/AG = EJ/JG and EJ/EG = 3/7.

D E

AF G

C

B

Since the areas of triangles that have a common altitude are proportional totheir bases, the ratio of the area of AEHJ to the area of AEHG is 3/7,and the ratio of the area of AEHG to that of AEFG is 3/5. Therefore theratio of the area of AEHJ to the area of AEFG is (3/5)(3/7) = 9/35.Thus the area of AEHJ is (9/35)(35/3) = 3.

23. ANSWER (A) If r and s are the integer zeros, the polynomial can bewritten in the form

P(x) = (x - r)(x - s)(x2 + ax + )

The coefficients of x3 and x2 are, respectively, a - (r + s) and, ,B - a (r +s)+rs. Because r and s are integers, so are a and ,B. Applying the quadratic

SIN

C1:


formula gives the remaining zeros as

22(-af a2-4,)=-2+i 2

Answer choices (A), (B), (C), and (E) require that a = -1, which impliesthat the imaginary parts of the remaining zeros have the form ± 4/B - 1/2.This is true only for choice (A).Note that choice (D) is not possible since this choice requires a = -2.This produces an imaginary part of the form ,B - 1, which cannot be 2.

24. ANSWER (D) Let E be a point on AD such that CE is perpendicular toAD, and draw BE. Since /ADC is an exterior angle of AADB, it followsthat

/ADC = /DAB + /ABD = 150 + 45° = 60°.

Thus ACDE is a 30°-60°-90° triangle and DE = 1CD = BD. HenceABDE is isosceles and /EBD = /BED = 30°. But /ECB is alsoequal to 30° and therefore ABEC is isosceles with BE = EC. On theother hand,

/ABE = /ABD - /EBD = 45°- 30° = 15° = /EAB.

Thus AABE is isosceles with AE = BE. Hence AE = BE = EC.Right triangle AEC is also isosceles with ZEAC = ZECA = 45°.Hence

/ACB = /ECA + /ECD = 45° + 30° = 75°.

25. ANSWER (D) If a, b, and c are three consecutive terms of such a se-quence, thenac - 1 = b, so c = (1 + b)/a. Applying this rule recursively and simpli-fying yields

l+b l+a+b l+a a, b, ...

76 Solutions

This shows that at most five different terms can appear in such a sequence.Moreover, the value of a is determined once the value 2000 is assigned to band the value 2001 is assigned to another of the first five terms. Thus thereare four such sequences that contain 2001 as a term, namely

201 , 2000, 1, 1000 ' 1000 ' 2001, .. ,

1, 2000, 2001, 1000' 1000' 1' '

2001 2000,4001999,2001 , 2002 2001 and4001999 4001999' 4001999 ....

2001 20024001999,2000 ,400 1999' 400 999' 2001, 4001999, ....

The four values of x are 2001, 1, 2001 and 4001999.4001999'

r-+

!.n

Vii

v",

v",

.-r

2002 AMC 12A Solutions 77

Problem Difficulty for 2002 12AA box indicates the correct answer choice.


1 37.98 2.74 16.01 23.51 2.08 17.66

2 93.73 0.36 0.39 0.26 3.75 1.49

3 33.70 31.93 4.35 2.10 1.12 26.80

4 1.94 65.75 2.13 4.49 0.95 24.73

5 1.81 1.94 65.96 12.72 1.30 16.26

6 3.22 0.78 1.05 0.61 71.25 23.06

7 21.35 21.12 1.11 26.59 6.06 23.75

8 76.00 3.19 2.46 2.77 1.61 13.94

9 20.22 39.97 7.62 21.60 2.00 8.57

10 11.61 10.34 5.99 54.70 7.18 10.17

11 0.96 25.61 44.95 2.86 1.05 24.56

12 14.71 17.14 4.78 1.40 10.08 51.86

13 8.38 3.85 10.89 2.02 3.71 71.15

14 3.52 1.54 5.12 27.24 6.10 56.46

15 5.31 19.72 4.28 8.53 12.32 49.83

16 18.19 7.56 8.15 5.71 1.60 58.77

17 2.45 10.37 13.75 6.33 4.12 62.98

18 6.86 3.64 5.02 2.15 2.80 79.52

19 22.32 11.17 3.42 13.34 1.09 48.66

20 10.22 5.84 8.52 5.42 3.69 66.31

21 1.61 8.53 5.55 5.61 1.68 77.02

22 3.76 3.94 10.75 2.85 3.17 75.53

23 1.03 2.20 2.85 6.49 2.75 84.67

24 2.72 1.63 2.04 3.98 0.82 88.80

25 8.84 2.64 5.01 2.73 3.28 77.49

L1.

78 Solutions

2002 AMC 12A SolutionsThe problems for the 2002 AMC 12A begin on page 8.

1. ANSWER (A) Factor to get (2x + 3)(2x - 10) = 0, so the two roots are-3/2 and 5, which sum to 7/2.

2. ANSWER (A) Let x be the number she was given. Her calculations pro-duce

x-93

so

= 43,

x-9=129 and x=138.The correct answer is

138 - 3 135= 15.

9 9

3. ANSWER (B) No matter how the exponentiations are performed, 222 al-ways gives 16. Therefore, the only other possible results are

(222)2 = 256, 2(222) = 65, 536, or (22)(22) = 256,

so there is one other possible value.

4. ANSWER (B) The appropriate angle x satisfies

190-x = 4(180-x), so 360-4x = 180-x.

Solving for x gives 3x = 180, so x = 60.

5. ANSWER (C) The large circle has radius 3, so its area is 7r . 32 = 9nThe seven small circles have a total area of 7 (yr . 12) = 7ir. So the shadedregion has area 9ir - 77r = 2,-r.

6. ANSWER (E) When n = 1, the inequality becomes m < 1 + m, whichis satisfied by all integers m. Thus there are infinitely many of the desiredvalues of m.


7. ANSWER (A) Let CA = 2nRA be the circumference of circle A, letCB = 2JrRB be the circumference of circle B, and let L the commonlength of the two arcs. Then

45

60CA

L36030

3

Therefore

CA 2 2 27rRA RAso =

CB 3 3 2JrRB RB

Thus, the ratio of the areas is

Area of Circle A 7rR2A RA 2 _ 4

Area of Circle B 7rRB RB 9

8. ANSWER (A) Draw additional lines to cover the entire figure with con-gruent triangles. There are 24 triangles in the blue region, 24 in the whiteregion, and 16 in the red region. Thus B = W.

9. ANSWER (B) First note that the amount of memory needed to store the30 files is

3(0.8) + 12(0.7) + 15(0.4) = 16.8 mb,

so the number of disks is at least

16.8144-11+3.

However, a disk that contains a 0.8-mb file can, in addition, hold only one0.4-mb file, so on each of these disks at least 0.24 mb must remain unused.Hence there is at least 3(0.24) = 0.72 mb of unused memory, which isequivalent to half a disk. Since

211+3 +2>12,

o0-

80 Solutions

at least 13 disks are needed.To see that 13 disks suffice, note that:Six disks could be used to store the 12 files containing 0.7 mb;Three disks could be used to store the three 0.8-mb files together with threeof the 0.4-mb files;Four disks could be used to store the remaining twelve 0.4-mb files.

10. ANSWER (D) After the first transfer, the first cup contains two ounces ofcoffee, and the second cup contains two ounces of coffee and four ouncesof cream. After the second transfer, the first cup contains 2 + (1/2)(2) = 3ounces of coffee and (1/2)(4) = 2 ounces of cream. Therefore, the fractionof the liquid in the first cup that is cream is 2/(2 + 3) = 2/5.

It. ANSWER (B) Let t be the number of hours Mr. Bird must travel toarrive precisely on time. Since three minutes is the same as 0.05 hours,40(t + 0.05) = 60(t - 0.05). Thus

40t + 2 = 60t - 3, so t = 0.25.

The distance from his home to work is 40(0.25 + 0.05) = 12 miles. There-fore, his average speed should be 12/0.25 = 48 miles per hour.

OR

Let d be the distance from Mr. Bird's house to work, and let s be the desiredaverage speed. Then the desired driving time is d/s. Since d/60 is threeminutes too short and d/40 is three minutes too long, the desired time mustbe the average, so

d _ 1 d d

S 2 60 + 40

1 1 1 1 5 1

Th -=- =48=+

usS 2 (60 40 240

sos .

48'

12. ANSWER (B) Let p and q be two primes that are roots of x2 -63x+k =0. Then

so p + q = 63 and p q = k. Since 63 is odd, one of the primes must be2 and the other 61. Thus there is exactly one possible value fork, namely

o-0

,.O

V'1


13. ANSWER (C) A number x differs by one from its reciprocal if and only ifx-1 = 1 /x or x+ 1 = 1 /x. These equations are equivalent to x2-x-1 =0 and x2 + x - 1 = 0. Solving these by the quadratic formula yields thepositive solutions

I 12 5and - 12 1

which are reciprocals of each other. The sum of the two numbers is 15.

14. ANSWER (D) We have

N = 1o92002 112 + 1o92002 132 + 1o92002 142 = 1o92002 112 . 132 142

= 1092002 (11 13 14)2.

Simplifying gives

N = 1o92002 (11 13 14)2 = 21o92002 2002 = 2. 1 = 2.

15. ANSWER (D) The values 6, 6, 6, 8, 8, 8, 8, 14 satisfy the requirementsof the problem, so the answer is at least 14. If the largest number were 15,the collection would have the ordered form 7, _, -, 8, 8, -, -, 15. But7 + 8 + 8 + 15 = 38, and a mean of 8 implies that the sum of all values is64. In this case, the four missing values would sum to 64 - 38 = 26, andtheir average value would be 6.5. This implies that at least one would beless than 7, which is a contradiction. Therefore the largest integer that canbe in the set is 14.

16. ANSWER (A) There are ten ways for Tina to select a pair of numbers. Thesums 9, 8, 4, and 3 can be obtained in just one way, and the sums 7, 6, and5 can each be obtained in two ways. The probability for each of Sergio'schoices is 1/10. Considering his selections in decreasing order, the totalprobability of Sergio's choice being greater is

10111+ 9+ 8+ 6+ 4+ 2+10+0+0+01=5.

17. ANSWER (B) First, observe that 4, 6, and 8 cannot be the units digit of anytwo-digit prime, so they must contribute at least 40 + 60 + 80 = 180 to the

...

V-.

82 Solutions

sum. The remaining digits must contribute at least 1 +2+3+5+7+9 = 27to the sum. Thus the sum must be at least 207, and we can achieve thisminimum only if we can construct a set of three one-digit primes and threetwo-digit primes. Using the facts that nine is not prime and neither twonor five can be the units digit of any two-digit prime, we can construct thesets {2, 3, 5, 41, 67, 89}, {2, 3, 5, 47, 61, 89}, or {2, 5, 7, 43, 61, 89}, each ofwhich yields a sum of 207.

18. ANSWER (C) The centers are at A = (10,0) and B = (-15,0), andthe radii are 6 and 9, respectively. Since the internal tangent is shorter thanthe external tangent, PQ intersects AB at a point D that divides AB intoparts proportional to the radii. Triangles APD and BQD are similar withratio of similarity 2 : 3. Therefore D = (0, 0), PD = 8, and QD = 12, soPQ = 20.

19. ANSWER (D) The equation f (f (x)) = 6 implies that f (x) = -2 orf (x) = 1. The horizontal line y = -2 intersects the graph of f twice,so f (x) = -2 has two solutions. Similarly, f (x) = 1 has 4 solutions, sothere are 6 solutions of f (f (x)) = 6.

20. ANSWER (C) Since 0.ab = ab/99, the denominator must be a factor of99 = 32 . 11. The factors of 99 are 1, 3, 9, 11, 33, and 99. Since a and bare not both nine, the reduced denominator cannot be 1. To see that each ofthe other denominators is possible, note that

1 1 1 1 1

0.01 = 99, 0.03 =33

, 0.09 =11

0.11 =9,

and 0.33 =3

21. ANSWER (B) Note that beginning with the third term of the sequence,each term depends only on the previous two. Writing out more terms of thesequence yields

4,7, 1, 8, 9, 7, 6, 3, 9, 2, 1,3,4,7, 1....

o.,

V')

O0.'

(_.

°X'


Therefore the sequence is periodic with period 12. The sequence repeatsitself, starting with the 13th term. Since S12 = 60, S12k = 60k for allpositive integers k. The largest k for which S12k < 10,000 is

k= [10'000] = 166,

60

and S12.166 = 60 166 = 9960. To have S, > 10,000, we need to addenough additional terms for their sum to exceed 40. This can be done byadding the next 7 terms of the sequence, since their sum is 42. Thus thesmallest value of n is 12. 166 + 7 = 1999.

22. ANSWER (C) Since AB is 10, we have BC = 5 and AC = 5,13. ChooseE on AC so that CE = 5. Then BE = 5,/2-. For BD to be greaterthan 5-,f2-, P has to be inside AABE. The probability that P is insideAABE is

AreaofAABE EA _ 5.-5 _ -1Area of AABC !CA BC AC 5f3-

3-/-3-3

5

C5

23. ANSWER (D) By the Angle Bisector Theorem, AB/BC = 9/7. LetAB = 9x and BC = 7x, let mLABD = m/CBD = 0, and let Mbe the midpoint of BC. Since M is on the perpendicular bisector of BC,we have BD = DC = 7. Then

7x2 'xcos0 = _7 2

84 Solutions

9 D 7

Applying the Law of Cosines to AABD yields

92 = (9x)2 +7 2 - 2(9x)(7) G

from which x = 4/3 and AB = 12. Apply Heron's formula to obtain thearea of triangle ABD as 14.2.5.7 = 14/.

24. ANSWER (E) Let z = a + bi, z = a - bi, and Izi = a2 -+b2. Thegiven relation becomes z2002 = Y. Note that

IZI2002 = IZ20021 = IZI =ZI

from which it follows that

IzI (IZ2001 - 1) = 0.

Hence IzI = 0 or zI = 1. The case IzI = 0 yields only the solutionz = 0. In the case I z I = 1, we have z2002 = z, which is equivalent to22003 = z z = Iz12 = 1. Since the equation z2003 = 1 has 2003 distinctsolutions, there are altogether 1 + 2003 = 2004 ordered pairs that meet therequired conditions.

25. ANSWER (B) The sum of the coefficients of P and the sum of the co-efficients of Q will be equal, so P(1) = Q(1). The only answer choicewith an intersection at x = 1 is (B). (The polynomials in graph B areP(x)=2x4-3x2-3x-4 andQ(x)=-2x4-2x2-2x-2.)

N

N

N

2002 AMC 12B Solutions 85

Problem Difficulty for 2002 12BA box indicates the correct answer choice.


1 77.17 1.30 1.01 1.10 1.77 17.64

2 19.96 0.64 1.11 76.76 0.19 1.32

3 20.04 34.96 11.16 4.97 6.77 22.08

4 1.29 1.16 3.42 2.59 66.11 25.41

5 13.23 2.14 5.31 36.97 2.23 40.12

6 8.26 5.48 49.10 1.84 4.54 30.76

7 1.27 70.33 1.34 1.56 0.93 24.56

8 1.84 10.95 7.05 57.59 9.34 13.18

9 1.74 1.78 42.41 1.68 1.53 50.85

10 18.56 4.83 5.37 6.19 25.69 39.34

11 24.42 3.35 1.40 1.91 45.55 23.34

12 11.36 11.11 26.73 7.83 2.20 40.78

13 3.53 62.56 2.42 4.72 4.10 22.67

14 6.18 2.69 8.80 51.72 3.57 27.03

15 3.15 2.01 2.32 9.19 2.92 80.39

16 8.95 12.70 28.50 12.62 3.18 34.04

17 4.07 39.48 5.54 13.07 19.58 18.24

18 10.18 12.21 24.24 4.79 8.45 40.13

19 1.06 1.56 1.58 36.08 0.56 59.15

20 3.10 14.26 3.56 3.43 1.97 73.69

21 12.60 3.90 1.48 2.83 2.19 76.98

22 1.46 16.3 6 2.10 1.45 1.46 77.16

23 3.18 3.84 5.28 2.33 7.77 77.59

24 2.38 1.14 2.20 1.20 4.49 88.59

25 1.84 1.51 1.96 2.42 2.33 89.94

86

2002 AMC 12B SolutionsThe problems for the 2002 AMC 12B begin on page 13.

1. ANSWER (A) The number M is equal to

1

9 (9 + 99 + 999 + + 999,999,999)

Solutions

= 123,456,789.

The number M does not contain the digit 0.

2. ANSWER (D) Since

(3x - 2)(4x + 1) - (3x - 2)4x + 1 = (3x - 2)(4x + 1 - 4x) + 1=3x-1,

when x = 4 the value is 3 4 - 1 = 11.

3. ANSWER (B) If n > 4, then

n2-3n+2=(n-1)(n-2)is the product of two integers greater than 1, and thus is not prime. Forn = 1, 2, and 3 we have, respectively,

(1 - 1)(1 - 2) = 0, (2- 1)(2-2) =0, and (3- 1)(3-2) = 2.

Therefore n2 - 3n + 2 is prime only when n = 3.

4. ANSWER (E) Note that

0< 1 + 1 + 1 + 1 41 + 1 < 41 +1 <2,2 3 7 n 42 n 42

so the expression is an integer if and only if it is equal to 1. Thus the onlypossible value of n is 42, which is divisible by 2, 3, 6, and 7. The onlyanswer choice that is not true is (E).

5. ANSWER (D) A pentagon can be partitioned into three triangles, so thesum of the degree measures of the angles of the pentagon is

3. 180 = 540 = v + w + x + y + z.

f'1

....O

+

'''


The terms of the arithmetic sequence can be expressed as x - 2d, x - d,x, x + d, and x + 2d, where d is the common difference, so

(x-2d)+(x-d)+x+(x+d)+(x+2d)=5x=540.

Thus x = 108.

6. ANSWER (C) The given conditions imply that

x2+ax+b = (x-a)(x-b) =x2-(a+b)x+ab,

so

a + b = -a and ab = b.

Since b 0, the second equation implies that a = 1. The first equationgives b = -2, so (a, b) = (1, -2).

7. ANSWER (B) Let n - 1, n, and n + 1 denote the three integers. Then

(n - 1)n(n + 1) = 8(3n).

Since n # 0, we have n2 - 1 = 24. It follows that n2 = 25 and n = 5.Thus the sum of the squares is

(n-1)2+n2+(n+1)2= 16+25+36=77.

8. ANSWER (D) Since July has 4. 7 + 3 = 31 days, Monday must be oneof the last three days of July. Therefore Thursday must be one of the firstthree days of August, which also has 31 days. So Thursday must occur fivetimes in August.

9. ANSWER (C) We have b = a + r, c = a + 2r, and d = a + 3r, wherer is a positive real number. Also, b2 = ad yields (a + r)2 = a(a + 3r),or r2 = ar. It follows that r = a and d = a + 3a = 4a. Hence d = 4.

10. ANSWER (A) Each number in the given set is one more than a multipleof 3. Therefore the sum of any three such numbers is itself a multiple of 3.It is easily checked that every multiple of 3 from 1 + 4 + 7 = 12 through13 + 16 + 19 = 38 is obtainable. There are 13 multiples of 3 between 12and 48, inclusive.

88 Solutions

11. ANSWER (E) The numbers A - B and A + B are both odd or both even.However, they are also both prime, so they must both be odd. Thereforeone of A and B is odd and the other even. Because A is a prime betweenA - B and A + B, A must be the odd prime. Because 2 is both the smallestprime and the only even prime, it follows that B = 2. So A - 2, A, andA + 2 are consecutive odd primes and thus must be 3, 5, and 7. The sumof the four primes 2, 3, 5, and 7 is the prime number 17.

12. ANSWER (D) If 20nn = k2, for some k > 0, then n = 20k2/(k2 + 1).Since k2 and k2 + 1 have no common factors and n is an integer, k2 +1 must be a factor of 20. This occurs only when k = 0, 1, 2, or 3. Thecorresponding values of n are 0, 10, 16, and 18.

13. ANSWER (B) Let n, n + 1, ... , n + 17 be the 18 consecutive integers.Then the sum is

18n + (1 + 2 + ... 17) = 18n +17218

= 9(2n + 17).

Since 9 is a perfect square, 2n + 17 must also be a perfect square. Thesmallest value of n for which this occurs is n = 4, so 9(2n + 17) _9.25 = 225.

14. ANSWER (D) Each pair of circles has at most two intersection points.There are (2) = 6 pairs of circles, so there are at most 6 x 2 = 12 pointsof intersection. The following configuration shows that 12 points of inter-section are indeed possible:

15. ANSWER (D) Let a denote the leftmost digit of N and let x denote thethree-digit number obtained by removing a. Then N = 1000a + x = 9x

._"

I0.'


and it follows that 1000a = 8x. Dividing both sides by 8 yields 125a = x.All the values of a in the range 1 to 7 result in three-digit numbers.

16. ANSWER (C) The product will be a multiple of 3 if and only if at leastone of the two rolls is a 3 or a 6. The probability that Juan rolls 3 or 6 is2/8 = 1/4. The probability that Juan does not roll 3 or 6, but Amal doesis (3/4)(1/3) = 1/4. Thus the probability that the product of the rolls is amultiple of 3 is

1 1 1

44 2

17. ANSWER (B) Let A be the number of square feet in Andy's lawn. ThenA/2 and A/3 are the areas of Beth's lawn and Carlos' lawn, respectively,in square feet. Let R be the rate, in square feet per minute, that Carlos'lawn mower cuts. Then Beth's mower and Andy's mower cut at rates of2R and 3R square feet per minute, respectively. Thus

AAndy takes

3Rminutes to mow his lawn,

and

A/2 ABeth takes

2R 4Rminutes to mow hers,

A/3 ACarlos takes =

R 3Rminutes to mow his.

Since R < 3R, Beth will finish first.

18. ANSWER (C) The point P is closer to (0, 0) than it is to (3, 1) if and onlyif P is within the trapezoid bounded by the lines y = 1, the x- and y- axes,and the perpendicular bisector of the segment joining (0, 0) and (3, 1). Theperpendicular bisector goes through the point (3/2, 1/2), which is the cen-ter of the square whose vertices are (1, 0), (2, 0), (2, 1), and (1, 1). Hence,the line cuts the square into two quadrilaterals of equal area 1/2. Thus thearea of the trapezoid is 3/2. Since the area of the rectangle is 2, the proba-bilityis 322 = 4'

!j,

Q.,

.Ti

C!1

90 Solutions

19. ANSWER (D) Adding the given equations gives 2(ab +bc +ea) = 484,so ab + be + ca = 242. Subtracting from this each of the given equationsyields be = 90, ca = 80, and ab = 72. It follows that a2b2c2 = 90. 8072 = 7202. Since abc > 0, we have abc = 720.

20. ANSWER (B) Let OM = a and ON = b. Then

192 = (2a)2 + b2 and 222 = a2 + (2b)2.

Hence

It follows that

5(a2 + b2) = 192 +22 2 = 845.

MN = Va2 + b2 = 169 = 13.

Since AXOY is similar to AMON and XO = 2 MO, we have XY =2 MN=26.

o..

...r


21. ANSWER (A) Since 2002 = 11 13 . 14, we have

11, ifn=13.14 i,where i=1,2,...,10;13, j=1,2,...,12;

an =14, ifn=11. 13 k, where k=1,2,...,13;0, otherwise.

Hence _200' a = 11.10 + 13.12 + 14.13 = 448.n=l n

22. ANSWER (B) We have an = Tog, 2002 = 1092002 n, so

b - c = (1o92002 2 + 1o92002 3 + 1o92002 4 + 1o92002 5)

- (1092002 10 + 1092002 11 + 1o92002 12 + 1092002 13 + 1o92002 14)

2.3.4.5 1

=1o9200210-11-12.13-14 = 1o92002 11 13 14

1 _= 1092002

2002

23. ANSWER (C) Let M be the midpoint of BC, let AM = 2a, and let0 = LAMB. Then cos LAMC = - cos 0. Applying the Law of Cosinesto AABM and to AAMC yields, respectively,

a2 + 4a2 - 4a2 cos 0 = 1

and

a2+4a2+4a2cos0=4.

Adding, we obtain 10a2 = 5, so a = J/2 and BC = 2a = Nf2-.

^y°

.Q'

"c,

0..

.-.

92 Solutions

OR

As above, let M be the midpoint of BC and AM = 2a. Put a rectangularcoordinate system in the plane of the triangle with the origin at M so thatA has coordinates (0, 2a). If the coordinates of B are (x, y), then the pointC has coordinates (-x, -y),

so

x2 + (2a - y)2 = 1 and x2 + (2a + y)2 = 4.

Combining the last two equations gives 2 (x2 + y2) + 8a2 = 5. But, x2 +y2 = a2, so 10a2 = 5. Thus a = /2/2 and BC = v.

24. ANSWER (E) We have

1

Area (ABCD) <2

AC BD,

with equality if and only if AC 1 BD. Since

1

2002 = Area(ABCD) <2AC BD

1 52.772 = 2002,

it follows that the diagonals AC and BD are perpendicular and intersectat P. Thus AB = 242 + 322 = 40, BC = /-282+ 322 = 4 113,


CD = 282 -+452 = 53, and DA = 452 + 242 = 51. The perimeterof ABCD is therefore

144 + 4113=4(36+ 113).


f(x)+ f(y) =x2+6x+y2+6y+2= (x+3)2+(y+3)2- 16

and

f(x)-f(y) =x2-y2+ 6(x - y) = (x - y)(x + y + 6).

The given conditions can be written as

(x + 3)2 + (y + 3)2 < 16 and (x-y)(x+y+6) <0.

The first inequality describes the region on and inside the circle of radius 4with center (-3, -3). The second inequality can be rewritten as

(x-y>Oandx+y+6<0) or (x-y <0 and x + y + 6 > 0).

Each of these inequalities describes a half-plane bounded by a line thatpasses through (-3, -3) and has slope 1 or -1. Thus the set R is the shadedregion in the following diagram, and its area is half the area of the circle,which is 8n 25.13.

..+57

1

O00

000

0

--j

NV

')

V')

N

Lei

.-rI'D

94 Solutions



1 4.29 14.28 5.95 55.37 4.03 16.06

2 0.31 90.21 1.08 6.02 0.18 2.19

3 3.21 24.42 2.81 52.59 2.07 14.87

4 56.36 1.31 1.81 29.62 0.62 10.27

5 2.27 0.86 1.2 1.57 78.45 15.64

6 11.81 19.28 37.32 4.4 5.65 21.51

7 15.69 25.99 8.35 16.46 2.97 30.53

8 8.49 3.74 2.6 2.1 74 9.05

9 3.45 6.73 13.86 30.48 1.82 43.65

10 4.31 8.63 4.1 34.19 1.77 46.98

11 6.72 6.2 17.7 4.06 7.96 57.35

12 1.08 2.39 5.2 4.22 64.27 22.83

13 4.25 4.12 6.47 11.54 46.45 27.16

14 7.08 4.13 4.32 48.4 9.69 26.37

15 3.13 2.9 10.2 1.63 3.24 78.9

16 5.77 4.64 38.49 3.68 9.14 38.27

17 6.57 10.33 3.33 6.64 2.19 70.93

18 1.45 4.89 2.16 1.75 2.27 87.47

19 5.75 5.36 9.74 11.83 4.61 62.72

20 4.63 2.83 6.1 11.53 1.68 73.21

21 8.07 3.54 2.36 7.37 13.45 65.2

22 5.48 4.97 3.53 6.04 2.03 77.94

23 2.19 4.17 2.52 1.87 2.01 87.22

24 1.47 17.01 7.81 1.69 1.96 70.05

25 5.16 3.6 1.67 0.85 15.81 72.91

sue..

d-0


2003 AMC 12A SolutionsThe problems for the 2003 AMC 12A begin on page 17

1. ANSWER (D) Each even counting number, beginning with 2, is onemore than the preceding odd counting number. Therefore the differenceis (1)(2003) = 2003.

2. ANSWER (B) The cost for each member is the cost of two pairs of socksand two shirts. This is 2.4+2(4+5) = 26 dollars, so there are 2366/26 =91 members.

3. ANSWER (D) The total volume of the eight removed cubes is 8 x 33 =216 cubic centimeters, and the volume of the original box is 15 x 10 x8 = 1200 cubic centimeters. Therefore the volume has been reduced by

2 6) (100%) = 18%.

4. ANSWER (A) Mary walks a total of 2 km in 40 minutes. Because 40minutes is 2/3 hr, her average speed, in km/hr, is 2/(2/3) = 3.

5. ANSWER (E) Since the last two digits of AMC10 and AMC12 sum to22, we have

AMC + AMC = 2(AMC) = 1234.

Hence AMC = 617,soA =6,M = 1,C =7,andA+M+C =6+1+7= 14.

6. ANSWER (C) For example, -1Z0 = I - 1 - 01 = 1 -1. All the otherstatements are true:(A)xOy= Ix - YJ = I - (Y - x)I = ly - xj = yC7xforallxandy.(B) 2(xpy) = 2Ix - yI = 12x - 2yI = (2x)V(2y) for all x and y.(D)xc7x = Ix - xI = 0forallx.(E) xc2y = Ix - yj > 0 if x 54 y.

7. ANSWER (B) The longest side cannot be greater than 3, since otherwisethe remaining two sides would not be long enough to form a triangle. Theonly possible triangles have side lengths 1-3-3 or 2-2-3.

CA

D

CA

D

,_,

.-r

GH

Q

CF,

°'o

.-.

96 Solutions

8. ANSWER (E) The factors of 60 are

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

Six of the twelve factors are less than 7, so the probability is 1/2.

9. ANSWER (D) The set S is symmetric about the line y = x and contains(2, 3), so it must also contain (3, 2). Also S is symmetric about the x-axis,so it must contain (2, -3) and (3, -2). Finally, since S is symmetric aboutthe y-axis, it must contain (-2, 3), (-3, 2), (-2, -3), and (-3, -2). Sincethe resulting set of 8 points is symmetric about both coordinate axes, it isalso symmetric about the origin.

10. ANSWER (D) Al, Bert, and Carl are to receive, respectively, 1/2, 1/3,and 1/6 of the candy. However, each believes he is the first to arrive. There-fore they leave behind, respectively, 1/2, 2/3, and 5/6 of the candy thatwas there when they arrived. The amount of unclaimed candy is

(1/2)(2/3)(5/6) = 5/18 of the original amount, regardless of the orderin which they arrive.

11. ANSWER (C) Rescaling to different units does not affect the ratio of theareas, so let the perimeter be 12. Each side of the square then has length3, and each side of the triangle has length 4. The diameter of the circlecircumscribing the square is the diagonal of the square, 3,/2. Thus A =7r(3,/2-/2)2 = 97r/2. The altitude of the triangle is 2,/3-, so the radius ofthe circle circumscribing the triangle is 4,13/3, and B = -r(4vf3-/3)2 =16,-r/3. Therefore

A 9ir 3 _ 27

B 2 16ir 32

12. ANSWER (E) Let Rl,..., R5 and B3,..., B6 denote the numbers on thered and blue cards, respectively. Note that R4 and R5 divide evenly intoonly B4 and B5, respectively. Thus the stack must be R4, B4,..., B5, R5,or the reverse. Since R2 divides evenly into only B4 and B6, we must haveR4, B4, R2, B6,..., B5, R5, or the reverse. Since R3 divides evenly intoonly B3 and B6, the stack must be R4, B4, R2, B6, R3, B3, R1, B5, R5, orthe reverse. In either case, the sum of the middle three cards is 12.


13. ANSWER (E) If the polygon is folded before the fifth square is attached,then edges a and a' must be joined, as must b and b'. The fifth face of thecube can be attached at any of the six remaining edges.

b' a'

ab, b'

14. ANSWER (D) Quadrilateral KLMN is a square because it has 90° rota-tional symmetry, which implies that each pair of adjacent sides is congruentand perpendicular. Since ABCD has sides of length 4 and K is 2J fromside AB, the length of the diagonal KM is 4 + 4,13. Thus the area is

1

2

(4+4f3-)2= 32 + 16,f3.

OR

r+.98 Solutions

Since KLMN is a square, its area is (NK)2. Note that m(LNAK) _150°. By the Law of Cosines,

(NK)2 = 42 +42 - 2(4)(4) - =2

32 + 16/.

15. ANSWER (C) First note that the area of the region determined by thetriangle topped by the semicircle of diameter 1 is

1 1 1 2 12 2 +2n 2 = 4 +8

The area of the lune results from subtracting from this the area of the sectorof the larger semicircle,

6rr(1)2 = -7r.6

So the area of the lune is

./ 1 1 _ 1

4 + 8 6" 4 24

2

Note that the answer does not depend on the position of the lune on thesemicircle.

16. ANSWER (C) Since the three triangles ABP, ACP, and BCP have equalbases, their areas are proportional to the lengths of their altitudes.

SD

.

,..


Let 0 be the centroid of AABC, and draw medians AOE and BOD.Any point above BOD will be farther from AB than from BC, and anypoint above AOE will be farther from AB than from AC. Therefore thecondition of the problem is met if and only if P is inside quadrilateralCDOE.

If CO is extended to F on AB, then AABC is divided into six congruenttriangles, of which two comprise quadrilateral CDOE. Thus CDOE hasone-third the area of AABC, so the required probability is 1/3.

OR

By symmetry, each of AABP, AACP, and ABCP is largest with thesame probability, so the probability must be 1/3 for each.

17. ANSWER (B) Place an xy-coordinate system with origin at D and pointsC and A on the positive x- and y-axes, respectively. Then the circle cen-tered at M has equation

(x - 2)2+y2=4,

and the circle centered at A has equation

x2+(y-4)2=16.

Solving these equations for the coordinates of P gives x = 16/5 andy = 8/5, so the answer is 16/5.

NIA

100 Solutions

H 11

i 10

DI M R C x

OR

Because AP and PM are radii of the larger and smaller circles, respec-tively, AP = AD = 4 and PM = MD = 2. Thus AADM is congruentto AAPM, and LAPM is a right angle. Draw PQ and PR perpendicu-lar to AD and CD, respectively. Note that LAPQ and ZMPR are bothcomplements of LQPM. Thus A A P Q is similar to AMPR, and

AQ _ AP _ 4

MR MP 2

LetMR=x.ThenAQ =2x,PR= QD =4-2x,andPQ =RD=x + 2. Therefore

2-AQ-PQ-x+2MR PR 4-2x'

so

6 6 16x=5 and PQ=5+2= 5.

OR

Let LMAD = a. Then

PQ = (PA) sin(LPAQ) = 4 sin(2a) = 8 sina cos a

= 8 ( 20) ( 20) 56.

.,.

CIO

Boa

v'.

C/]


18. ANSWER (B) Note that n = 100q + r = q + r + 99q. Hence q + ris divisible by I 1 if and only if n is divisible by 11. Since 10, 000 < n <99, 999, there are

99999 9999

11 11= 9090 - 909 = 8181

such numbers.

19. ANSWER (D) The original parabola has equation y = a(x - h)2 + k,for some a, h, and k, with a # 0. The reflected parabola has equationy = -a (x - h)2 - k. The translated parabolas have equations

f (x) = a(x - h + 5)2 + k and g(x) = -a(x - h T 5)2 - k,

so

(f + g)(x) = ±20a(x - h).

Since a 0, the graph is a non-horizontal line.

20. ANSWER (A) Since the first group of five letters contains no A's, it mustcontain k B's and (5 - k) C's for some integer k with 0 < k < 5. Sincethe third group of five letters contains no C's, the remaining k C's must bein the second group, along with (5 - k) A's.

Similarly, the third group of five letters must contain k A's and (5 - k) B's.Thus each arrangement that satisfies the conditions is determined uniquelyby the location of the k B's in the first group, the k C's in the second group,and the k A's in the third group.

For each k, the letters can be arranged in (k) 3 ways, so the total number ofarrangements is

.

5

o (k1)

3

21. ANSWER (D) Since P(0) = 0, we have e = 0 and P(x) = x(x4 +ax3 + bx2 + cx + d). Suppose that the four remaining x-intercepts are atp, q, r, and s. Then

x4+ax3+bx2+cx+d = (x - p)(x - q)(x - r)(x - s),

and d = pqrs # 0.

'L7

102 Solutions

Any of the other constants could be zero. For example, consider

P1(x) = x5 - 5x3 + 4x = x(x + 2)(x + 1)(x - 1)(x - 2)

and

P2(x) = xs - 5x4 + 20x2 - 16x = x(x + 2) (x - 1) (x - 2) (x - 4).

OR

Since P(0) = 0, we must have e = 0, so

P(x) = x(x4 + ax3 + bx2 + cx + d).

If d = 0, then

P(x) = x(x4 + ax3 + bx2 + cx) = x2(x3 + axe + bx + c),

which has a double root at x = 0. Hence d 0.

OR

There is also a calculus-based solution. Since P(x) has five distinct zerosand x = 0 is one of the zeros, it must be a zero of multiplicity one. This isequivalent to having P(0) = 0, but P'(0) 0. Since

P'(x) = 5x4+4ax3+3bx2+2cx+d, we must have 0 54 P'(0) = d.

22. ANSWER (C) Since there are twelve steps between (0, 0) and (5, 7), Aand B can meet only after they have each moved six steps. The possiblemeeting places are Po = (0, 6), Pi = (1, 5), P2 = (2, 4), P3 = (3, 3),P4 = (4, 2), and P5 = (5, 1). Let a; and b; denote the number of pathsto P; from (0, 0) and (5, 7), respectively. Since A has to take i steps to theright and B has to take i + 1 steps down, the number of ways in which Aand B can meet at P; is

ai - bi =(6) ( 6 1 ).

Los

`-'+

-+

(II


Since A and B can each take 26 paths in six steps, the probability that theymeet is

5 6 6 6 6 6 6 6 666(a6) (bi (0) (6) + (1) (2) + (2) (3)2 (3) (4) + (4) (5) + (5)(6)

ii =026

99ti 0.20.

512

OR

Consider the 12) walks that start at (0, 0), end at (5, 7), and consist of 12steps, each one either up or to the right. There is a one-to-one correspon-dence between these walks and the set of (A, B)-paths where A and Bmeet. In particular, given one of the 12) walks from (0, 0) to (5, 7), thepath followed by A consists of the the first six steps of the walk, and thepath followed by B is obtained by starting at (5, 7) and reversing the lastsix steps of the walk. There are 26 paths that take 6 steps from (0, 0) and26 paths that take 6 steps from (5, 7), so there are 212 pairs of paths that Aand B can take. The probability that they meet is

1 (12) 99P212 5 29.

23. ANSWER (B) We have

II .2i . 3! ... 91 = (1)(1 -2)(1.2.3) ... (1.2...9)

192837465564738291 = 2303135573

The perfect square divisors of that product are the numbers of the form

22a32b52c72d

with 0<a < 15, 0<b <6, 0<c <2,and0<d < 1. Thus there are(16)(7)(3)(2) = 672 such numbers.


logob

+ logob

a= logo a - logo b + logo b - logo a

= 1 - logo b + 1 - logo a

=2-logab-logba.

104 Solutions

Let c = log,, b, and note that c > 0 since a and b are both greater than 1.Thus

a b 1 _ c2 - 2c + 1 _ (c - 1)2logo - + logo - = 2 - c -

c -c -c < 0.

This expression is 0 when c = 1, that is, when a = b.

OR

As above,a b 1logab+logb - =2-c--.

From the Arithmetic-Geometric Mean Inequality we have

c+1/c>c c

and

logo b + logo a = 2- (c + c) < 0,

with equality when c = c , that is, when a \= b. l

25. ANSWER (C) The domain off is {x I ax2 + bx > 0}. If a = 0, thenfor every positive value of b, the domain and range of f are each equal tothe interval [0, oo), so 0 is a possible value of a.If a f 0, the graph of y = ax2 + bx is a parabola with x-intercepts atx = 0 and x = -b/a. If a > 0, the domain off is (-oo, -b/a] U [0, oo),but the range of f cannot contain negative numbers. If a < 0, the domainof f is [0, -b/a]. The maximum value of f occurs halfway between thex-intercepts, at x = -b/2a, and

z

f -2 ) = a(4 z) +b -2a =

2

b iHence the range of f is [0, b/2]. For the domain and range to beequal, we must have

-b bso 2 _ -a.

a 2

The only solution is a = -4. Thus there are two possible values of a, andthey are a = 0 and a = -4.

V'1

V"1

--+

2003 AMC 12B Solutions


105


1 0.17 0.71 98 0.53 0.17 0.42

2 0.35 0.61 11.6 83.47 2.02 1.94

3 89.49 1.81 1.76 0.41 1.45 5.06

4 3.03 2.98 51.81 7.77 10.28 24.11

5 6.83 10.11 2.07 65.01 3.75 12.22

6 1.36 33.07 2.79 6.45 2.29 54.03

7 3.47 2.37 4.08 19.3 41.82 28.94

8 25.75 6.5 3.25 10.69 28.21 25.58

9 0.7 2.62 19.63 56.79 1.22 19.04

10 4.13 35.47 7.47 5.63 10.7 36.57

11 13.5 33.94 22.93 5.06 4.51 20.05

12 2.64 10.25 2.33 26.78 19.3 38.67

13 3.08 21.97 7.15 3.24 1.48 63.07

14 1.91 3.6 4.34 33.16 11.06 45.92

15 3.37 3.43 8.58 17.67 5.5 61.45

16 2.87 3.23 4.04 5.01 11.33 73.52

17 0.92 3.02 3.93 22.89 15.1 54.13

18 1.37 4.16 2.43 1.45 0.9 89.7

19 9.61 12.04 1.9 1.5 8.37 66.57

20 3.78 19.1 4.76 7.02 1.38 63.95

21 3.12 2.7 3.41 14.71 3.79 72.27

22 2.66 2.99 3.12 2.33 6.31 82.59

23 5.22 2.64 2.83 1.72 1.53 86.06

24 2.77 4.51 4.11 2.22 0.91 85.48

25 6.61 2.95 3.97 3.83 4.5 78.14

C/1

106 Solutions


1. ANSWER (C) We have

2-4+6-8+ 10- 12+ 14 _ 2(1 -2+3-4+5-6+7) 2

3-6+9-12+15-18+21 3(1-2+3-4+5-6+7) 3.

2. ANSWER (D) The cost of each day's pills is 546/14 = 39 dollars. If xdenotes the cost of one green pill, then x + (x - 1) = 39, so x = 20.

3. ANSWER (A) To minimize the cost, Rose should place the most expensiveflowers in the smallest region, the next most expensive in the second small-est, etc. The areas of the regions are shown in the figure, so the minimaltotal cost, in dollars, is

(3)(4) + (2.5)(6) + (2)(15) + (1.5)(20) + (1)(21) = 108.

4 7

5 20

4

1 6

6

21

15

5

4. ANSWER (C) The area of the lawn is

90. 150 = 13,500 ft2.

3

3

Moe cuts about two square feet for each foot he pushes the mower for-ward, so he cuts 2(5000) = 10,000 ft2 per hour. Therefore, it takes about1x,500 = 1.35 hours.10,000 -

5. ANSWER (D) The height, length, and diagonal are in the ratio 3 : 4 : 5.

The length of the diagonal is 27, so the horizontal length is

4-(27) = 21.6 inches.

C/1

SIN

...

ago 'Z

3

'-a

'ZS

.';


6. ANSWER(B) Let the sequence be denoted a,ar,ar2,ar3,...,with ar =2 and ar3 = 6. Then r2 = 3 and r = or r = -Nf3-. Thereforea = 23 ora = 23

7. ANSWER (D) Let n represent the number of nickels in the bank, d thenumber of dimes, and q the number of quarters. Then n + d + q = 100and 5n + 10d + 25q = 835. Further, n, d, and q must all be nonnegativeintegers. Dividing the second equation by 5 yields n + 2d + 5q = 167.Subtracting the first equation from this gives d + 4q = 67. Since q cannotbe negative, d is at most 67, and we check that 67 dimes and 33 nickelsindeed produces $8.35. On the other hand, d cannot be 0, 1, or 2 becausethen q would not be an integer. Thus the smallest d can be is 3, leavingq = 16. We check that 16 quarters, 3 dimes, and 81 nickels also produces$8.35. Thus the largest d can be is 67, the smallest is 3, and the differenceis 64.

8. ANSWER (E) Let y = 4(x). Since each digit of x is at most 9, we havey < 18. Thus if *(y) = 3, then y = 3 or y = 12. The 3 values of xfor which 4(x) = 3 are 12, 21, and 30, and the 7 values of x for which4(x) = 12 are 39, 48, 57, 66, 75, 84, and 93. There are 10 values in all.

9. ANSWER (D) Since f is a linear function, its slope is constant. Therefore

f (6) - f (2)-

f(12) - f(2) 12 f(12) - f(2)6-2 12-2 '

so 4 = 10

and f(12) - f(2) = 30.

OR

Since f is a linear function, it has a constant rate of change, given by

f(6)-f(2) 12 =3.6-2 4

Therefore f (12) - f (2) = 3(12 - 2) = 30.

OR

If f (x) = mx + b, then

12= f(6)-f(2) = 6m +b-(2m +b) = 4m,

108

so m = 3. Hence

f(12) - f(2) = 12m + b - (2m + b) = 10m = 30.

Solutions

10. ANSWER (B) We may assume that one of the triangles is attached to sideA B. The second triangle can be attached to BC or CD to obtain two non-congruent figures. If the second triangle is attached to A E or to DE, thefigure can be reflected about the vertical axis of symmetry of the pentagonto obtain one of the two already counted. Thus the total is two.

11. ANSWER (C) When it is 1:00 PM it is 60 minutes after noon, but Cas-sandra's watch has recorded only 57 minutes and 36 seconds, that is, 57.6minutes. Thus when her watch has recorded t minutes since noon, the ac-tual number of minutes passed is

60 25-t = -t.57.6 24

In particular, when her watch reads 10:00 PM it has recorded 600 minutespast noon, so the actual number of minutes past noon is

2524(600) = 625,

or 10 hours and 25 minutes. Therefore the actual time is 10:25 PM.

12. ANSWER (D) Among five consecutive odd numbers, at least one is divisi-ble by 3 and exactly one is divisible by 5, so the product is always divisibleby 15. The cases n = 2, n = 10, and n = 12 demonstrate that no largercommon divisor is possible, since 15 is the greatest common divisor of3.5.7.9.11,11. 13. 15. 17. 19, and 13. 15. 17. 19.21.

13. ANSWER (B) Let r be the radius of the sphere and cone, and let h be theheight of the cone. Then the conditions of the problem imply that

4

(34= 37rr2h, so h = 3r.

Therefore, the ratio of h to r is 3 : 1.


14. ANSWER (D) Let H be the foot of the perpendicular from E to DC.Since CD = AB = 5, it follows that FG = 2. Because AFEG is similarto AAEB, we have

EH _ 2so 5EH = 2EH + 6,

EH + 3 5'

and EH = 2. Hence the area of AAEB is

1 252.

5

OR

Let I be the foot of the perpendicular from E to AB. Since AEIA issimilar to AADF and AEIB is similar to ABCG, we have

AI 1 5-AI BI 2and =EI 3 EI EI 3

110 Solutions

Adding gives 5/El = 1, so E I = 5. The area of the triangle is15.5= i5

15. ANSWER (D) Let 0 be the intersection of the diagonals of ABEF. Sincethe octagon is regular, AAOB has area 1/8. Since 0 is the midpoint ofAE, A OAB and A BOE have the same area. Thus A ABE has area 1/4,so ABEF has area 1/2.

OR

Let 0 be the intersection of the diagonals of the square IJKL. RectanglesABJI, JCDK, KEFL, and LGHI are congruent. Also IJ = AB =AH, so the right isosceles triangles A A I H and AJOI are congruent. Bysymmetry, the area in the center square IJKL is the sum of the areas ofAAIH, ACJB, AEKD, and AGLF. Thus the area of rectangle ABEFis half the area of the octagon.

16. ANSWER (E) The area of the larger semicircle is

n2

(2)2 = 2n.

NIA


The region deleted from the larger semicircle consists of five congruentsectors and two equilateral triangles. The area of each of the sectors is

6ir (1)2 6

and the area of each triangle is

1 /"3-/3-2 1 2 4 '

so the area of the shaded region is

n 7 v27r - 5 6- 2. 4= 6 Jr -2


1 = log(xy3) = log x + 3 log y and 1 = log(xy y) = 2 log x + logy.

Solving yields log x =s

and log y = 5 . Thus

3log(xy) = log x + log y = 5 .

OR

The given equations imply that xy3 = 10 = x2y. Thus

10 l0 3

y = X2and x X2 = 10.

It follows that x = 102/5 and y = 10'5, so log(xy) = log(103/5) = 3/5.

OR

Since log(xy3) = log(x2y), we have xy3 = x2y, sox = y2. Hence

1 = log(xy3) = log(Y5) = S log y, and logy =1

.5

So log(xy) = log(y3) = 3/5.

112 Solutions


I1y13 = 7x5 = 7(acbd)5 = 7a5cb5d

The minimum value of x is obtained when neither x nor y contains primefactors other than 7 and 11. Therefore we may assume that a = 7 andb = 11, so x = 7c 11d and 7x5 = 75c+1115d. Letting y = 7' 11n weobtain l ly13 = 713m 1113n+1 Hence 75c+1 115d = 713m 1113n+1 The

smallest positive integer solutions are c = 5, d = 8, m = 2, and n = 3.Thus a+b+c+d =7+11+5+8=31.

19. ANSWER (E) Since the first term is not 1, the probability that it is 2 is1/4. If the first term is 2, then the second term cannot be 2. If the first termis not 2, there are four equally likely values, including 2, for the secondterm. Thus the probability that the second term is 2 is

1 3 1 3

+_

4 4 4 16'

so a + b = 3 + 16 = 19.

OR

The set S contains (4)(4!) = 96 permutations, since there are 4 choicesfor the first term, and for each of these choices there are 4! arrangements ofthe remaining terms. The number of permutations in S whose second termis 2 is (3)(3!) = 18, since there are 3 choices for the first term, and foreach of these choices there are 3! arrangements of the last 3 terms. Thusthe requested probability is 18/96 = 3/16, and a + b = 19.


0= f(-1)=-a+b-c+d and 0= f(1)=a+b+c+d,so b + d = 0. Also d = f (0) = 2, so b = -2.

OR

The polynomial is divisible by (x + 1)(x - 1) = x2 - 1, its leading termis ax3, and its constant term is 2, so

P x) _ (x2 - 1)(ax - 2) = ax3 - 2x2 - ax + 2 and b=-2.

1.)

SIN

C/1


21. ANSWER (D) Let -r - a. Apply the Law of Cosines to AABC toobtain

(AC)2 = 82 +5 2 - 2(8)(5) cos 8 = 89 - 80cos,8.

Thus AC < 7 if and only if

89 - 80 cos P < 49, that is, if and only if cos P >2

Therefore we must have 0 < ,B < 3 , and the requested probability isit/3 - IJr 3

22. ANSWER (C) Let 0 be the point of intersection of AC and BD. ThenAOB is a right triangle with legs OA = 8 and OB = 15. QuadrilateralOPNQ is a rectangle because it has right angles at 0, P, and Q. ThusPQ = ON, because the diagonals of a rectangle are of equal length. Theminimum value of PQ is the minimum value of ON. This is achieved ifand only if N is the foot of the altitude from 0 in triangle AOB. Writingthe area of AAOB in two different ways yields

1 1

2 2

Hence the minimum value of PQ is

ON =OA OB _ OA OB _ 8.15 _ 120

ti 7.06.AB OA2+OB2 17 17

23. ANSWER (A) The intercepts occur where sin(1/x) = 0, that is, wherex = 1/(kir) and k is a nonzero integer. Solving

1

0.0001 <k n

< 0.001

C'1

CID

a.,

.1]

t-.

114 Solutions

yields

1000< k <

10,000

n it

Thus the number of x intercepts in (0.0001, 0.001) is

10'000] - 11000] = 3183 - 318 = 2865,7r 7r

which is closest to 2900.

24. ANSWER (C) Since the system has exactly one solution, the graphs of thetwo equations must intersect at exactly one point. If x < a, the equationy=Ix-aI+Ix-bi+Ix-ciisequivalent toy=-3x+(a+b+ c).By similar calculations we obtain

-3x+(a+b+c), ifx <a-x+(-a+b+c), ifa <x <b

Y= x+(-a-b+c), ifb <x <c3x + (-a - b - c), if c < x.

Thus the graph consists of four lines with slopes -3, -1, 1, and 3, and ithas corners at (a, b + c - 2a), (b, c - a), and (c, 2c - a - b).

On the other hand, the graph of 2x + y = 2003 is a line whose slope is-2. If the graphs intersect at exactly one point, that point must be (a, b +c - 2a). Therefore

2003 = 2a + (b + c - 2a) = b + c.

Since b < c, the minimum value of c is 1002.

25. ANSWER (D) We can assume that the circle has its center at (0, 0) and aradius of 1. Call the three points A, B, and C, and let a, b, and c denote thelength of the counterclockwise arc from (1, 0) to A, B, and C, respectively.Rotating the circle if necessary, we can also assume that a = it/3. Since band c are chosen at random from [0, 2ir), the ordered pair (b, c) is chosenat random from a square with area 4ir2 in the bc-plane. The condition ofthe problem is met if and only if

0<b< 3, 0<c<23, and lb-cl<3.

AIM


This last inequality is equivalent to b - 3 < c < b + 3 .

C4

2ir

37r

The graph of the common solution to these inequalities is the shaded regionshown. The area of this region is

6 2n2

8 3-'r2/3'

so the requested probability is

n2/3 _ 1

47r2 12

<`c

.-.

.-.

.-.

.-.

(-A

--J

0000

V"1

V"1

V"1

V')

11D

V'1

116


Solutions


1 0.71 1.31 41.41 3.21 52.12 1.23

2 1.65 2.48 89.23 0.92 5.22 0.49

3 3.89 37.09 20.78 4.75 5.62 27.85

4 1.05 0.56 4.31 2.93 84.29 6.83

5 11.79 61.7 0.59 9.06 2.35 14.51

6 44.09 3.87 23.78 1.33 7.44 19.47

7 11.31 32.93 15.34 16.51 7.63 16.26

8 11.83 48.15 2.43 4.15 1.7 31.73

9 3.09 14.37 40.76 4.59 5.76 31.43

10 1.36 3.36 65.01 5.38 0.68 24.19

11 21.63 5.92 5.85 4.63 2.85 59.09

12 3.6 33.07 7.6 1.69 2.92 51.11

13 8.06 30.52 4.56 3.37 6.49 46.99

14 9.37 1.88 8.22 11.24 2.46 66.81

15 9.62 13.24 15.77 7.08 6.61 47.66

16 11.09 3.55 1.36 1.15 6.95 75.89

17 5.85 5.21 22.12 14.52 1.29 50.99

18 5 3.44 5.73 11.41 3.68 70.73

19 2.77 4.95 2.63 2.69 3.9 83.05

20 9.52 6.81 11.12 5.13 6.81 60.6

21 1 1.88 2.13 3.03 1.44 90.52

22 1.43 2.7 1.3 1.41 4.85 88.3

23 2.56 2.34 2.04 2.1 3.01 87.94

24 1.86 1.47 1.53 1.26 1.48 92.4

25 1.27 1.51 2.32 1.61 0.82 92.47

C17



1. ANSWER (E) Since $20 is 2000 cents, she pays (0.0145)(2000) = 29cents per hour in local taxes.

2. ANSWER (C) The 8 unanswered problems are worth (2.5)(8) = 20points, so Charlyn must earn at least 80 additional points. The smallestmultiple of 6 that is at least 80 is (6)(14) = 84, so Charlyn must have atleast 14 correct answers.

3. ANSWER (B) The value of x = 100 - 2y is a positive integer for eachpositive integer y with 1 < y < 49.

4. ANSWER (E) Bertha has 30 - 6 = 24 granddaughters, none of whomhave any daughters. The granddaughters are the children of 24/6 = 4of Bertha's daughters, so the number of women having no daughters is30-4= 26.

5. ANSWER (B) The y-intercept of the line is between 0 and 1, so 0 < b <1. The slope is between-1 and 0, so-1 <m <0.Thus -1 <mb <0.

6. ANSWER (A) Since none of V, W, X, Y, or Z exceeds 20042005 thedifference U - V = 20042005 is the largest.

7. ANSWER (B) After three rounds the players A, B, and C have 14, 13,and 12 tokens, respectively. Every subsequent three rounds of play reduceseach player's supply of tokens by one. After 36 rounds they have 3, 2, and1 token, respectively, and after the 37`h round Player A has no tokens.

8. ANSWER (B) Let x, y, and z be the areas of AADE, ABDC, andAABD, respectively. The area of AABE is (1/2)(4)(8) = 16 = x + z,and the area of ABAC is (1/2)(4)(6) = 12 = y + z. The requesteddifference is

x-y=(x+z)-(y+z)=16-12=4.

--h

118 Solutions

9. ANSWER (C) Let r, h, and V, respectively, be the radius, height, andvolume of the jar that is currently being used. The new jar will have aradius of 1.25r and volume V. Let H be the height of the new jar. Then

7rr2h = V = r(1.25r)2H, so 0 .64.h (1.25)2

Thus H is 64% of h, so the height must be reduced by (100-64)% = 36%.

OR

Multiplying the diameter by 5/4 multiplies the area of the base by (5/4)2 =25/16, so in order to keep the same volume, the height must be multipliedby 16/25. Thus the height must be decreased by 9/25, or 36%.

10. ANSWER (C) The sum of a set of real numbers is the product of the meanand the number of real numbers, and the median of a set of consecutiveintegers is the same as the mean. So the median must be 75/49 = 73.

11. ANSWER (A) If n is the number of coins in Paula's purse, then theirtotal value is 20n cents. If she had one more quarter, she would have n + 1coins whose total value in cents could be expressed both as 20n + 25 andas 21(n + 1). Therefore

20n+25=21(n+1), so n=4.

Since Paula has four coins with a total value of 80 cents, she must havethree quarters and one nickel, so the number of dimes is 0.

12. ANSWER (B) Line AC has slope -2 and y-intercept (0,9), so its equa-tion is

1y=-2x+9.

Since the coordinates of A' satisfy both this equation and y = x, it followsthat A' = (6, 6). Similarly, line BC has equation y = -2x + 12, andB' = (4, 4). Thus

A'B' = (6-4)2+(6-4)2=2'.

CIA

coo

G1.

G1.

v°'


YA

B(0,12)

A

(0,9)

13. ANSWER (B) There are (2) = 36 pairs of points in S, and each pairdetermines a line. However, there are three horizontal, three vertical, andtwo diagonal lines that pass through three points of S, and these lines areeach determined by three different pairs of points in S. Thus the numberof distinct lines is 36 - 2 8 = 20.

OR

There are 3 vertical lines, 3 horizontal lines, 3 each with slopes 1 and -1,and 2 each with slopes 2, -2, 1/2, and -1/2, for a total of 20.

14. ANSWER (A) The terms of the arithmetic progression are 9, 9 + d, and9 + 2d for some real number d. The terms of the geometric progressionare 9, 11 + d, and 29 + 2d. Therefore

(11+d)2=9(29+2d) so d2+4d-140=0.Thus d = 10 or d = -14. The corresponding geometric progressions are9, 21, 49 and 9, -3, 1, so the smallest possible value for the third term ofthe geometric progression is 1.

15. ANSWER (C) When they first meet, they have run a combined distanceequal to half the length of the track. Between their first and second meet-ings, they run a combined distance equal to the full length of the track.Because Brenda runs at a constant speed and runs 100 meters before theirfirst meeting, she runs 2(100) = 200 meters between their first and secondmeetings. Therefore the length of the track is 200 + 150 = 350 meters.

16. ANSWER (B) The given expression is defined if and only if

109200300920020092001 x)) > 0,

'2]

120 Solutions

that is, if and only if

10920020092001 x) > 2003° = 1.

This inequality in turn is satisfied if and only if

1092001 x > 2002,

that is, if and only if x > 20012002.


f(21) _ .f(2) _ f(2.1) = 1 -.f(1) = 2° 2° = 2°,

f (22) _ .f (4) _ f (2.2) = 2 .f (2) = 21.2° = 2',

f (23) _ .f (8) _ f (2.4) = 4 .f (4) = 22.2 ' 20 = 2(1 +2),

f(24) _ .f(16) _ .f(2.8) = 8 f(8) = 23.22.21.20 = 2(1+2+3)

and in general

f(2n) = 2n(n-1)/2

It follows that f(2100) = 2(100)(99)/2 = 24950

18. ANSWER (D) Let F be the point at which CE is tangent to the semicircle,and let G be the midpoint of AB. Because CF and CB are both tangentsto the semicircle, CF = CB = 2. Similarly, EA = EF. Let x = EA.The Pythagorean Theorem applied to ACDE gives

(2-x)2+22=(2+x)2.

It follows that x = 1/2 and CE = 2 + x = 5/2.

C

19. ANSWER (D) Let E, H, and F be the centers of circles A, B, and D,respectively, and let G be the point of tangency of circles B and C. Letx = FG and y = GH. Because circle A has radius 1 and passes through

-ti


the center of circle D, the radius of circle D is 2. Applying the PythagoreanTheorem to right triangles EGH and FGH gives

(1 + y)2 = (1 + x)2 + y2 and


(2-y)2=x2+ y2,

x2 x2y=x+ 2 and y=1- 4.The solutions of this system are (x, y) = (2/3, 8/9) and (x, y) = (-2, 0).The radius of circle B is the positive solution for y, which is 8/9.

20. ANSWER (E) The conditions under which A + B = C are as follows.

(i) If a + b < 1/2, then A = B = C = 0.

(ii) If a > 1/2 and b < 1/2, then B = 0 and A = C = 1.

(iii) If a < 1/2 and b > 1/2, then A = 0 and B = C = 1.

(iv) If a + b > 3/2, then A = B = 1 and C = 2.

These conditions correspond to the shaded regions of the graph shown. Thecombined area of those regions is 3/4, and the area of the entire square is1, so the requested probability is 3/4.

1

0

b

a1

122 Solutions

21. ANSWER (D) The given series is geometric with an initial term of 1 anda common ratio of cos2 0, so its sum is

00

5 = COS2n 0 =

n=0

1 _ 1

I - COS2 0 sin2 0

Therefore sin2 0 =s

, and

cos20= 1-2sin20= 1- 2 3-=-.5 5

22. ANSWER (B) Let A, B, C and E be the centers of the three small spheresand the large sphere, respectively. Then AABC is equilateral with sidelength 2. If D is the intersection of the medians of AABC, then E isdirectly above D. Because AE = 3 and AD = 2//3, it follows that

DE = 322f3- 2

3 3

Because D is 1 unit above the plane and the top of the larger sphere is 2units above E, the distance from the plane to the top of the larger sphere is

23. ANSWER (E) Since zi = 0, it follows that co = P(0) = 0. The non-004 bk = 0 andreal zeros of P must occur in conjugate pairs, so y2

k=i2004 a = 0 also. The coefficient c2003 is the sum of the zeros of P,

which is2004 2004 2004

Zk= Yak+i >bk=0.k=i k=1 k=1

KID


Finally, since the degree of P is even, at least one of z2, ... , Z2004 must bereal, so at least one of b2, ... , b2004 is 0 and consequently b2b3 b2004 =0. Thus the quantities in (A), (B), (C), and (D) must all be 0.

Note that the polynomial

2003

P(x) = x(x - 2)(x - 3) ... (x - 2003) x+Y' k)k=2

satisfies the given conditions, and 2004Ck P(l) 0-

24. ANSWER (C) The center of the disk lies in a region R, consisting ofall points within 1 unit of both A and B. Let C and D be the points ofintersection of the circles of radius 1 centered at A and B. Because A A B Cand AABD are equilateral, arcs CAD and CBD are each 120°. Thus thesector bounded by BC, BD, and arc CAD has area it/3, as does the sectorbounded by AC, AD, and arc CBD. The intersection of the two sectors,which is the union of the two triangles, has area vf3-/2, so the area of R is

The region S consists of all points within 1 unit of R. In addition to Ritself, S contains two 60° sectors of radius 1 and two 120° annuli of outerradius 2 and inner radius 1. The area of each sector is it/6, and the area ofeach annulus is

3 (22 - 12) = r.

Therefore the area of S is

2 n

32 +2(6 +r) =37r- 2

124 Solutions

Region S

25. ANSWER (E) Note that n3an = 133.133n = an + n2 + 3n + 3, so

n2+3n+3 (n+1)3-1an =

n3 - 1 n(n3 - 1)

Therefore

a4a5 ... a9953-1 l ( 63-1 1003-1

4(43-1))5(53-1))...(99(993-1)/

(99!) (1403 11) -(996 ) 99(1002

6100 + 1)

_ (2)(10,101) _ 962 vv

(21)(98!) 98!

Vii

Vii

Vii

Vii

.-r

.-r



125


1 96.87 1.28 0.38 0.43 0.07 0.96

2 0.15 1.61 7.14 89.08 0.75 1.25

3 81.91 1.25 1.11 0.47 0.8 14.44

4 12.09 63.06 11.64 2.23 0.38 10.59

5 52.43 8.86 3.32 2.14 1.95 31.29

6 78.44 2.21 1.2 1.26 3.03 13.85

7 4.6 49.85 4.38 12.43 0.92 27.8

8 0.71 0.95 1.95 89 9 3.85 3.34

9 10.55 6.24 7.02 18.19 45.51 12.48

10 40.85 1.58 1.83 12.32 3.31 40.1

11 3.55 4.65 27.4 50.05 1.89 12.43

12 6.57 11.03 31.75 9.16 2.35 39.12

13 9.53 2.3 12.36 5.91 5.84 64.02

14 2.42 3.79 3.84 18.48 2.72 68.74

15 2.8 62.47 2.45 2.59 1.54 28.14

16 5.71 4.02 7.34 1.89 0.38 80.66

17 3.65 2.34 2.67 2.4 1.36 87.57

18 3.19 2.26 2.27 1.66 8.26 82.34

19 3.61 1.95 2.7 3.49 1.76 86.48

20 10.97 8.58 6.98 2.39 5.12 65.94

21 1.09 0.96 1.95 1.41 0.92 93.67

22 4.19 3.51 2.8 3.56 2.6 83.32

23 0.92 1.57 2.08 1.31 0.98 93.13

24 1.03 1.3 3 3 1.65 1.03 91.96

25 2.91 2.65 3.83 2.5 7.81 80.3

'ZS

126 Solutions


1. ANSWER (A) At Jenny's fourth practice she made 2(48) = 24 freethrows. At her third practice she made 12, at her second practice she made6, and at her first practice she made 3.

2. ANSWER (D) If d 0 0, the value of the expression can be increased byinterchanging 0 with the value of d. Therefore the maximum value mustoccur when d = 0. If a = 1, the value is c, which is 2 or 3. If b = 1, the

value is is 23 = 8or32 = 9.Thus the maximum value is 9.

3. ANSWER (A) Note that 1296 = 64 = 2434, so x = y = 4 and x +y = 8.

4. ANSWER (B) There are 90 possible choices for x. Ten of these have a tensdigit of 7, and nine have a units digit of 7. Because 77 has been countedtwice, there are 10 + 9 - 1 = 18 choices of x for which at least one digitis a 7. Therefore the probability is 90 = s

5. ANSWER (A) Isabella received 10d/7 Canadian dollars at the borderand spent 60 of them. Thus l0d/7 - 60 = d, from which it follows thatd = 140, and the sum of the digits of d is 5.

6. ANSWER (A) Let downtown St. Paul, downtown Minneapolis, and theairport be located at S, M, and A, respectively. Then AMAS has a rightangle at A, so by the Pythagorean Theorem,

MS = 102+82= /-16-4 ti 169=13.

7. ANSWER (B) The areas of the regions enclosed by the square and thecircle are 102 = 100 and ir(10)2 = 100mr, respectively. One quarter of thesecond region is also included in the first, so the area of the union is

100 + 1007r - 257r = 100 + 75mr.

BC

D

CA

D

mo=

t

CA

DC

AD

+


8. ANSWER (D) If there are n rows in the display, the bottom row contains2n -1 cans. The total number of cans is therefore the sum of the arithmeticseries

1+3+5+-+(2n-1),which is

2 [(2n - 1) + 1] = n2.

Thus n2 = 100, son = 10.

9. ANSWER (E) The rotation takes (-3, 2) into B = (2, 3), and the reflec-tion takes B into C = (3, 2).

B(2,3),/

'C(3,2)

10. ANSWER (A) The area of the annulus is the difference between the ar-eas of the two circles, which is 7rb2 - irc2. Because the tangent X Z isperpendicular to the radius OZ, b2 - c2 = a2, so the area is 7ra 2.

11. ANSWER (D) Each score of 100 is 24 points above the mean, so the fivescores of 100 represent a total of (5) (24) = 120 points above the mean.Those scores must be balanced by scores totaling 120 points below themean. Since no student scored more than 76 - 60 = 16 points below themean, the number of other students in the class must be an integer no lessthan 120/16. The smallest such integer is 8, so the number of students inthe class is at least 13. Note that the conditions of the problem are met if 5students score 100 and 8 score 61.

OR

If there are k students in the class, the sum of their scores is 76k. If the fivescores of 100 are excluded, the sum of the remaining scores is 76k - 500.

128 Solutions

Since each student scored at least 60, the sum is at least 60(k - 5). Thus

76k - 500 > 60(k - 5),

so k > 12.5. Since k must be an integer, k > 13.

12. ANSWER (C) Let ak be the kth term of the sequence. Fork > 3,

ak+1 = ak-2 + ak-i - ak, so ak+1 - ak-1 = -(ak - ak-2)

Because the sequence begins

2001, 2002, 2003, 2000, 2005, 1998, ...

it follows that the odd-numbered terms and the even-numbered terms eachform arithmetic progressions with common differences of 2 and -2, re-spectively. The 2004"h term of the original sequence is the 1002"d term ofthe sequence 2002, 2000, 1998,..., and that term is 2002+ 1001(-2) = 0.

13. ANSWER (A) Since f (f -I (x)) = x, it follows that a (bx + a) + b = x,so ab = 1anda2+b= 0. Hence a = b = -1, so a + b = -2.

14. ANSWER (D) Because AABC, ANBK, and AAMJ are similar righttriangles whose hypotenuses are in the ratio 13 : 8 : 1, their areas are in theratio 169: 64 : 1.

The area of AABC is 1(12)(5) = 30, so the areas of ANBK and AAMJ(30),respectively.are i69(30)and i1-69

Thus the area of pentagon CMJKN is (1 - 169 - 169)(30) = 240/13.

15. ANSWER (B) Let Jack's age be lOx + y and Bill's age be 10y + x. Infive years Jack will be twice as old as Bill. Therefore

lOx + y + 5 = 2(10y + x + 5),

so 8x = 19y + 5. The expression l9y + 5 = 16y + 8 + 3(y - 1) is amultiple of 8 if and only if y - 1 is a multiple of 8. Since both x and y are9 or less, the only solution is y = 1 and x = 3. Thus Jack is 31 and Bill is13, so the difference between their ages is 18.

OR

Let Bill's age be B, and let Jack's age be B + D. Then (B + D) + 5 =2(B + 5), so D = B + 5. Because the two-digit number B + D is obtained

G1.


by reversing the digits of B, D is a multiple of 9, so B leaves a remainder of4 upon division by 9. Therefore the only possible ordered pairs (B, B + D)are (13, 31), (49, 94), (58, 85) and (67, 76). Only in the first case will Jackbe twice as old as Bill in five years.

16. ANSWER (C) From the definition of f,

f(x+iy)=i(x-iy)=y+ix

for all real numbers x and y, so the numbers that satisfy f (z) = z arethe numbers of the form x + ix. The set of all such numbers is a linethrough the origin in the complex plane. The set of all numbers that satisfyI z I = 5 is a circle centered at the origin of the complex plane. The numberssatisfying both equations correspond to the points of intersection of the lineand circle, of which there are two.

17. ANSWER (A) Let r1, r2, and r3 be the roots. Then

5 = loge r1 + loge r2 + loge r3 = loge r1 r2r3,

so r1r2r3 = 25 = 32. Since

8x3 + 4ax2 + 2bx + a = 8(x - ri)(x - r2)(x - r3),

it follows that a = -8r1 r2r3 = -256.

18. ANSWER (E) Let B = (a, b) and A = (-a, -b). Then

4a2 + 7a - 1 =b and 4a2-7a- 1 = -b.

Subtracting gives b = 7a, so 4a2 + 7a - 1 = 7a. Thus

a2 = 4 and b2 = (7a)2 = 49

so

AB=2 a2+b2=2 4 =5/.

19. ANSWER (A) Let AB and DC be parallel diameters of the bottom andtop bases, respectively. A great circle of the sphere is tangent to all four

130 Solutions

sides of trapezoid ABCD. Let E, F, and G be the points of tangency onAB, BC, and CD, respectively. Then

FB=EB=18 and FC=GC=2,

so BC = 20. If H is on AB such that LCHB is a right angle, then HB =18-2= 16. Thus

CH = 202-162=12,

and the radius of the sphere is (1/2)(12) = 6.

2H CG F

12

E H

16

18

20. ANSWER (B) If the orientation of the cube is fixed, there are 26 = 64possible arrangements of colors on the faces. There are

arrangements in which all six faces are the same color and

2(6) = 125

arrangements in which exactly five faces have the same color. In each ofthese cases the cube can be placed so that the four vertical faces have thesame color. The only other suitable arrangements have four faces of onecolor, with the other color on a pair of opposing faces. Since there are three

.-ti

(,U

viiN

00

'Z3

a)

'ZS

,-.

V'1


pairs of opposing faces, there are 2(3) = 6 such arrangements. The totalnumber of suitable arrangements is therefore 2 + 12 + 6 = 20, and theprobability is 20/64 = 5/16.

21. ANSWER (C) A line y = mx intersects the ellipse in 0, 1, or 2 points.The intersection consists of exactly one point if and only if m = a orm = b. Thus a and b are the values of m for which the system

2x2+xy+3y2-11x-20y+40=0y=mx

has exactly one solution. Substituting mx for y in the first equation gives

2x2 + mx2 + 3m2x2 - l lx - 20mx + 40 = 0,

or, by rearranging the terms,

(3m2+m+2)x2-(20m+11)x+40=0.

The discriminant of this equation is

(20m + 11)2 -4.40. (3m2 + m + 2) = -80m2 + 280m - 199,

which must be zero if m = a or m = b. Thus a + b is the sum of the rootsof the equation -80m2 + 280m - 199 = 0, which is go = I.

22. ANSWER (C) All the unknown entries can be expressed in terms of b.Since 100e = beh = ceg = def, it follows that h = 100/b, g = 100/c,and f = 100/d. Comparing rows l and 3 then gives

100 10050bc=2b c,from which c = 20/b. Thus the product of the entries of the first row is1000, so 100e = 1000 also, and e = 10.

Comparing columns 1 and 3 gives

50dloo=2e100

C d

from which d = c/5 = 4/b.

Finally, f = 100/d = 25b, g = 100/c = 5b, and e = 10. All the entriesare positive integers if and only if b = 1, 2, or 4. The corresponding valuesfor g are 5, 10, and 20, and their sum is 35.

arc

N17

v^,

CA

D

ti?

AID

AID AID

132 Solutions

23. ANSWER (C) Let a denote the zero that is an integer. Because the coeffi-cient of x3 is 1, there can be no other rational zeros, so the two other zerosmust be 1 ± r for some irrational number r. The polynomial is then

(x-a)Lx-(2+r)][x (2-r)](1a2= x3 - 2ax2 + 5 5a2 - r2 x - a - r2

Therefore a = 1002 and the polynomial is

x3 - 2004x2 + (5(501)2 - r2)x - 1002((501)2 - r2).

All coefficients are integers if and only if r2 is an integer, and the zeros arepositive and distinct if and only if 1 < r2 < 5012 - 1 = 251,000. Becauser cannot be an integer, there are 251,000 - 500 = 250,500 possible valuesof n.

24. ANSWER (B) Let LDBE = a and LDBC = 3. Then LCBE = a -,Band LABE = a +,8, so tan(a -,B) tan(a + B) = tan 2 a. Thus

tan a - tan tan a + tan 2= tan a,

1 + tan a tan tan a tan


tan2 a -tan2 Y = tan2 a (1 - tan 2 a tan 2 ,B).

Upon simplifying, tan 2 ,B (tan 4 a - 1) = 0, so tan a = 1 and a = I.. LetDC = a and BD = b. Then cot/DBC = Q. Because LCBE =

a-,B

and LABE = a +,8, it follows that+b =b +a

4 1-b b-aThus the numbers 1, b+a , and

aform an arithmetic progression, so a =

b as . Setting b = ka yields k2-2k-3 = 0, and the only positive solutionis k = 3. Hence b = = 5., a =3, and the area of AABC isab = so

25. ANSWER (B) The smallest power of 2 with a given number of digits hasa first digit of 1, and there are elements of S with n digits for each positiveinteger n < 603, so there are 603 elements of S whose first digit is 1.Furthermore, if the first digit of 2k is 1, then the first digit of 2k+1 is either

t!7


2 or 3, and the first digit of 2k+2 is either 4,5,6, or 7. Therefore there are603 elements of S whose first digit is 2 or 3, 603 elements whose first digitis 4,5,6, or 7, and 2004 - 3(603) = 195 whose first digit is 8 or 9. Finally,note that the first digit of 2k is 8 or 9 if and only if the first digit of 2k-1 is4, so there are 195 elements of S whose first digit is 4.

C7,

--j

.-.

.~-i

134


Solutions


1 0.56 0.66 1.22 94.05 1.18 2.32

2 0.24 93.83 1.74 2.84 0.31 1.04

3 3.69 43.23 11.18 4.29 4.8 32.79

4 69.65 3.3 24.85 0.66 0.36 1.16

5 2.09 69.3 17.99 1.06 0.6 8.96

6 5.01 50.76 2.75 3.2 7.79 30.47

7 5.76 1.93 69.24 0.93 0.9 21.22

8 3.42 4.59 3.47 23.3 2.3 62.91

9 37.34 3.94 8.87 3.02 1.34 45.48

10 3.61 30.99 2.04 5.86 1.13 56.36

11 9.75 4.25 3.67 4.69 33.33 44.29

12 10.39 3.16 2.29 32.51 12.23 39.4

13 1.62 2.38 4.38 29.76 3.46 58.38

14 3.43 4.26 31.28 10.54 3.92 46.56

15 2.41 3.57 15.37 8.57 7.67 62.38

16 2.55 3.18 8.25 7.28 3.36 75.37

17 17.24 8.69 11.86 3.28 4 54.91

18 3747 4.01 2.96 2.79 2.37 84.38

19 8.69 20.8 8.98 4.83 7.93 48.74

20 3.17 1.72 1.94 1.39 2.29 89.48

21 4.02 4.71 4.08 2.13 0.7 84.35

22 3.91 5.63 2.84 2.74 1.99 82.86

23 4.76 6.03 2.39 2.61 4.61 79.57

24 1.1 1.67 1.94 2.14 0.62 92.53

25 3.17 2.45 1.94 2.04 1.02 89.38

C27

0.-p '

a)



1. ANSWER (D) It is given that 0.1x = 2 and 0.2y = 2, so x = 20 andy = 10. Thus x - y = 10.

2. ANSWER (B) Since 2x + 7 = 3 we have x = -2. Hence

-2 = bx - 10 = -2b - 10, so 2b = -8, and b = -4.

3. ANSWER (B) Let w be the width of the rectangle. Then the length is 2w,and

x2 = w2 +(2 W)2 = 5w2.

Thus w2 = x2/5 and the area is w(2w) = 2w2 = 3x2.

4. ANSWER (A) If Dave buys seven windows separately he will purchasesix and receive one free, for a cost of $600. If Doug buys eight windowsseparately, he will purchase seven and receive one free, for a total cost of$700. The total cost to Dave and Doug purchasing separately will be $1300.If they purchase fifteen windows together, they will need to purchase only12 windows, for a cost of $1200, and will receive 3 free. This will result ina savings of $100.

5. ANSWER (B) The sum of the 50 numbers is 20. 30 + 30. 20 = 1200.Their average is 1200/50 = 24.

6. ANSWER (B) Because (rate)(time) = (distance), the distance Josh rodewas (4/5)(2) = 8/5 of the distance that Mike rode. Let m be the numberof miles that Mike had ridden when they met. Then the number of milesbetween their houses is

8 1313=m+Sm= Sm.

Thus m = 5.

7. ANSWER (C) The symmetry of the figure implies that AABH, ABCE,ACDF, and ADAG are congruent right triangles. So

BH = CE = BC2 - BE2 50-1=7,

+

t+1

136 Solutions

and EH = BH - BE = 7 - 1 = 6. Hence the square EFGH has area62=36.

OR

As in the first solution, BH = 7. Thus

Area(EFGH) = Area(ABCD)-4Area(AABH) = 50-4 (z . 1 7) = 36.

8. ANSWER (D) Since A, M, and C are digits we have

0<A+M+C <9+9+9= 27.and

0< 100A+1OM+C < 100.9+10.9+9=999.The only ways to express 2005 as the product of two positive integers areas 2005. 1 or 401-5. Because neither A + M + C nor 100A + 1OM + Ccan be 2005, it follows that 100A + 10M +C = 401 and A +M +C = 5.Hence A = 4, M = 0, and C = 1.

9. ANSWER (A) The quadratic formula yields

-(a+8)f (a+8)2-4 4 9x-2.4

The equation has only one solution precisely when the value of the dis-criminant, (a + 8)2 - 144, is 0. This implies that a = -20 or a = 4, andthe sum is -16.

OR

The equation has one solution if and only if the polynomial is the squareof a binomial with linear term ± 4x2 = ±2x and constant term + =+3. Because (2x±3)2 has a linear term f 12x, it follows that a+8 = +12.Thus a is either -20 or 4, and the sum of those values is -16.

10. ANSWER (B) There are n3 unit cubes with a total of 6n3 faces. Thesurface area of the large cube is 6n2, so 6n2 faces of the unit cubes are red.Therefore

1 _6n2=

1

'so n = 4.

4 6n3 n

..z


11. ANSWER (E) The first and last digits must be both odd or both even fortheir average to be an integer. There are 5 5 = 25 odd-odd combinationsfor the first and last digits. There are 4 5 = 20 even-even combinationsthat do not use zero as the first digit. Hence the total is 45.

12. ANSWER (D) The slope of the line is

1000 - 1 111

so all points on the line have the form (1 + lit, 1 + l l lt). Such a pointhas integer coordinates if and only if t is an integer, and the point is strictlybetween A and B if and only if 0 < t < 9. Thus there are 8 points with therequired property.

13. ANSWER (D) Each number appears in two sums, so the sum of the se-quence is

2(3+5+6+7+9)=60.

The middle term of a five-term arithmetic sequence is the mean of its terms,so 60/5 = 12 is the middle term.

The figure shows an arrangement of the five numbers that meets therequirement.

14. ANSWER (D) A standard die has a total of 21 dots. For 1 < n < 6,a dot is removed from the face with n dots with probability n/21. Thusthe face that originally has n dots is left with an odd number of dots withprobability n/21 if n is even and 1 - (n/21) if n is odd. Each face is the

in,

all

WIN

P0-

MIN

138 Solutions

top face with probability 1/6. Therefore the top face has an odd number ofdots with probability

6

6 ((1 21)+21 +(1 21)+21 +I215

1 3 _1\66

_ 11

=6 3+21 6 21 21

OR

The probability that the top face is odd is 1/3 if a dot is removed froman odd face, and the probability that the top face is odd is 2/3 if a dot isremoved from an even face. Because each dot has the probability 1/21 ofbeing removed, the top face is odd with probability

Cal\1+21+51(3)(2+i4 +6) 633 11

3 21

15. ANSWER (C) Let 0 be the center of the circle. Each of ADCE andAABD has a diameter of the circle as a side. Thus the ratio of their areasis the ratio of the two altitudes to the diameters. These altitudes are DCand the altitude from C to DO in ADCE. Let F be the foot of this secondaltitude. Since ACFO is similar to AD CO,

CF _ CO _ AO - AC 'AB - 'AB _ 1

DC DO DOZAB

3'

which is the desired ratio.

an.


OR

Because AC = AB/3 and AO = AB/2, we have CO = AB/6. TrianglesDCO and DAB have a common altitude to AB so the area of ODCO is6 the area of AADB. Triangles DCO and ECO have equal areas sincethey have a common base CO and their altitudes are equal. Thus the ratioof the area of ADCE to the area of AABD is 1/3.

16. ANSWER (D) Consider a right triangle as shown. By the PythagoreanTheorem,

(r + s)2 = (r - 3s)2 + (r - s)2

so

and

But r

r2+2rs+s2 = r2-6rs+9s2+r2-2rs+s2

0 = r2 - lOrs + 9s2 = (r - 9s) (r - s).

s, so r = 9s and r/s = 9.

r-3s

3s

OR

Because the ratio r/s is independent of the value of s, assume that s = 1and proceed as in the previous solution.

17. ANSWER (A) The piece that contains W is shown. It is a pyramid withvertices V, W, X, Y, and Z. Its base WX YZ is a square with sides oflength 1/2 and its altitude VW is 1. Hence the volume of this pyramidis

3 (2)2 (1) 12

.--.

'C7

b10

+-.

v-;

140 Solutions

18. ANSWER (A) Of the positive integers less than 1000, 499 of them aredivisible by two, 333 are divisible by 3, and 199 are divisible by 5. Thereare 166 multiples of 6, 99 multiples of 10, and 66 multiples of 15. Andthere are 33 numbers that are divisible by 30. So by the Inclusion-ExclusionPrinciple there are

499+333+ 199- 166-99-66+33 = 733

numbers that are divisible by at least one of 2, 3, or 5. Of the remaining999 - 733 = 266 numbers, 165 are primes other than 2, 3, or 5. Note that1 is neither prime nor composite. This leaves exactly 100 prime-lookingnumbers.

19. ANSWER (B) The number of miles traveled is the same as the numberof integers between 1 and 2005, inclusive, that do not contain the digit 4.First consider the integers less than 2000. There are two choices for thefirst digit, including 0, and 9 choices for each of the other three. Becauseone combination of choices is 0000, there are 2 93 - 1 = 1457 positiveintegers less than 2000 that do not contain the digit 4. There are 5 integersbetween 2000 and 2005, inclusive, that do not have a 4 as a digit, so the cartraveled 1457 + 5 = 1462 miles.

OR

Because the odometer uses only 9 digits, it records mileage in base-9 nu-merals, except that its digits 5, 6, 7, 8, and 9 represent the base-9 digits 4,5, 6, 7, and 8. Therefore the mileage is

2004base 9 = 2 .93 +4=2-729+4= 1462.


20. ANSWER (E) The graphs of y = f (x) and y = f 121(x) are shownbelow. For n > 2 we have

=f In 11 (2x), if 0 < x < 2,

x x.f1nlO = f[n-i1 (.fO) f[n-,1 (2 - 2x), if 2 < x < 1.

Let g(n) be the number of values of x in [0, 1] for which f 1n1(x) = 1/2.Then f lnl(x) = 1/2 for g(n - 1) values of x in [0, 1/2] and g(n - 1)values of x in [1/2, 1]. Furthermore f ln1(1/2) = f [n-i](1) = 0 1/2for n > 2. Hence g(n) = 2g(n - 1) for each n > 2. Because g(1) = 2, itfollows that g(2005) = 22005.

1/4 1/2 3/4 1

21. ANSWER (C) The two equations are equivalent to b = a(,2005)and

c=2005-b-a, so

c = 2005 -a(C2005) - a.

If c > 1, then

b >2(22005)

> 2005 > 2005 - a - c = b,

which is a contradiction. For c = 0 and for c = 1, the only solutions arethe ordered triples (2004, 1, 0) and (1002, 1002, 1), respectively. Thus thenumber of solutions is 2.

22. ANSWER (B) Let the dimensions of P be x, y, and z. The sum of thelengths of the edges of P is 4(x + y + z), and the surface areaof P is 2xy + 2yz + 2xz, so

x + y + z = 28 and 2xy + 2yz + 2xz = 384.

,-"

L..

.-.

142 Solutions

Each internal diagonal of P is a diameter of the sphere, so

(2r)2 = (x2+y2+z2) _ (x+y+z)2-(2xy+2xz+2yz) = 282-384 = 400.

So 2r = 20 and r = 10.

NOTE: There are infinitely many positive solutions of the system x +y + z = 28, 2xy + 2yz + 2xz = 384, so there are infinitely many non-congruent boxes meeting the given conditions, but each can be inscribed ina sphere of radius 10.

23. ANSWER (B) Let a = 2-' and b = 2k. Then

log, b loge; 2klog 2k k log 2 k

log 2J j log 2 j '

so log, b is an integer if and only if k is an integer multiple of j. For eachj, the number of integer multiples of j that are at most 25 is L5 J . Because

j k, the number of possible values of k for each j is L5 J - 1. Hencethe total number of ordered pairs (a, b) is

25zs

([]_1)=24+h1+7+5+4+3+2(2)+41)=62.

Since the total number of possibilities for a and b is 25.24, the probabilitythat log, b is an integer is

62 _ 31

25.24 300

24. ANSWER (B) The polynomial P(x) R(x) has degree 6, so Q(x) musthave degree 2. Therefore Q is uniquely determined by the ordered triple(Q(1), Q(2), Q(3)). When x = 1, 2, or 3, we have 0 = P(x) R(x) =P (Q(x)). It follows that (Q(1), Q(2), Q(3)) is one of the 27 orderedtriples (i, j, k), where i, j, and k can be chosen from the set {1, 2, 3}.However, the choices (1, 1, 1), (2, 2, 2), (3, 3, 3), (1, 2, 3), and (3, 2, 1)lead to polynomials Q(x) defined by Q(x) = 1, 2, 3, x, and 4 - x, re-spectively, all of which have degree less than 2. The other 22 choices for(Q(1), Q(2), Q(3)) yield non-collinear points, so in each case Q(x) is aquadratic polynomial.

7-i

...


25. ANSWER (C) Let A(xi, yi, zi), B(x2, y2, z2), and C(x3, y3, z3) be thevertices of such a triangle. Let

(Oxk, AYk, AZk) = (xk+1 - Xk, Yk+1 - Yk, Zk+1 - Zk), for 1 < k < 3,

where (x4,Y4,z4) = (xI,yI,zI). Then (IAXkI, IAYkI, IOzkD is a per-mutation of one of the ordered triples (0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 1, 2),(0, 2, 2), (1, 1, 1), (1, 1, 2), (1, 2, 2), or (2, 2, 2). Since A A B C is equilat-eral, AB, BC, and CA correspond to permutations of the same orderedtriple (a, b, c). Because

3 3 3

Y' AXk=YAYk=L AZk=0,k=1 k=1

k=11

the sums3 3 3

JAXkl, IAYk1, and E JAZk1k=1

k=1

1 k=1

are all even. Therefore (J Axk l , I AYk I , I Azk l) is a permutation of one ofthe triples (0, 0, 2), (0, 1, 1), (0, 2, 2), (1, 1, 2), or (2, 2, 2).

If (a, b, c) = (0, 0, 2), each side of AABC is parallel to one of the coor-dinate axes, which is impossible.

If (a, b, c) = (2, 2, 2), each side of A A B C is an interior diagonal of the2 x 2 x 2 cube that contains S, which is also impossible.

If (a, b, c) = (0, 2, 2), each side of AA BC is a face diagonal of the 2 x 2x 2cube that contains S. The three faces that join at any vertex determine sucha triangle, so the triple (0, 2, 2) produces a total of 8 triangles.

If (a, b, c) = (0, 1, 1), each side of A A B C is a face diagonal of a unitcube within the larger cube that contains S. There are 8 such unit cubesproducing a total of 8 - 8 = 64 triangles.

There are two types of line segments for which (a, b, c) = (1, 1, 2). Onetype joins the center of the face of the 2 x 2 x 2 cube to a vertex on theopposite face. The other type joins the midpoint of one edge of the cubeto the midpoint of another edge. Only the second type of segment canbe a side of AABC. The midpoint of each of the 12 edges is a vertexof two suitable triangles, so there are 12 2/3 = 8 such triangles.

The total number of triangles is 8 + 64 + 8 = 80.

.-.

571

v'1

NN

0000

N

.Mr

fir

144


Solutions


1 92.18 1.7 2.71 0.94 1.93 0.53

2 0.74 1.27 0.86 92.12 1.39 3.6

3 0.72 0.52 91.41 1.32 2.72 3.29

4 1.04 89.91 1.61 1.6 1.68 4.15

5 78.34 0.92 6.58 0.37 0.5 13.27

6 31.31 8.28 5.52 3.13 2.68 49.08

7 10.09 14.44 2.19 23.99 4.59 44.7

8 11.02 15.62 43.47 0.46 9.14 20.25

9 3.62 76.41 3.21 3.45 2.48 10.8

10 1.59 5.16 1.46 19.87 35 36.9

11 11.66 10.57 7.13 42.64 4.08 23.89

12 2.84 6.06 7.72 10.1 0.93 72.35

13 3.93 2.88 2.44 18.97 1.91 69.87

14 7.13 9.62 6.44 3.44 11.12 62.24

15 5.28 2.88 3.29 48.57 4.46 35.51

16 2.32 2.15 7.42 5.56 2.2 80.35

17 3.6 5.98 2.42 1.36 6.13 80.49

18 12.94 4.79 5.03 2.97 2.62 71.65

19 2.34 1.89 1.86 2.74 21.68 69.48

20 1.07 2.91 42.83 6.09 17.16 29.92

21 1.79 2.35 4.98 3.36 1.77 85.75

22 3.69 6.48 1.59 1.63 1.3 85.31

23 1.16 2.66 1.63 1.16 0.68 92.71

24 2.66 2.16 2.79 2.27 1.87 88.24

25 2.37 2.64 1.91 2.09 3.13 87.86



1. ANSWER (A) The scouts bought 1000/5 = 200 groups of 5 candy barsat a total cost of 200. 2 = 400 dollars. They sold 1000/2 = 500 groups of2 candy bars for a total of 500. 1 = 500 dollars. Their profit was $500 -$400 = $100.


x- x = 4, so100

x2 = 400.

Because x > 0, it follows that x = 20.

3. ANSWER (C) The number of CDs that Brianna will finally buy is threetimes the number she has already bought. The fraction of her money thatwill be required for all the purchases is (3)(1/5) = 3/5. The fraction shewill have left is 1 - (3/5) = 2/5.

4. ANSWER (B) To earn an A on at least 80% of her quizzes, Lisa needsto receive an A on at least (0.8)(50) = 40 quizzes. Thus she must earn anA on at least 40 - 22 = 18 of the remaining 20. So she can earn a gradelower than an A on at most 2 of the remaining quizzes.

5. ANSWER (A) The four white quarter circles in each tile have the samearea as a whole circle of radius 1/2, that is, -r(1/2)2 = 7r/4 square feet.So the area of the shaded portion of each tile is 1- (7r/4) square feet. Sincethere are 8 10 = 80 tiles in the entire floor, the area of the total shadedregion in square feet is

80 (1 - 4) = 80 - 20ir.

6. ANSWER (A) Let CH be an altitude of AABC. Applying the Pytha-gorean Theorem to ACHB and to ACHD produces

82 - (BD + 1)2 = CH2 = 72 - 12 = 48, so (BD + 1)2 = 16.

146 Solutions

Thus BD = 3.

7. ANSWER (D) The graph is symmetric with respect to both coordinateaxes, and in the first quadrant it coincides with the graph of the line 3x +4y = 12. Therefore the region is a rhombus, and the area is

Area = 4 (--(4.3)) = 24.

8. ANSWER (C) The vertex of the parabola is (0, a2). The line passesthrough the vertex if and only if a2 = 0 + a. There are two solutionsto this equation, namely a = 0 and a = 1.

9. ANSWER (B) The percentage of students getting 95 points is

100-10-25-20-15= 30,

so the mean score on the exam is

0.10(70) + 0.25(80) + 0.20(85) + 0.15(90) + 0.30(95) = 86.

in'

..o

1-1


Since fewer than half of the scores were less than 85, and fewer than halfof the scores were greater than 85, the median score is 85. The differencebetween the mean and the median score on this exam is 86 - 85 = 1.

10. ANSWER (E) The sequence begins 2005, 133, 55, 250, 133, .... Thusafter the initial term 2005, the sequence repeats the cycle 133, 55, 250.Because 2005 = 1 + 3 668, the 2005th term is the same as the last termof the repeating cycle, 250.

11. ANSWER (D) There are

18 _ 8!= 28

2 - 6! . 2!

ways to choose the bills. A sum of at least $20 is obtained by choosingboth $20 bills, one of the $20 bills and one of the six smaller bills, or both$10 bills. Hence the probability is

1 +2.6+ 1 _ 14 _ 1

28 28 2

12. ANSWER (D) Let r1 and r2 be the roots of x2 + px + m = 0. Sincethe roots of x2 + mx + n = 0 are 2r1 and 2r2, we have the followingrelationships:

m = rlr2, n = 4r1r2, p = -(r1 + r2), and m = -2(rl + r2).

So

1 n 4mn=4m, p=2m, andp Zm

OR

The roots of( llz

+ p \ll + m = 0\2/ 21

are twice those of x2 + px + m = 0. Since the first equation is equivalentto x2 + 2 px + 4m = 0, we have

nm = 2p and n = 4m, so -8.8.

P

.-h

o-0

,a?

(1)

148 Solutions

13. ANSWER (D) Since 4x1 = 5, 512 = 6, ... , 127x124 = 128, we have

47/2 = 128 = 127x124 = (126-123)x124 =

So -1x2 ... X124 = 7/2.

We haveOR

-1x2 ... X124 = 1094 5 .1095 6...109127 128

_ log 5 log 6 log 128 _ log 128

log 4 log 5 log 127 log 4

_ log 27 _ 7 log 2 _ 7

log 22 2 log 2 2

14. ANSWER (E) Let 0 denote the origin, P the center of the circle, and rthe radius. A radius from the center to the point of tangency with the liney = x forms a right triangle with hypotenuse OP. This right triangle isisosceles since the line y = x forms a 45° angle with the y-axis. So

6r/=r+6 and r= =6/+6.2_1

Y

OR

Let the line y = -x intersect the circle and the line y = 6 at M and K,respectively, and let the line y = x intersect the circle and the line y = 6at N and L, respectively. Quadrilateral PMON has four right angles andMP = PN, so PMON is a square. In addition, MK = KJ = 6 andKO = 6,/2-. Hence

r=MO =MK +KO =6+6h.

LY

E

d-0

d-0


15. ANSWER (D) The sum of the digits 1 through 9 is 45, so the sum ofthe eight digits is between 36 and 44, inclusive. The sum of the four unitsdigits is between 1 + 2 + 3 + 4 = 10 and 6 + 7 + 8 + 9 = 30, inclusive,and also ends in 1. Therefore the sum of the units digits is either 11 or 21.If the sum of the units digits is 11, then the sum of the tens digits is 21,so the sum of all eight digits is 32, an impossibility. If the sum of the unitsdigits is 21, then the sum of the tens digits is 20, so the sum of all eightdigits is 41. Thus the missing digit is 45 - 41 = 4. Note that the numbers13, 25, 86, and 97 sum to 221.

OR

Each of the two-digit numbers leaves the same remainder when dividedby 9 as does the sum of its digits. Therefore the sum of the four two-digitnumbers leaves the same remainder when divided by 9 as the sum of alleight digits. Let d be the missing digit. Because 221 when divided by 9leaves a remainder of 5, and the sum of the digits from 1 through 9 is 45,the number (45 - d) must leave a remainder of 5 when divided by 9. Thusd = 4.

16. ANSWER (D) The centers of the unit spheres are at the 8 points withcoordinates (± 1, ± 1, ± 1), which are at a distance

12+12+12

from the origin. Hence the maximum distance from the origin to any pointon the spheres is 1 + 3-.

17. ANSWER (B) The given equation is equivalent to

log10 (2a 3b 5` . 7d ) = 2005, so 2a 1 02005 = 22005.52005

.,p

150 Solutions

Let M be the least common denominator of a, b, c and d. It follows that

2Ma . 3Mb SMc . 7Md = 22005M . 52005M

Since the exponents are all integers, the Fundamental Theorem of Arith-metic implies that

Ma = 2005M, Mb = 0, Mc = 2005M, and Md = 0.

Hence the only solution is (a, b, c, d) = (2005, 0, 2005, 0).

18. ANSWER (C) For LABC to be acute, all angles must be acute. ForLA to be acute, point C must lie above the line passing through A andperpendicular to AB. The segment of that line in the first quadrant liesbetween P (4, 0) and Q (0, 4). For L B to be acute, point C must lie belowthe line through B and perpendicular to AB. The segment of that line inthe first quadrant lies between S(14, 0) and T (O, 14). For LC to be acute,point C must lie outside the circle U that has A B as a diameter. Let 0denote the origin. Region R, shaded below, has area equal to

Area(A0ST) - Area(AOPQ) - Area(Circle U)2

1.142- 1.42-,r =90-2ir 51.2 2 2 2

19. ANSWER (E) By the given conditions, it follows that x > y. Let x =l0a + b and y = 10b + a, where a > b. Then

m2 = x2-y2 = (10a+b)2-(10b+a)2 = 99a2-99b2 = 99(a2-b2).

Since 99(a2 - b2) must be a perfect square,

a2 -b 2 = (a + b)(a - b) = 11k2,

F1.

..j

+

Ins

'1^

..,


for some positive integer k. Because a and b are distinct digits, we havea-b<9-1 = 8 and a + b < 9 + 8 = 17. It follows that a+b= 11,a - b = k2, and k is either 1 or 2.

If k = 2, then (a, b) = (15/2, 7/2), which is impossible. Thus k = 1 and(a, b) = (6, 5). This gives x = 65, y = 56, m = 33, and x+y+m = 154.

20. ANSWER (C) Note that the sum of the elements in the set is 8. Letx = a + b + c + d, so e + f +g+h=8-x. Then

(a+b+c+d)2+(e+f +g+h)2=x2+(8-x)2= 2x2 - 16x + 64 = 2(x - 4)2 + 32 > 32.

The value of 32 can be attained if and only if x = 4. However, it may beassumed without loss of generality that a = 13, and no choice of b, c, andd gives a total of 4 for x. Thus (x - 4)2 > 1, and

(a+b+c+d)2+(e+f +g+h )2 = 2(x - 4)2 + 32 > 34.

A total of 34 can be attained by letting a, b, c, and d be distinct elementsin the set {-7, -5, 2, 13}.

21. ANSWER (C) Let n = 7k Q, where Q is the product of primes, none ofwhich is 7. Let d be the number of divisors of Q. Then n has (k + 1)ddivisors. Also 7n = 7k+1 Q, so 7n has (k + 2)d divisors. Thus

(k + 2)d - 80 - 4 and 3(k + 2) = 4(k + 1).(k + 1)d 60 3

Hence k = 2. Note that n = 21972 meets the conditions of the problem.


iZn iZn iZnZn+1 = _

Zn ZnZn IZnI2

Since Izol = 1, the sequence satisfies

Z1 = iZp, Z2 = iZ1 = i (IZ0 2 = -lZ0,

and, in general, when k > 2,

kZk = -I ZO

y,,,

152 Solutions

)Hence z0 satisfies the equation 1 = -iz0

),so z

(22005° = i. Because

every nonzero complex number has n distinct nth roots, this equation has22005 solutions. So there are 22005 possible values for z°.

OR

Definecis 0 = cos 0 + i sin 0.

Then if z = r cis 0 we have

cis (0 + 90°)zn+i

= = cis (20 + 90°).cis (-0)

The first terms of the sequence are z0 = cisa, zI = cis (2a + 90°) = izo,

Z2 = cis (4a + 270°) = cis (4a - 90°) = , z3 = cis (8a - 90°)and, in general,

Z(2")

forn>2.

So(22005)

Z2005 = Z°i

= 1 and2(22005) - i.

As before, there are 22005 possible solutions for z0.

23. ANSWER (B) From the given conditions it follows that

x+y = 102, x2+y2 = 10.10z and 102z = (x+)2 = x2+2xy+y2.

Thus

xy = 2 (102z - 10. 10z).

Also

(x + y)3 = 103z and x3 + y3 = (x + y)3 - 3xy(x + y),

which yields

x3 + y3 = 103z -2

(102z - 10. 102)(102)

= 103z-3(1032 -10.102z)=-2103z+15.102z,

and a+b=-2+15=29/2.


No other value of a + b is possible for all members of S, because the triple

((1 + 19)/2, (1 - 19)/2, 0) is in S, and for this ordered triple, the

equation x3 + y3 = a 103z + b 102z reduces to a + b = 29/2.

24. ANSWER (A) Suppose that the triangle has vertices A(a, a2), B(b, b2)and C(c, c2). The slope of line segment A -B is

b2 - a2 =b +a,b-aso the slopes of the three sides of the triangle have a sum

n

The slope of one side is 2 = tan 0, for some angle 0, and the two remainingsides have slopes

tan O fn tan 0± tan(ir/3)

=2± 8 f 5/

( -) =13 1 + tan 0 tan(ir/3) 1 + 2f3- 1

Therefore

m 1 8+5J 8-5f3- _ 32-11 11 ) 11'n

and m + n = 14.

Such a triangle exists. The x-coordinates of its vertices are (11 ± 5J)/11and -19/11.

OR

Define the vertices as in the first solution, with the added stipulations thata < b and A B has slope 2. Then

b2 - a22= b-a =b+a, so a=1-kandb=l+k,

for some k > 0. If D is the midpoint of AB, then

D-(1 (1-k)2+(1+k)2)=(1,1+k2).2

t-,

ti,

.-.

ce.

F-'

`ti

154 Solutions

The slope of the altitude CD is -1/2, so

1 - c = 2(c2 - 1 - k2).

Therefore

CD2 = (1 - c)2 + (c2 - 1 -k 2 ) 2= 4 (1 - c)2.

Because A A B C is equilateral, we also have

CD2 = 3AB2 = 4 ((2k)2 + (4k)2) = 15k2.

Hence

5 2

4(1 -c)2 = 15k2, so k2_

= (1

12 )

Substitution into the equation 1 - c = 2(c2 - 1 - k2) yields c = 1 orc = -19/11. Because c < 1, it follows that

19 3 ma+b+c=2-1111 '

so m+n=14.n

25. ANSWER (A) Because each ant can move from its vertex to any of fouradjacent vertices, there are 46 possible combinations of moves. In the fol-lowing, consider only those combinations in which no two ants arrive at thesame vertex. Label the vertices as A, B, C, A', B' and C, where A', B'and C' are opposite A, B and C, respectively. Let f be the function thatmaps each ant's starting vertex onto its final vertex. Then neither of f (A)nor f (A') can be either A or A', and similar statements hold for the otherpairs of opposite vertices. Thus there are 4.3 = 12 ordered pairs of valuesfor f (A) and f (A'). The vertices f (A) and f (A') are opposite each otherin four cases and adjacent to each other in eight.

Suppose that f (A) and f (A') are opposite vertices, and, without loss ofgenerality, that f (A) = B and f (A') = B'. Then f (C) must be either Aor A' and f (C) must be the other. Similarly, f (B) must be either C orC and f (B') must be the other. Therefore there are 4 2 2 = 16 combi-nations of moves in which f (A) and f (A') are opposite each other.

Suppose now that f (A) and f (A') are adjacent vertices, and, without lossof generality, that f (A) = B and f (A') = C. Then one of f (B) andf (B') must be C' and the other cannot be B'. So there are four possible

NIA


ordered pairs of values for f (B) and f (B'). For each of those there aretwo possible ordered pairs of values for f (C) and f (C'). Therefore thereare 8 4 2 = 64 combinations of moves in which f (A) and f (A') areadjacent to each other.

Hence the probability that no two ants arrive at the same vertex is

16+64 5.24 5

46 = 212 = 256

--`

v'1

u';N

.--

.-.

v')

.-r

156


Solutions


1 99.35 0.1 0.23 0.2 0.03 0.09

2 13.59 1.5 51.57 0.76 7.53 25.03

3 0.44 88.66 0.32 0.29 9.61 0.66

4 18.72 1.74 0.51 0.86 74.53 3.62

5 1.47 6.97 8.74 78.48 1.37 2.95

6 75.12 0.99 3.28 2.19 2.61 15.8

7 6.85 53.53 7.53 5.98 3.83 22.28

8 9.25 12.06 67.23 2.53 1.26 7.65

9 82.06 1.79 0.78 0.44 2.76 12.16

10 2.47 1.5 6.69 35.34 20.04 33.93

11 15.28 9.83 27.26 9.15 6.18 32.28

12 14.2 54.07 1.74 6.87 3.26 19.84

13 4.24 1.17 2.1 1.02 59.3 32.14

14 0.64 5.16 50.61 24 4.14 15.43

15 35.67 15.04 3.09 5.51 14.12 26.55

16 3.25 40.88 1.59 2.06 5.47 46.75

17 3.08 6.25 4.79 3.47 0.98 81.42

18 13.15 1.04 3.13 6.52 4.64 71.52

19 1.26 1.5 1.99 1.67 3.05 90.52

20 17.25 5.8 5.57 2.68 1.69 66.99

21 0.92 1.21 3.77 1.43 2.78 89.88

22 1.44 1.42 1.58 1.61 0.79 93.16

23 1.01 1.52 1.65 1.06 1.81 92.94

24 10.26 1.78 2.05 2.72 1.38 81.79

25 1.24 1.85 2.01 1.56 1.42 91.9

0.l

C/)


2006 AMC 12A Solutions

The problems for the 2006 AMC 12A begin on page 47.

1. ANSWER (A) Five sandwiches cost 5 3 = 15 dollars and eight sodascost 8 - 2 = 16 dollars. Together they cost 15 + 16 = 31 dollars.

2. ANSWER (C) By the definition we have

h®(h(& h)=h®(h3-h)=h3-(h3-h)=h.

3. ANSWER (B) Mary is (3/5)(30) = 18 years old.

4. ANSWER (E) The largest possible sum of the two digits representingthe minutes is 5 + 9 = 14, occurring at 59 minutes past each hour. Thelargest possible single digit that can represent the hour is 9. This exceedsthe largest possible sum of two digits that can represent the hour, which is1 + 2 = 3. Therefore, the largest possible sum of all the digits is 14 + 9 =23, occurring at 9:59.

5. ANSWER (D) Each slice of plain pizza cost $1. Dave paid $2 for theanchovies in addition to $5 for the five slices of pizza that he ate, for a totalof $7. Doug paid only $3 for the three slices of pizza that he ate. HenceDave paid 7 - 3 = 4 dollars more than Doug.

6. ANSWER (A) Let E represent the end of the cut on DC, and let Frepresent the end of the cut on AB. For a square to be formed, we musthave

DE=y=FB and DE+y+FB=18, so y=6.

The rectangle that is formed by this cut is indeed a square, since the originalrectangle has area 8. 18 = 144, and the rectangle that is formed by this cuthas a side of length 2 - 6 = 12 = 144.

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158 Solutions

DYE E C D

F F Y B

B

F

7. ANSWER (B) Let Danielle be x years old. Sally is 40% younger, so sheis 0.6x years old. Mary is 20% older than Sally, so Mary is 1.2(0.6x) =0.72x years old. The sum of their ages is 23.2 = x + 0.6x + 0.72x =2.32x years, so x = 10. Therefore Mary's age is 0.72x = 7.2 years, andshe will be 8 on her next birthday.

8. ANSWER (C) First note that, in general, the sum of n consecutive integersis n times their median. If the sum is 15, we have the following cases:

if n = 2, then the median is 7.5 and the two integers are 7 and 8;

if n = 3, then the median is 5 and the three integers are 4, 5, and 6;

if n = 5, then the median is 3 and the five integers are 1, 2, 3, 4,and 5.

Because the sum of four consecutive integers is even, 15 cannot be writtenin such a manner. Also, the sum of more than five consecutive integers mustbe more than 1 + 2 + 3 + 4 + 5 = 15. Hence there are 3 sets satisfying thecondition.

Note: It can be shown that the number of sets of two or more consecutivepositive integers having a sum of k is equal to the number of odd positivedivisors of k, excluding 1.

9. ANSWER (A) Let p be the cost (in cents) of a pencil, and let s be the cost(in cents) of a set of one pencil and one eraser. Because Oscar buys 3 setsand 10 extra pencils for $1.00, we have

3s+10p=100.

Thus 3s is a multiple of 10 that is less than 100, so s is 10, 20, or 30. Thecorresponding values of p are 7, 4, and 1. Since the cost of a pencil is morethan half the cost of the set, the only possibility is s = 10.

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10. ANSWER (E) Suppose that k = 120 - is an integer. Then 0 < k <120, and because k is an integer, we have 0 < k < 10. Thus there are

11 possible integer values of k. For each such k, the corresponding valueof x is (120 - k2)2. Because (120 - k2)2 is positive and decreasing for0 < k < 10, the 11 values of x are distinct.

11. ANSWER (C) The equation (x + y)2 = x2 + y2 is equivalent to x2 +2xy + y2 = x2 + y2, which reduces to xy = 0. Thus the graph of theequation consists of the two lines that are the coordinate axes.

12. ANSWER (B) The top of the largest ring is 20 cm above its bottom. Thatpoint is 2 cm below the top of the next ring, so it is 17 cm above the bottomof the next ring. The additional distances to the bottoms of the remainingrings are 1 6 cm, 15 cm, ... , 1 cm. Thus the total distance is

17.1820+(17+ 1) = 20+ 2 = 20+ 17.9= 173 cm.

OR

The required distance is the sum of the outside diameters of the 18 ringsminus a 2-cm overlap for each of the 17 pairs of consecutive rings. Thisequals

20.21_ - 37 = 173 cm.

2

13. ANSWER (E) Let r, s, and t be the radii of the circles centered at A, B,and C, respectively. Then r + s = 3, r + t = 4, ands + t = 5, from whichr = 1, s = 2, and t = 3. Thus the sum of the areas of the circles is

n(12+22+32)= 14r.

14. ANSWER (C) If a debt of D dollars can be resolved in this way, thenintegers p and g must exist with

D = 300p + 210g = 30(10p + 7g).

As a consequence, D must be a multiple of 30, so no positive debt of lessthan $30 can be resolved. A debt of $30 can be resolved since

30 = 300(-2) + 210(3).

This is done by giving 3 goats and receiving 2 pigs.

160 Solutions

15. ANSWER (A) Because cos x = 0 and cos(x + z) = 1/2, it follows thatx = m,-r/2 for some odd integer m and x + z = 2n7r ± it/3 for someinteger n. Therefore

m7r 7r 7r 7rz=2n,-r- 2 f 3 =kir+2 ± 3

for some integer k. The smallest value of k that yields a positive value forz is 0, and the smallest positive value of z is 7r/2 - 7r/3 = 7r/6.

OR

Let 0 denote the center of the unit circle. Because cos x = 0, the termi-nal side of an angle of measure x, measured counterclockwise from thepositive x-axis, intersects the circle at A = (0, 1) or B = (0, -1).

Because cos(x + z) = 1/2, the terminal side of an angle of measure x + z

intersects the circle at C = (2 , 2 ) or D = (2 , - 2 ). Thus all angles of

positive measure z = (x + z) - x can be measured counterclockwise fromeither OA or OB to either OC or OD. The smallest such angle is LBOD,which has measure it/6 and is attained, for example, when x = -,-r/2 andx + z = -7r/3.

16. ANSWER (B) Radii AC and BD are each perpendicular to CD. By thePythagorean Theorem,

CE = 52-32=4.

Because AACE and ABDE are similar,

DE BD BD 8 32=4-- _'

so DE = CECE AC AC 3 3


Therefore

CD=CE+DE=4+ 32= 34

17. ANSWER (B) Let B = (0, 0), C = (s, 0), A = (0, s), D = (s, s), andE = (s+ s + ). Apply the Pythagorean Theorem to AAFE to

2 2obtain

l2 r ll2

r2+(9+5,2) r_(s+72 ) +(r) ,

from which 9 + 5 = s2 + rs vf2-. Because r and s are rational, it followsthat s2 = 9 and rs = 5, so r/s = 5/9.

OR

Extend AD past D to meet the circle at G 54 D. Because E is collinearwith B and D, AEDG is an isosceles right triangle. Thus DG =By the Power of a Point Theorem,

9 + 5n = AF2 = AD AG = AD (AD + DG)

=s(s+rV )=s2+rsN12.

As in the first solution, conclude that r/s = 5/9.

18. ANSWER (E) The conditions on f imply that both

x=f(x)+f(I)xand x=.f (1)x+.f (_-)lx_f (1)+x.

Thus if x is in the domof f , then x = 1 /x, so x = ± 1.

The conditions are satisfied if and only if f (1) = 1/2 and f (-1) = -1 /2.

19. ANSWER (E) The slope of the line I containing the centers of the circlesis 5/12 = tan 0, where 0 is the acute angle between the x-axis and line1. The equation of line I is y - 4 = (5/12)(x - 2). This line and the twocommon external tangents are concurrent. Because one of these tangentsis the x-axis, the point of intersection is the x-intercept of line 1, which is(-38/5, 0). The acute angle between the x-axis and the other tangent is

+-+

162 Solutions

20, so the slope of that tangent is

tan 20 = 25/12 _ 120

1-(5/12)2 119

Thus the equation of that tangent is y = (120/119) (x + (38/5)), and

b120 38 912

119 5 119

20. ANSWER (C) At each vertex there are three possible locations that thebug can travel to in the next move, so the probability that the bug will visitthree different vertices after two moves is 2/3. Label the first three verticesthat the bug visits as A, B, and C, in that order. In order to visit everyvertex, the bug must travel from C to either G or D.

B

The bug travels to G with probability 1/3, and from there the bug mustvisit the vertices F, E, H, and D in that order. Each of these choices hasprobability 1/3 of occurring. So the probability that the path continues inthe order

C->G--F--E-*H-*Dis(3)5

Alternatively, the bug can travel from C to D and then from D to H. Eachof these occurs with probability 1/3. From H the bug could go either to Gor to E, with probability 2/3, and from there to the two remaining vertices,each with probability 1/3. So the probability that the path continues in oneof the orders

/EFGC ->D -*H " GFE

is (3) (3)4.

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CD

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Hence the bug will visit every vertex in seven moves with probability

(3) [(3)5 +(32)

\3/4l -(23)

\3 + 3/(3)4

243.

2

J

OR

From a given starting point there are 37 possible walks of seven moves forthe bug, all of them equally likely. If such a walk visits every vertex exactlyonce, there are three choices for the first move and, excluding a return tothe start, two choices for the second. Label the first three vertices visited asA, B, and C, in that order, and label the other vertices as shown. The bugmust go to either G or D on its third move. In the first case it must then visitvertices F, E, H, and D in order. In the second case it must visit eitherH, E, F, and G or H, G, F, and E in order. Thus there are 3 2 3 = 18walks that visit every vertex exactly once, so the required probability is18/37 = 2/243.

21. ANSWER (E) For j = 1 and 2, the given inequality is equivalent to

j +x2+y2 < 10'(x+y),

or to

(___)2 ( )2 102j+ y- 2 < 2 -j,provided that x + y > 0. These inequalities define regions bounded bycircles. For j = 1 the circle has center (5, 5) and radius 7. For j = 2the circle has center (50, 50) and radius 4998. In each case the centeris on the line y = x in the first quadrant, and the radius is less than thedistance from the center to the origin. Thus x + y > 0 at each interiorpoint of each circle, as was required to ensure the equivalence of the in-equalities. The squares of the radii of the circles are 49 and 4998 for j = 1and 2, respectively. Therefore the ratio of the area of S2 to that of S1 is(4998 r)/(497r) = 102.

22. ANSWER (D) Place the hexagon in a coordinate plane with center atthe origin 0 and vertex A at (2, 0). Let B, C, D, E, and F be the othervertices in counterclockwise order.

NIA

NIA

...

164 Solutions

Corresponding to each vertex of the hexagon, there is an arc on the cir-cle from which only the two sides meeting at that vertex are visible. Thegiven probability condition implies that those arcs have a combined degreemeasure of 180°, so by symmetry each is 30°. One such arc is centeredat (r, 0). Let P be the endpoint of this arc in the upper half-plane. ThenLPOA = 15°. Side BC is visible from points immediately above P, soP is collinear with B and C. Because the perpendicular distance from 0to BC is /3-, we have

= r sin 15° = r sin(45°-30°) = r (sin 450 cos 30° - sin 30° cos 45°) .

So

-r - (./3 1 r _4

_2 2 2)

Therefore

r= 4f3- 4,/3- 16-+18+/6- =3/-2- +16.

OR

Call the hexagon ABCDEF. Side AB is visible from point X if and onlyif X lies in the half-plane that is in the exterior of the hexagon and thatis determined by the line AB. The region from which the three sides AB,BC, and CD are visible is the intersection of three such half-planes.

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CT

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Let rays AB and DC intersect the circle at R and Q, respectively. ThenQR is one of the six arcs of the circle from which three sides are visible.Symmetry implies that the six arcs are congruent, and because the givenprobability is 1/2, the measure of each arc is 30°. Let 0 be the center ofthe hexagon and the circle. Then LQOR = 30°, so

LQOC = LQOR+2 (LBOC - ZQOR) = 30°+2 (60° - 30°) = 45°.

Thus

LOQC = 180° - LQOC - LOCQ = 15°.

Apply the Law of Sines in AOQC to obtain

r - 2 and then r -sin 120° sin 15°

,sin 15°

Then proceed as in the first solution.

23. ANSWER (B) For every sequence S = (a 1 , a2, ... , an) of at least threeterms,

A2(S) Cal + 2a2 + a3 a2 + 2a3 + a44 4

an-2 + 2an-1 + an

4

Thus for m = I and 2, the coefficients of the terms in the numerator ofA' (S) are the binomial coefficients (o), ( T )

. . . . . ( ) ,and the denomina-

tor is 2'n. Because (m) + (,+1) = ("r+1) for all integers r > 0, the coefficients of the terms in the numerators of Am+l (S) are (m+1) (m+1)

m+10 1

(m+1) for 2 < m < n - 2. The definition implies that the denominator of

166 Solutions

each term in A'+' (S) is 2m+i For the given sequence, the sole term inA'00(S) is

1i

(00)1

1 1

mam+, = 2ioo L, m x = 200 (x + 1)

m=0 m=0

Therefore

(_i) = A'oo(S) _250

C(1 +x)100)2100

so (1 + x)'00 = 250, and because x > 0, we have x = - 1.

24. ANSWER (D) The given expression is equal to

2006!

a!b!c!(xaybzc +x°(-Y)b(-z)°

where the sum is taken over all non-negative integers a, b, and c with a +b + c = 2006. If a is odd,then b and c have opposite parity, and the termis 0. Thus there is exactly one term in the simplified expression for everymonomial of the form xa ybz`, where a, b, and c are non-negative integers,a is even, and a + b + c = 2006. There are 1004 even values of a with0 < a < 2006. For each such value, b can assume any of the 2007 - ainteger values between 0 and 2006 - a, inclusive, and the value of c is thenuniquely determined as 2006 - a - b. Thus the number of terms in thesimplified expression is

(2007 - 0) + (2007 - 2) + + (2007 - 2006) = 2007 + 2005 + + 1.

This is the sum of the first 1004 odd positive integers, which is 10042 =1,008,016.

OR

Because the number of non-negative integer solutions of a + b + c = k isk+2),2), the sum in the first solution is taken over /2008) terms, but those for

which b and c have opposite parity have a sumo`f

zero. If b is odd and c iseven, then a is odd, so a = 2A+ 1, b = 2B+ 1, and c = 2C for some non-negative integers A, B, and C. Therefore 2A + 1 + 2B + 1 + 2C = 2006,so A + B + C = 1002. Because the last equation has (1004) non-negativeinteger solutions, there are (1004) terms for which b is odd and c is even.


The number of terms for which b is even and c is odd is the same. Thus thenumber of terms in the simplified expression is

2008) -2 (1004) = 1004.2007 - 1004.1003 = 10042 = 1,008,016.2 2

25. ANSWER (E) For 1 < k < 15, the k-element sets with properties (1) and(2) are the k-element subsets of U k = {k, k + 1, ... , 15} that contain notwo consecutive integers. If {a 1 , a2, ... , ak } is such a set, with its elementslisted in increasing order, then {a 1 + k -1, a2 + k - 2, ... , ak_ 1 + 1, ak } isa k-element subset of U2k_1. Conversely, if {b1, b2, ... , bk } is a k-elementsubset of U2k_1, with its elements listed in increasing order, then {b1-k +1, b2 -k +2,... , bk_1-1, bk } is a set with properties (1) and (2). Thus foreach k, the number of k-element sets with properties (1) and (2) is equalto the number of k-element subsets of the (17 - 2k)-element set U2k_1.Because k < 17 - 2k only if k < 5, the total number of such sets is

kE

( 17 k 2k/

115) \ 2/ \31/ \4/ \5/

= 15+78+ 165+ 126+21 = 405.

....

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168


Solutions


1 3.61 3.11 79.75 8.46 2.97 2.09

2 77.14 3.98 0.62 0.9 2.71 14.65

3 85.29 1.93 0.25 8.4 2.57 1.55

4 91.62 2.88 1.01 2.16 0.48 1.84

5 71.8 7.54 2.91 2.11 1.67 13.97

6 0.7 87.24 1.02 0.77 5.6 4.66

7 4.52 83.47 3.68 3.57 1.14 3.6

8 1.55 2.49 2.15 3.73 70.45 19.59

9 4.36 45.94 3.42 7.94 4.22 34.11

10 35.87 5.19 11.28 1.36 24.55 21.72

11 6.85 1.39 12.15 8.28 46.55 24.77

12 6.1 5.09 4.15 12.38 5.66 66.61

13 5.86 4.89 28.01 2.21 4.06 54.96

14 2.07 4.71 18.92 21.26 1.12 51.91

15 2.12 23.04 8.03 3.41 1.95 61.43

16 2.12 2.08 12.67 2.32 1.94 78.87

17 6 2.19 15.14 13.37 3.87 59.43

18 4.63 8.13 11.11 3.14 10.9 71.09

19 1.36 21.98 1.52 13.56 2.91 58.65

20 1.69 2.68 2.81 1.49 5.57 85.75

21 2.2 1.37 2.31 1.14 1.26 91.72

22 1.4 3.3 2.8 1.42 0.74 90.33

23 1.83 2.15 3.76 2.36 1.58 88.31

24 1.67 1.47 1.71 1.87 0.96 92.33

25 1.04 2.19 2.05 2.78 0.96 90.97

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2006 AMC 1 2B SolutionsThe problems for the 2006 AMC 12B begin on page 53.

1. ANSWER (C) Because

k 1, if k is even,

(-1) -1, if k is odd,

the sum can be written as

2. ANSWER (A) Because 4* 5 = (4 + 5)(4 - 5) = -9, it follows that

3 6 (4 6 5) = 3 6 (-9) = (3 + (-9)) (3 - (-9)) = (-6)(12) = -72.

3. ANSWER (A) Let c and p represent the number of points scored by theCougars and the Panthers, respectively. The two teams scored a total of 34points, so c + p = 34. The Cougars won by 14 points, so c - p = 14. Thesolution is c = 24 and p = 10, so the Panthers scored 10 points.

4. ANSWER (A) The five items cost approximately 8 + 5 + 3 + 2 + 1 =19 dollars, so Mary's change is about $1.00, which is 5 percent of her$20.00.

5. ANSWER (A) In order to catch up to John, Bob must walk 1 mile fartherin the same amount of time. Because Bob's speed exceeds John's speed by5 - 3 = 2 miles per hour, the time required for Bob to catch up to John is1/2 hour, or 30 minutes.

6. ANSWER (B) Francesca's 600 grams of lemonade contains 25 + 386 =411 calories, so 200 grams of her lemonade contains 411/3 = 137 calories.

7. ANSWER (B) There are only two possible occupants for the driver's seat.After the driver is chosen, any of the remaining three people can sit in thefront, and there are two arrangements for the other two people in the back.Thus there are 2. 3 2 = 12 possible seating arrangements.

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Aim

KID

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170 Solutions

8. ANSWER (E) Substituting x = 1 and y = 2 into the equations gives

2 11=4+a and 2=j+b.It follows that

a+b I 1-41+I2-41 =3-4 = 4.

OR

Because

x 3a = x -4

and b=y--, we have a + b = 4 (x + y).4

Since x = 1 when y = 2, this implies that a + b =a

(1 + 2) = 2

9. ANSWER (B) Let the integer have digits a, b, and c, read left to right.Because 1 < a < b < c, none of the digits can be zero and c cannotbe 2. If c = 4, then a and b must each be chosen from the digits 1, 2,and 3. Therefore there are (3) = 3 choices for a and b, and for each choicethere is one acceptable order. Similarly, for c = 6 and c = 8 there are,respectively, (5) = 10 and (z) = 21 choices for a and b. Thus there arealtogether 3 + 10 + 21 = 34 such integers.

10. ANSWER (A) The sides of the triangle are x, 3x, and 15 for some posi-tive integer x. By the Triangle Inequality, these three numbers are the sidesof a triangle if and only if x + 3x > 15 and x + 15 > 3x. Because x isan integer, the first inequality is equivalent to x > 4, and the second in-equality is equivalent to x < 7. Thus the greatest possible perimeter is7 + 21 + 15 = 43.

11. ANSWER (E) Joe has 2 ounces of cream in his cup. JoAnn has drunk2 ounces of the 14 ounces of coffee-cream mixture in her cup, so she hasonly 12/14 = 6/7 of her 2 ounces of cream in her cup. Therefore the ratioof the amount of cream in Joe's coffee to that in JoAnn's coffee is

2 _ 7

2 6


12. ANSWER (D) A parabola with the given equation and with vertex (p, p)must have equation y = a (x - p)2 + p. Because the y-intercept is (0, -p)and p 0, it follows that a = -2/p. Thus

y=-?(x2-2px+p2)+p=_2x2+4x-p,P P

sob=4.

13. ANSWER (C) Since LBAD = 60°, isosceles ABAD is also equilat-eral. As a consequence, AAEB, AAED, ABED, ABFD, ABFC, andACFD are congruent. These six triangles have equal areas and their unionforms rhombus ABCD, so each has area 24/6 = 4. Rhombus BFDE isthe union of ABED and ABFD, so its area is 8.

B

OR

Let the diagonals of rhombus ABCD intersect at O. Since the diagonalsof a rhombus intersect at right angles, AABO is a 30 - 60 - 90 triangle.Therefore AO = Nf3- BO. Because A O and BO are half the length of thelonger diagonals of rhombi ABCD and BFDE, respectively, it followsthat

Area(BFDE) _ BO 2-

1

Area(ABCD) - k AO ) 3

Thus the area of rhombus BFDE is (1/3)(24) = 8.

'C7

172 Solutions

14. ANSWER (D) The total cost of the peanut butter and jam is N(4B +5J) = 253 cents, so N and 4B + 5J are factors of 253 = 11 .23. BecauseN > 1, the possible values of N are 11, 23, and 253. If N = 253, then4B + 5J = 1, which is impossible since B and J are positive integers.If N = 23, then 4B + 5J = 11, which also has no solutions in positiveintegers. Hence N = 11 and 4B + 5J = 23, which has the unique positiveinteger solution B = 2 and J = 3. So the cost of the jam is 11(3)(5¢) _$1.65.

15. ANSWER (B) Through 0 draw a line parallel to AD intersecting PDat F.

Then AOFD is a rectangle and OPF is a right triangle. Thus DF = 2,FP = 2, and OF = 4,/2. The area of trapezoid AOPD is 12,/2, and thearea of hexagon AOBCPD is 2. 12,/ = 24'.

OR

Lines AD, BC, and OP intersect at a common point H.

Because LPDH = LOAH = 90°, triangles PDH and OAH are similarwith ratio of similarity 2. Thus 2HO = HP = HO + OP = HO + 6, so

\^J

NIA


HO = 6 and AH = HO2 - OA2 = 4V. Hence the area of AOAHis (l/2)(2)(4h) = 4,vf2-, and the area of A P D H is (22)(4,) = 16/.The area of the hexagon is twice the area of APDH minus twice the areaof AOAH, so it is 24/.

16. ANSWER (C) Diagonals AC, CE, EA, AD, CF, and EB divide thehexagon into twelve congruent 30-60-90° triangles, six of which make upequilateral AACE. Because AC = 72 + 12 = 50, the area of AACE

2is a (_150) = 2 /J. The area of hexagon ABCDEF is 2 (Zs ) _

25/. \ J

OR

Let 0 be the center of the hexagon. Then triangles A BC, CDE, and EFAare congruent to triangles AOC, COE, and EOA, respectively. Thus thearea of the hexagon is twice the area of equilateral AACE. Then proceedas in the first solution.

17. ANSWER (C) On each die the probability of rolling k, for 1 < k < 6, is

k _ k

1+2+3+4+5+6 21

There are six ways of rolling a total of 7 on the two dice, represented bythe ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). Thus theprobability of rolling a total of 7 is

1 6+2.5+3.4+4.3+5.2+6.1 56 8

212 212 63

18. ANSWER (B) Each step changes either the x-coordinate or the y-coordi-nate of the object by 1. Thus if the object's final point is (a, b), then a + bis even and la I + Jhi < 10. Conversely, suppose that (a, b) is a lattice pointwith lal + Jhi = 2k < 10. One ten-step path that ends at (a,b) beginswith Ja I horizontal steps, to the right if a > 0 and to the left if a < 0. Itcontinues with IbI vertical steps, up if b > 0 and down if b < 0. It hasthen reached (a, b) in 2k steps, so it can finish with 5 - k steps up and5 - k steps down. Thus the possible final points are the lattice points thathave even coordinate sums and lie on or inside the square with vertices(± 10, 0) and (0, ± 10). There are 11 such points on each of the 11 linesx + y = 2k, -5 < k < 5, for a total of 121 different points.

.-a)

V-.

NO'

174 Solutions

19. ANSWER (B) The 4-digit number on the license plate has the form aabbor abab or baab, where a and b are distinct integers from 0 to 9. BecauseMr. Jones has a child of age 9, the number on the license plate is divisibleby 9. Hence the sum of the digits, 2(a+b), is also divisible by 9. Because ofthe restriction on the digits a and b, this implies that a + b = 9. Moreover,since Mr. Jones must have either a 4-year-old or an 8-year-old, the licenseplate number is divisible by 4. These conditions narrow the possibilitiesfor the number to 1188, 2772, 3636, 5544, 6336, 7272, and 9900. The lasttwo digits of 9900 could not yield Mr. Jones's age, and none of the othersis divisible by 5, so he does not have a 5-year-old.

Note that 5544 is divisible by each of the other eight non-zero digits.

20. ANSWER (C) The given condition is equivalent to [log10 xj = [log10 4xj.Thus the condition holds if and only if

n <loglox <loglo4x <n+

for some negative integer n. Equivalently,

10"<x<4x<10n+1

This inequality is true if and only if

ion+110n<x<4

1

Hence in each interval [10n, lon+1) the given condition holds with prob-ability

(ton+1/4) - Ion _ Ion((10/4) - 1) _ 1

lon+1 - 1o" 10"(10-1) 6

Because each number in (0, 1) belongs to a unique interval ion, Ion+1)

and the probability is the same on each interval, the required probability isalso 1/6.

21. ANSWER (C) Let 2a and 2b, respectively, be the lengths of the majorand minor axes of the ellipse, and let the dimensions of the rectangle bex and y. Then x + y is the sum of the distances from the foci to pointA on the ellipse, which is 2a. The length of a diagonal of the rectangleis the distance between the foci of the ellipse, which is 2 a2 - b2. Thusx + y = 2a and x2 + y2 = 4a2 - 4b2. The area of the rectangle is

2006 = xy = 2 [(x + y)2 - (x2 + y2)] = 2 [(2a)2 - (4a2 -4b 2)] = 2b2,

WIN 5

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y_,

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aryl


so b = 1003. Thus the area of the ellipse is

2006ir = crab = Ira 1003,

175

so a = 2 1003, and the perimeter of the rectangle is 2(x + y) = 4a =8 1003.

22. ANSWER (B) Note that n is the number of factors of 5 in the producta!b!c!, and 2006 < 55. Thus

4

n = E (La/Sk] + Lb/Ski + Lc/5ki) .

k=1

Because LxJ + Lyi + Lzi > Lx + y + zJ - 2 for all real numbers x, y,and z, it follows that

4

n > (L(a + b + c)/5kj - 2)

k=1

4

_ Y (L2oo6/Skj - 2)

k=i

=401+80+16+3-4.2=492.

The minimum value of 492 is achieved, for example, when a = b = 624and c = 758.

23. ANSWER (E) Let D, E, and F be the reflections of P about AB, BC,and CA, respectively. Then ZFAD = LDBE = 90°, and LECF =180°. Thus the area of pentagon ADBEF is twice that of AABC, so it iss2.

0.11

176 Solutions

Observe that DE = 7V, EF = 12, and FD = 11,/2. Furthermore,(7,)2 + 122 = 98 + 144 = 242 = (11)2, so ADEF is a righttriangle. Thus the pentagon can be tiled with three right triangles, two ofwhich are isosceles, as shown.

It follows that

7,12

so a + b = 127.

OR

Rotate AABC 90° counterclockwise about C, and let B' and P be theimages of B and P, respectively.

Then CP' = CP = 6, and LPCP' = 90°, so APCP' is an isosce-les right triangle. Thus PP' = 6V, and BP' = AP = 11. Because(6' ) + 72 = 112, the converse of the Pythagorean Theorem implies

that LBPP' = 90°. Hence LBPC = 135°. Applying the Law of Cosinesin A BP C gives

BC2 = 62 +72 85+42n,

and a + b = 127.

WIC

4..

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24. ANSWER (C) For a fixed value of y, the values of sin x for whichsin2 x - sin x sin y + sin2 y = 4 can be determined by the quadratic for-mula. Namely,

sin x =

sin y ± Vsin2 y - 4(sin2 y - 4)

2= 1 sin y ± cos y.

2 2

Because cos (3) =2

and sin (3) = 2 , this implies that

sinx = cos(3)siny ±sin(3)cosy =sin Y± 3

Within S, sin x = sin(y - 11) implies x = y - 3 . However, the casesin x = sin(y + 3) implies x = y + 3 when y < 6 , and x = -y + 3when y > 6 . Those three lines divide the region S into four subregions,within each of which the truth value of the inequality is constant. Testingthe points (0, 0), (2 , 0), (0,

2), and (2 , 2) shows that the inequality is true

only in the shaded subregion. The area of this subregion is

2 2 3 2 6()2 ()2)2 7r2

6

It3

25. ANSWER (B) The condition l a,+1 - a, l implies that a,, anda,+3 have the same parity for all n > 1. Because a2006 is odd, a2 is alsoodd. Because a2006 = 1 and a, is a multiple of gcd(a 1, a2) for all n,it follows that 1 = gcd(a 1, a2) = gcd(33 37, a2). There are 499 oddintegers in the interval [ 1, 998], of which 166 are multiples of 3, 13 are

C3.

178 Solutions

multiples of 37, and 4 are multiples of 3 37 = 111. By the Inclusion-Exclusion Principle, the number of possible values of a2 cannot exceed499-166-13+4=324.To see that there are actually 324 possibilities, note that for n > 3, a <max(an_2, ai_,) whenever and are both positive. Thus aN = 0for some N < 1999. If gcd(al,a2) = 1, then aN_2 = aN_1 = 1, and forn > N the sequence cycles through the values 1, 1, 0. If in addition a2 isodd, then a3k+2 is odd fork > 1, so a2006 = I

.-.

.-.

.-.

.-.

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2007 AMC 12A Solutions


179


1 0.33 1.45 95.55 1.69 0.58 0.39

2 2.96 2.65 2.28 75.9 3.26 12.93

3 89.23 1.02 3.75 0.86 0.91 4.21

4 75.41 1.86 12.43 2.63 3.6 4.04

5 0.5 1.33 14.74 76.88 0.73 5.81

6 7.61 3.09 4.49 68.37 2.43 13.98

7 30.42 1.13 32.31 0.57 8.52 26.98

8 2.76 4.5 43.69 14.03 13.76 21.23

9 4.89 40.13 2.63 2.53 14.83 34.95

10 32.11 8.9 3.44 5.3 3 47.22

11 10.33 4.86 6.23 25.54 1.36 51.66

12 6.24 5.16 19.08 7.19 15.23 47.07

13 7.28 41.11 15.32 8.35 4.72 23.18

14 5.95 18.86 29.52 3.16 2.12 40.37

15 10.97 4.99 14.77 5.26 31.3 32.67

16 11.7 5.22 9.17 8.52 3.52 61.85

17 4.64 11.21 3.03 4.05 3.9 73.14

18 4.43 2.63 4.14 8.73 2.81 77.24

19 3.34 3.05 4.16 3.88 3.86 81.69

20 2 5.17 3.44 3.99 4.07 81.3

21 6.14 4.97 8.29 6.24 5.74 68.59

22 7.56 6.1 7.8 4.3 1.43 72.79

23 4.93 3.1 5.71 3.48 3.63 79.13

24 1.72 3.51 3.22 2.07 2.25 87.19

25 4.7 4.2 3.2 3.36 235 81.97

-..

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180 Solutions


1. ANSWER (C) Susan pays (4)(0.75)(20) = 60 dollars. Pam pays (5)(0.70)(20) = 70 dollars, so she pays 70 - 60 = 10 more dollars than Susan.

2. ANSWER (D) The brick has a volume of 40 20 10 = 8000 cubiccentimeters. Suppose that after the brick is placed in the tank, the waterlevel rises by h centimeters. Then the additional volume occupied in theaquarium is 100.40 h = 4000h cubic centimeters. Since this must be thesame as the volume of the brick, we have

8000 = 4000h and h = 2 centimeters

3. ANSWER (A) Let the smaller of the integers be x. Then the larger isx + 2. So x + 2 = 3x, from which x = 1. Thus the two integers are 1 and3, and their sum is 4.

4. ANSWER (A) Kate rode for 30 minutes = 1/2 hour at 16 mph, so sherode 8 miles. She walked for 90 minutes = 3/2 hours at 4 mph, so shewalked 6 miles. Therefore she covered a total of 14 miles in 2 hours, so heraverage speed was 7 mph.

5. ANSWER (D) After paying the federal taxes, Mr. Public had 80% of hisinheritance money left. He paid 10% of that, or 8% of his inheritance, instate taxes. Hence his total tax bill was 28% of his inheritance, and hisinheritance was $10,500/0.28 = $37,500.

6. ANSWER (D) Because AABC is isosceles,

LBAC =2

(180°- /ABC) = 70°.

+S

Fr

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boo

coo

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Similarly,

ZDAC = 2 (180° - /ADC) = 20°.

Thus LBAD = LBAC - LDAC = 50°.

OR

Because A A B C and A A D C are isosceles triangles, applying the ExteriorAngle Theorem to AABD gives LBAD = 70° - 20° = 50°.

7. ANSWER (C) Let D be the difference between consecutive terms of thesequence. Then a = c - 2D, b = c - D, d = c+ D, and e = c + 2D, so

a+b+c+d+e = (c-2D)+(c-D)+c+(c+D)+(c+2D) = 5c.

Thus 5c = 30, so c = 6.

To see that the values of the other terms cannot be found, note that thesequences 4, 5, 6, 7, 8 and 10, 8, 6, 4, 2 both satisfy the given conditions.

8. ANSWER (C) Consider the two chords with an endpoint at 5. The arcsubtended by the angle determined by these chords extends from 10 to 12,so the degree measure of the arc is (2/12)(360) = 60. By the CentralAngle Theorem, the degree measure of this angle is (1/2)(60) = 30. Bysymmetry, the degree measure of the angle at each vertex is 30.

9. ANSWER (B) Let w be Yan's walking speed, and let x and y be thedistances from Yan to his home and to the stadium, respectively. The timerequired for Yan to walk to the stadium is y/w, and the time required forhim to walk home is x/w. Because he rides his bicycle at a speed of 7w,the time required for him to ride his bicycle from his home to the stadiumis (x + y)/(7w). Thus

y x +x+y - 8x+yW w 7w 7w

As a consequence, 7y = 8x + y, so 8x = 6y. The required ratio is x/y =6/8 = 3/4.

OR

Because we are interested only in the ratio of the distances, we may assumethat the distance from Yan's home to the stadium is 1 mile. Let x be his

(IQ

...

182 Solutions

present distance from his home. Imagine that Yan has a twin, Nay. WhileYan walks to the stadium, Nay walks to their home and continues 1/7 ofa mile past their home. Because walking 1/7 of a mile requires the sameamount of time as riding 1 mile, Yan and Nay will complete their trips atthe same time. Yan has walked 1 - x miles while Nay has walked x + 1miles, so 1 - x = x + 1. Thus x = 3/7, 1 - x = 4/7, and the requiredratio is x/(1 - x) = 3/4.

10. ANSWER (A) Let the sides of the triangle have lengths 3x, 4x, and 5x.The triangle is a right triangle, so its hypotenuse is a diameter of the circle.Thus 5x = 2 3 = 6, so x = 6/5. The area of the triangle is

1 1 18 24 216-. 3x 4x=

_2 5 5 25

= 8.64.2

OR

A right triangle with side lengths 3, 4, and 5 has area (1/2)(3)(4) = 6.Because the given right triangle is inscribed in a circle with diameter 6, thehypotenuse of this triangle has length 6. Thus the sides of the given triangleare 6/5 as long as those of a 3-4-5 triangle, and its area is (6/5)2 timesthat of a 3-4-5 triangle. The area of the given triangle is

(6 2 216= 8.64.5 (6) =

25

11. ANSWER (D) A given digit appears as the hundreds digit, the tens digit,and the units digit of a term the same number of times. Let k be the sumof the units digits in all the terms. Then S = 11 lk = 3 37k, so S mustbe divisible by 37. To see that S need not be divisible by any larger prime,note that the sequence 123, 231, 312 gives S = 666 = 2. 32 - 37.

12. ANSWER (E) The number ad - be is even if and only if ad and be areboth odd or are both even. Each of ad and be is odd if both of its factorsare odd, and even otherwise. Exactly half of the integers from 0 to 2007are odd, so each of ad and be is odd with probability (1/2) (1/2) = 1/4and are even with probability 3/4. Hence the probability that ad - be iseven is

1 1 3 3_ 54 4+4 4 8

v'.


13. ANSWER (B) The point (a, b) is the foot of the perpendicular from(12, 10) to the line y = -5x + 18. The perpendicular has slope 5, soits equation is

1 1 38y=10+5(x-12)=5x+ 5.The x-coordinate at the foot of the perpendicular satisfies the equation

1 385 x + 5 = -5x + 18,

sox=2andy=-5.2+18=8.Thus(a, b)=(2,8),anda+b= 10.

OR

If the mouse is at (x, y) = (x, 18 - 5x), then the square of the distancefrom the mouse to the cheese is

(x-12)2+(8-5x)2=26(x2-4x+8) =26((x-2)2+4).

The value of this expression is smallest when x = 2, so the mouse isclosest to the cheese at the point (2, 8), and a + b = 2 + 8 = 10.

14. ANSWER (C) If 45 is expressed as a product of five distinct integer fac-tors, the absolute value of the product of any four is at least 1(-3)(-1)(1)(3)I = 9, so no factor can have an absolute value greater than 5. Thusthe factors of the given expression are five of the integers ±1, ±3, and ±5.The product of all six of these is -225 = (-5)(45), so the factors are -3,-1, 1, 3, and 5. The corresponding values of a, b, c, d, and e are 9, 7, 5, 3,and 1, and their sum is 25.

15. ANSWER (E) The mean of the augmented set is (28 + n)/5. If n < 6,the median of that set is 6, so 28 + n = 5 6, and n = 2. If 6 < n < 9,the median is n, so 28 + n = 5n, and n = 7. If n > 9, the median is 9,so 28 + n = 5 9, and n = 17. Thus the sum of all possible values of n is2 + 7 + 17 = 26.

16. ANSWER (C) The set of the three digits of such a number can be arrangedto form an increasing arithmetic sequence. There are 8 possible sequenceswith a common difference of 1, since the first term can be any of the digits0 through 7. There are 6 possible sequences with a common difference of2, 4 with a common difference of 3, and 2 with a common difference of 4.

sue..

184 Solutions

Hence there are 20 possible arithmetic sequences. Each of the 4 sets thatcontain 0 can be arranged to form 2 2! = 4 different numbers, and the 16sets that do not contain 0 can be arranged to form 3! = 6 different numbers.Thus there are a total of 4 4 + 16 6 = 112 numbers with the requiredproperties.

17. ANSWER (B) Square both sides of both given equations to obtain

sine a+2 sin a sin b+sin2 b = 5/3 and cost a+2 cos a cos b+cos2 b = 1.

Then add corresponding sides of the resulting equations to obtain

(sin2 a + cos2 a) + (sin2 b + cos2 b) + 2(sin a sin b + cos a cos b) =8

3

Because sin2 a + cos2 a = sin2 b + cos2 b = 1, it follows that

cos(a - b) = sin a sin b + cos a cos b = .3

One ordered pair (a, b) that satisfies the given condition is approximately(0.296, 1.527).

18. ANSWER (D) Because f(x) has real coefficients and 2i and 2 + i arezeros, so are their conjugates -2i and 2 - i. Therefore

f(x) = (x + 2i) (x - 2i) (x - (2 + i))(x - (2 - i))

=(x2+4)(x2-4x+5)=x4-4x3+9x2-16x+20.

Hencea+b+c+d =-4+9- 16+20= 9.

OR

As in the first solution,

f(x)=(x+2i)(x-2i)(x-(2+i))(x-(2-i)),so

a + b + c + d = f(1) - 1 = (1 + 2i)(1 - 2i)(-1 - i)(-l + i) - 1=(1+4)(1+1)-1 =9.

tin


19. ANSWER (E) Let h be the length of the altitude from A in AABC. Then

2007=-.BC

h = 18. Thus A is on one of the lines y = 18 or y = -18. Line DE hasequation x - y - 300 = 0. Let A have coordinates (a, b). By the formulafor the distance from a point to a line, the distance from A to line DE isla - b - 300J/1h. The area of AADE is

i7002 -

la-b-3001 . DE - Ija ± 18 - 3001.9

.

Thus a = ± 18 ± 1556 + 300, and the sum of the four possible values ofais4.300= 1200.

OR

As above, conclude that A is on one of the lines y = ±18. By similarreasoning, A is on one of two particular lines 11 and 12 parallel to DE.Therefore there are four possible positions for A, determined by the inter-sections of the lines y = 18 and y = -18 with each of l1 and 12. Let theline y = 18 intersect 11 and 12 in points (xi, yi) and (x2, y2), and let theline y = -18 intersect 11 and 12 in points (x3, y3) and (x4, y4). The fourpoints of intersection are the vertices of a parallelogram, and the center ofthe parallelogram has x-coordinate (1/4) (xi + x2 + x3 + x4). The cen-ter is the intersection of the line y = 0 and line DE. Because line DE hasequation y = x - 300, the center of the parallelogram is (300, 0). Thus thesum of all possible x-coordinates of A is 4. 300 = 1200.

20. ANSWER (B) Removing the corners removes two segments of equallength from each edge of the cube. Call that length x. Then each octagonhas side length 2_x, and the cube has edge length 1 = (2 + 1/2-)x, so

1 2-,/2-X _2+/2- 2

Each removed comer is a tetrahedron whose altitude is x and whose baseis an isosceles right triangle with leg length x. Thus the total volume of theeight tetrahedra is

l 1 2 1 3 10-71/23 2 6 3

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186 Solutions

21. ANSWER (A) The product of the zeros off is c/a, and the sum of thezeros is -b/a. Because these two numbers are equal, c = -b, and the sumof the coefficients is a + b + c = a, which is the coefficient of x2. To seethat none of the other choices is correct, let f (x) = -2x2 - 4x + 4. Thezeros of f are -1 + /3-, so the sum of the zeros, the product of the zeros,and the sum of the coefficients are all -2. However, the coefficient of x is-4, the y-intercept is 4, the x-intercepts are -1 ± /3-, and the mean of thex-intercepts is -1.

22. ANSWER (D) If n < 2007, then S(n) < S(1999) = 28. If n < 28, thenS(n) < S(28) = 10. Therefore if n satisfies the required condition it mustalso satisfy

n > 2007 - 28 - 10 = 1969.

In addition, n, S(n), and S(S(n)) all leave the same remainder when di-vided by 9. Because 2007 is a multiple of 9, it follows that n, S(n), andS(S(n)) must all be multiples of 3. The required condition is satisfied by4 multiples of 3 between 1969 and 2007, namely 1977, 1980, 1983, and2001.

Note: There appear to be many cases to check, that is, all the multiplesof 3 between 1969 and 2007. However, for 1987 < n < 1999, we haven + S(n) > 1990 + 19 = 2009, so these numbers are eliminated. Thus weneed only check 1971, 1974, 1977, 1980, 1983, 1986, 2001, and 2004.

23. ANSWER (A) Let A = (p, log,, p) and B = (q, 2log, q). Then AB =6 = p - q 1. Because A _B is horizontal, log, p = 2 log,, q = log, q2, sop = q2. Thus Iq2 - qI = 6, and the only positive solution is q = 3. Notethat C = (q, 3 log, q), so BC = 6 = log,, q, from which a6 = q = 3 anda =.

24. ANSWER (D) Note that F(n) is the number of points at which the graphsof y = sin x and y = sin nx intersect on [0, ir]. For each n, sin nx > 0on each interval [(2k - 2)ir/n, (2k - 1)7r/n] where k is a positive integerand 2k - 1 < n. The number of such intervals is n/2 if n is even and(n + 1)/2 if n is odd. The graphs intersect twice on each interval unlesssin x = 1 = sin nx at some point in the interval, in which case the graphsintersect once. This last equation is satisfied if and only if n ° 1 (mod 4)and the interval contains -r/2. If n is even, this count does not include thepoint of intersection at (ir, 0). Therefore F(n) = 2(n/2) + I = n + 1 if n


is even, F(n) = 2(n + 1)/2 = n + 1 if n - 3 (mod 4), and F(n) = n ifn - 1 (mod 4). Hence

2007 200 _ 2007 - 1F(n) _ >2 (n + 1)

4 Jn=2 n=2

(2006)(3 + 2008)_ 501 = 2,016,532.

2

25. ANSWER (E) For each positive integer n, let Sn = {k : 1 < k < n},and let en be the number of spacy subsets of Sn. Then cl = 2, c2 = 3,and c3 = 4. For n > 4, the spacy subsets of S, can be partitioned into twotypes: those that contain n and those that do not. Those that do not contain nare precisely the spacy subsets of Sn_1. Those that contain n do not containeither n - 1 or n - 2 and are therefore in one-to-one correspondence withthe spacy subsets of Sn_3. It follows that cn = cn_3 + cn_1. Thus the firsttwelve terms in the sequence (cn) are 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88,129, and there are c12 = 129 spacy subsets of S12.

OR

Note that each spacy subset of S12 contains at most 4 elements. For eachsuch subset a1, a2, ... , ak, let b1 = a1 - 1, bj = aj - aj_1 - 3 for2 < j < k, and bk+1 = 12 - ak. Then b; > 0 for 1 < j < k + 1, and

is 14-2k) for 0 < k < 4,The number of solutions for (bl, b2, ... , bk+1) ( k

so the number of spacy subsets of S12 is

14) +(12)1

+( 0)+(3) +(4) = 1+12+45+56+15= 129.

..-

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v"1

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188


Solutions


1 0.67 1.19 2.71 2.99 85.66 6.78

2 0.44 48.79 47.05 0.62 0.25 2.84

3 1.26 2.83 1.49 75.18 4.02 15.2

4 1.3 91.51 1.27 2.98 1.58 1.36

5 0.17 1.31 2.66 84.55 10.69 0.58

6 0.52 2.09 6.56 78.56 3.09 9.18

7 18.28 2.64 15.65 2.4 42.54 18.49

8 10.89 4.93 3.71 44.29 4.08 32.08

9 50.82 1.42 14.83 0.61 20.03 12.25

10 1.19 3.14 78.85 4.83 3.85 8.14

11 1.95 9.67 1.17 62.8 2.21 22.2

12 14.4 2.57 72.59 2.09 0.58 7.76

13 4.11 40.81 6.24 32.03 2.38 14.43

14 3.59 6.41 5.75 15.75 1.33 67.15

15 3.01 2.98 2.62 4.83 10.54 76

16 5.61 6.19 5.56 3.63 1.73 77.27

17 2.56 9.98 11.43 10.5 10.05 55.46

18 1.76 2.87 27.05 3.24 1.22 63.83

19 8.84 3.78 4.49 3.26 8.01 71.6

20 1.86 4.46 2.76 3.66 1.22 86.03

21 5.47 9.82 3.63 3.11 1.79 76.16

22 7.44 3.49 4.41 6.15 11.64 66.85

23 4.76 3.57 3.11 1.65 1.36 85.54

24 2.29 1.94 1.67 1.16 13.78 79.15

25 2.84 4.05 9.07 2.85 3.34 77.84

V'1



1. ANSWER (E) The perimeter of each bedroom is 2(12 + 10) = 44 feet, sothe surface to be painted in each bedroom has an area of 44.8 - 60 = 292square feet. Since there are 3 bedrooms, Isabella must paint 3 292 =square feet.

876

2. ANSWER (B) The student used 120/30 = 4 gallons on the trip home and120/20 = 6 gallons on the trip back to school. So the average gas mileagefor the round trip was

240 miles

10 gallons= 24 miles per gallon.

3. ANSWER (D) Since OA = OB = OC, triangles AOB, BOC, and COAare all isosceles. Hence

LABC = LABO + LOBC =180°

2140° + 1800

2120° = 50°.

Since

OR

LAOC = 360° - 140° - 1200 = 100°,

the Central Angle Theorem implies that

LABC = 1LAOC = 50°.2

4. ANSWER (B) Because 3 bananas cost as much as 2 apples, 18 bananascost as much as 12 apples. Because 6 apples cost as much as 4 oranges, 12apples cost as much as 8 oranges. Therefore 18 bananas cost as much as 8oranges.

5. ANSWER (D) Sarah will receive 4.5 points for the three questions sheleaves unanswered, so she must earn at least 100 - 4.5 = 95.5 points onthe first 22 problems. Because

95.515 < 6 <16,

v,"

CA

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L1.

III

190 Solutions

she must solve at least 16 of the first 22 problems correctly. This wouldgive her a score of 100.5.

6. ANSWER (D) The perimeter of the triangle is 5 + 6 + 7 = 18, sothe distance that each bug crawls is 9. Therefore A B + BD = 9, andBD = 4.

7. ANSWER (E) Because AB = BC = EA and LA = LB = 90°,quadrilateral ABCE is a square, so LAEC = 90°.

Also CD = DE = EC, so LCDE is equilateral and LCED = 60°.Therefore

LE = LAEC + LCED = 90° + 60° = 150°.

8. ANSWER (D) Tom's age N years ago was T - N. The sum of his threechildren's ages at that time was T - 3N. Therefore T - N = 2(T - 3N),so 5N = T and T/N = 5. The conditions of the problem can be met, forexample, if Tom's age is 30 and the ages of his children are 9, 10, and 11.In that case T = 30 and N = 6.

9. ANSWER (A) Let u = 3x - 1. Then x = (u + 1)/3, and

u+1 2 u+1 u2+2u+1 a+1f(u)( 3 ) + 3 +1= 9 + 3 +1

u2+5u+139

In particular,52+5.5+13 63

M)9 9

+


OR

The value of 3x - 1 is 5 when x = 2. Thus

f(5)=f(3.2-1)=22+2+1=7.

10. ANSWER (C) Let g be the number of girls and b the number of boysinitially in the group. Then g = 0.4(g + b). After two girls leave andtwo boys arrive, the size of the entire group is unchanged, so g - 2 =0.3(g + b). The solution of the system of equations

g=0.4(g+b) and g-2=0.3(g+b)

is g = 8 and b = 12, so there were initially 8 girls.

OR

After two girls leave and two boys arrive, the size of the group is un-changed. So the two girls who left represent 40% - 30% = 10% of thegroup. Thus the size of the group is 20, and the original number of girlswas 40% of 20, or 8.

11. ANSWER (D) Let x be the degree measure of LA. Then the degree mea-sures of angles B, C, and D are x/2, x/3, and x/4, respectively. Thedegree measures of the four angles have a sum of 360, so

x x x 25x360=x+++ _2

34 12

Thus x = (12 .360)/25 = 172.8 .: 173.

12. ANSWER (C) Let N be the number of students in the class. Then thereare 0.1 N juniors and 0.9N seniors. Let s be the score of each junior. Thescores totaled 84N = 83 (0.9N) + s (0.1 N), so

84N - 83(0.9N) -s= =93.OA N

Note: In this problem, we could assume that the class has one junior andnine seniors. Then

9-83+s = 10-84= 9.84+84 and s=9(84-83)+84=93.9(84-83)

'-C

V'1

192 Solutions

13. ANSWER (D) The light completes a cycle every 63 seconds. Leah sees thecolor change if and only if she begins to look within three seconds beforethe change from green to yellow, from yellow to red, or from red to green.Thus she sees the color change with probability (3 + 3 + 3)/63 = 1/7.

14. ANSWER (D) Let the side length of AABC be s. Then the areas ofAAPB, ABPC, and ACPA are, respectively, s/2, s, and 3s/2. The areaof AABC is the sum of these, which is 3s. The area of AABC may also beexpressed as (V/4)s2, so 3s = (.//4)s2. The unique positive solutionfor s is 4..

15. ANSWER (E) The terms involving odd powers of r form the geometricseries

a

a1-r

r = 3/4. It follows that a/(1/4) = 7, so a = 7/4 and

7 3 5a+r=4+4=2.

ORThe sum of the terms involving even powers of r is 7 - 3 = 4. Therefore

3=ar+ar3+ar5+

so r = 3/4. As in the first solution, a = 7/4 and a + r = 5/2.

16. ANSWER (A) Let r, w, and b be the number of red, white, and bluefaces, respectively. Then (r, w, b) is one of 15 possible ordered triples,namely one of the three permutations of (4, 0, 0), (2, 2, 0), or (2, 1, 1), orone of the six permutations of (3, 1, 0). The number of distinguishable col-orings for each of these ordered triples is the same as for any of its per-mutations. If (r, w, b) = (4, 0, 0), then exactly one coloring is possible.If (r, w, b) = (3, 1, 0), the tetrahedron can be placed with the white face

CA

D

^.a

-0n0

f")

WIC

G1,

s.:.-.

c},

r'+

..a


down. If (r, w, h) = (2, 2, 0), the tetrahedron can be placed with one whiteface down and the other facing forward. If (r, w, b) = (2, 1, 1), the tetra-hedron can be placed with the white face down and the blue face forward.Therefore there is only one coloring for each ordered triple, and the totalnumber of distinguishable colorings is 15.

17. ANSWER (D) Because b < job for all b > 0, it follows that loglo b < b.If b > 1, then 0 < (log10 b) /b2 < 1, so a cannot be an integer. Therefore0 < b < 1,solog10b <0anda = (log10b)/b2 <0.Thus a <0 <b <1 < 1/b, and the median of the set is b.

Note that the conditions of the problem can be met with b = 0.1 anda = -100.

18. ANSWER (C) Let N2 be the smaller of the two squares. Then the differ-ence between the two squares is (N + 1)2 - N2 = 2N + 1. The givenconditions state that

100a+10b+c = N2+2N3 I and 100a+10c+b = N2+2(2N3+ 1)

Subtraction yields 9(c-b) = (2N+1)13, from which 27(c-b) = 2N+1.If c - b = 0 or 2, then N is not an integer. If c - b > 3, then N > 40, soN2 is not a three-digit integer. If c - b = 1, then N = 13. The numbersthat are, respectively, one third of the way and two thirds of the way from132 to 142 are 178 and 187,soa+b+c = 1 +7+8 = 16.

19. ANSWER (A) Let 0 = LA BC. The base of the cylinder is a circle withcircumference 6, so the radius of the base is 6/(2ir) = 3/7r. The heightof the cylinder is the altitude of the rhombus, which is 6 sin 0. Thus thevolume of the cylinder is

2

6 = 7r (-) 3(6 sin 0) =14

sin 0,7r 7r

so sin 0 = it/9.

20. ANSWER (D) Two vertices of the first parallelogram are at (0, c) and(0, d). The x-coordinates of the other two vertices satisfy ax +c = bx +d

194 Solutions

and ax + d = bx + c, so the x-coordinates are f(c - d)/(b - a). Thusthe parallelogram is composed of two triangles, each of which has area

1 c-db-a

It follows that (c-d)2 = 181b-a I. By a similar argument using the secondparallelogram, (c + d )2 = 721b - al. Subtracting the first equation fromthe second yields 4cd = 541b - a 1, so 2cd = 27lb - a 1. Thus l b - a I is

even, and a + b is minimized when {a, b} = { 1, 3}. Also, cd is a multipleof 27, and c + d is minimized when {c, d } = {3, 9}. Hence the smallestpossible value of a + b + c + d is I + 3 + 3 + 9 = 16. Note that therequired conditions are satisfied when (a, b, c, d) = (1, 3, 3, 9).

21. ANSWER (A) Because 36 = 729 < 2007 < 2187 = 37, it is convenientto begin by counting the number of base-3 palindromes with at most 7digits. There are two palindromes of length 1, namely 1 and 2. There arealso two palindromes of length 2, namely 11 and 22. For n > 1, eachpalindrome of length 2n + 1 is obtained by inserting one of the digits0, 1, or 2 immediately after the nth digit in a palindrome of length 2n.Each palindrome of length 2n + 2 is obtained by similarly inserting oneof the strings 00, 11, or 22. Therefore there are 6 palindromes of each ofthe lengths 3 and 4, 18 of each of the lengths 5 and 6, and 54 of length7. Because the base-3 representation of 2007 is 2202100, that integer isless than each of the palindromes 2210122, 2211122, 2212122, 2220222,2221222, and 2222222. Thus the required total is 2 + 2 + 6 + 6 + 18 +18+54-6= 100.

22. ANSWER (A) Imagine a third particle that moves in such a way thatit is always halfway between the first two. Let D, E, and F denote themidpoints of BC, CA, and AB, respectively, and let X, Y, and Z denotethe midpoints of AD, BE, and CF, respectively. When the first particle isat A, the second is at D and the third is at X. When the first particle is at F,the second is at C and the third is at Z. Between those two instants, bothcoordinates of the first two particles are linear functions of time. Becausethe average of two linear functions is linear, the third particle traversesXZ. Similarly, the third particle traverses ZY as the first traverses FBand the second traverses CE. Finally, as the first particle traverses BD andthe second traverses EA, the third traverses -)7-X-. As the first two particlesreturn to A and D, respectively, the third makes a second circuit of AX YZ.

NIA

ono

?C`

.O.

NIA


If 0 is the center of AABC, then by symmetry 0 is also the center ofequilateral AX YZ. Note that

2 1 1

OZ = OC - ZC = 3-CF - -CF = 6-CF,

so the ratio of the area of A X Y Z to that of A A B C is

OZ 2 6CF2

1

(OC) k3CF) 16

23. ANSWER (A) Let the triangle have leg lengths a and b, with a < b. Thegiven condition implies that

2ab=3(a+b+ a2+b2),

so

ab-6a-6b=6 a2+b2.Squaring both sides and simplifying yields

ab (ab - 12a - 12b + 72) = 0,

from which(a - 12) (b - 12) = 72.

The positive integer solutions of the last equation are (a, b) = (3, 4),(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), and (20, 21). However, thesolution (3, 4) is extraneous, and there are six right triangles with the re-quired property.

Query: Why do the given conditions imply that the hypotenuse also hasinteger length?

24. ANSWER (A) Let u = a/b. Then the problem is equivalent to finding allpositive rational numbers u such that

14u ku+9 =

for some integer k. This equation is equivalent to 9u2 - 9uk + 14 = 0,whose solutions are

_ 9k f 8 1 _k2- 504 k 1

'5 6U18 =2±6 9k

196 Solutions

Hence u is rational if and only if 9k2 - 56 is rational, which is true ifand only if 9k2 - 56 is a perfect square. Suppose that 9k2 - 56 = s2 forsome positive integer s. Then (3k - s)(3k + s) = 56. The only factorsof 56 are 1, 2, 4, 7, 8, 14, 28, and 56, so (3k - s, 3k + s) is one of theordered pairs (1, 56), (2, 28), (4, 14), or (7, 8). The cases (1, 56) and (7, 8)yield no integer solutions. The cases (2, 28) and (4, 14) yield k = 5 andk = 3, respectively. If k = 5, then u = 1/3 or u = 14/3. If k = 3, thenu = 2/3 or u = 7/3. Therefore there are four pairs (a, b) that satisfy thegiven conditions, namely (1, 3), (2, 3), (7, 3), and (14, 3).

Rewrite the equation

OR

a 14b

b + 9a = k

in two different forms. First, multiply both sides by b and subtract a toobtain

14b2 =bk -a.9a

Because a, b, and k are integers, 14b2 must be a multiple of a, and be-cause a and b have no common factors greater than 1, it follows that 14 isdivisible by a. Next, multiply both sides of the original equation by 9a andsubtract 14b to obtain

9a2

b= 9ak - 14b.

This shows that 9a2 is a multiple of b, so 9 must be divisible by b. Thus if(a, b) is a solution, then b = 1, 3, or 9, and a = 1, 2, 7, or 14. This givesa total of twelve possible solutions (a, b), each of which can be checkedquickly. The only such pairs for which

a 14b

b 9a

is an integer are when (a, b) is (1, 3), (2, 3), (7, 3), or (14, 3).

25. ANSWER (C) Introduce a coordinate system in which D = (-1, 0, 0),E = (1, 0, 0), and AABC lies in a plane z = k > 0. Because LCDEand LDEA are right angles, A and C are located on circles of radius 2centered at E and D in the planes x = 1 and x = -1, respectively. ThusA = (1,yi,k) and C = (-1, Y2, k), where yj = f 4-k2 for j = 1

CT

'

`.i°2007 AMC 12B Solutions 197

and 2. Because AC = 2v/2, it follows that (1 - (_1))2 + (yi - y2)2 =(2v/-2-)2. If y, = Y2, there is no solution, so yi _ -y2. It may be assumed

without loss of generality that yl > 0, in which case yl = I and y2 = -1.It follows that k = vf3-, so A = (1, 1,.J), C = (-1, -1, V-3), and B isone of the points (1, -1, ,f3-) or (-1, 1, ,r3-). In the first case, BE = 2 andBE 1 DE. In the second case, BD = 2 and BD 1 DE. In either case,the area of ABDE is (1/2)(2)(2) = 2.

OV

A

Index of ProblemsAlgebra:

algebraic fractions: 2002B-4 (13, 86)

AM-GM inequality: 2003A-24 (21, 104)

completing the square: 2005B-20 (45, 151)

discriminant: 2004B-21(36,131), 2005A-9(39,136),2007B-24(64,196)

distance-rate-time: 2002A-11(9,80), 2005A-6(38,135),2007A-9 (58, 182)

exponential equations: 2005B-13(44,148)

factoring: 2002A-1(8,78), 2002A-12(9,80), 2002B-3(13,86),2003B-20 (26, 112)

integer solutions: 2001-7 (2,68), 2001-21 (6,74), 2002A-6 (8,78),2002A-12(9,80), 2002B-3(13,86), 2002B-4(13,86),2002B-12(14,88), 2002B-15(14,89), 2003B-7(23,107),2003B-18(26,112), 2004A-3(28,117), 2004B-11(34,128),2004B-15(35,129), 2005B-19(45,151), 2006A-9(48,158),2006A-10(48, 159), 2006A-14(49, 159), 2006B-14(54, 172),2007B-17(63,193), 2007B-18(63,193), 2007B-20(63,194),2007B-23(64,195)

linear equations: 2001-3(1,66), 2001-4(1,67), 2002A-2(8,78),2002A-4(8,78), 2002A-11(9,80), 2003A-5(17,95), 2003B-2(22,106),2004A-11 (29,118), 2004B-5(33,126), 2005A-2(38,135),2006A-7(47,158), 2006B-8(53,170), 2007A-3(57,180)

linear graphs: 2004A-5(28,117), 2005A-12(39,137)

199

0000

200 The Contest Problem Book IX

linear inequalities: 2004B-11 (34, 128)

modeling: 2001-1(1,66), 2001-2(1,66), 2001-3(1,66), 2001-4(1,67),2002A-2(8,78), 2002A-4(8,78), 2002A-11(9,80), 2002A-13(10,81),2002B-7(13,87), 2002B-15(14,89), 2002B-17(14,89),2003B-2(22,106), 2003B-7(23,107), 2003B-11(24,108),2005A-22(42,142), 2006A-7(47,158), 2006A-9(48,158),2006B-3(53,169), 2006B-14(54,172),2007A-9(58,182), 2007B-8(62,190), 2007B-10(62,191),2007B-12(62,191)

nonlinear inequalities: 2002A-6(8,78)

parabolas: 2005B-8(44,146), 2005B-24(46,154), 2006B-12(54,171)

parametric equations of lines: 2001-20(5,73), 2005A-12(39,137)

percentages: 2001-3(1,66), 2005A-1(38,135), 2005B-2(43,145),2006A-7(47, 158), 2007B-10(62, 191), 2007B-12(62, 191)

polynomial equations: 2002B-7 (13, 87)

quadratic equations: 2001-23(6,75), 2002A-1(8,78),2002A-12(9,80), 2002A-13(10,81), 2002B-6(13,87),2005A-9(39,136), 2005B-8(44,146), 2005B-12(44,147),2006B-12(54, 171), 2007B-24(64, 196)

quadratic formula: 2001-23(6,75), 2002A-13(10,81),2005A-9(39,136), 2007B-24(64,196)

radical equations: 2006A-10 (48, 159)

rational solutions: 2007B-24 (64, 196)

ratios: 2002B-17(14,89), 2003B-5(23,106), 2003B-11(24,108),2003B-13(24,108), 2004B-6(33,126), 2007A-9(58,182),2007B-8 (62, 190)

slope-intercept form: 2003B-9(24,108), 2004A-5(28,117),2004A-12 (29, 119)

slope: 2001-20(5,73), 2003B-9(24,108), 2003B-24(27,114),2004A-5(28,117), 2004A-12(29,119), 2005A-12(39,137),2005B-24(46,154), 2006A-19(50,162), 2007A-13(58,183)

CJ!

C!!

r+'

tin

NN

Index of Problems

substitution in formulas: 2002B-2 (13, 86), 2004B-2 (33, 126)

sum and product of roots: 2002A-12 (9, 80), 2002B-6 (13, 87),2005B-12(44, 147)

systems of inequalities: 2002B-25 (16, 93), 2004B-18 (35, 129)

201

systems of linear equations: 2003B-7 (23, 107), 2006A-13 (49, 159),2006B-3(53,169), 2007B-10(62,191), 2007B-18(63,193)

systems of nonlinear equations: 2001-4(1,67), 2002B-19(15,90),2004B-21(36,131), 2004B-22(36,131), 2005A-22(42,142),2005B-23 (45, 153), 2006B-21 (56, 175)

vertex: 2005B-8(44,146), 2006B-12(54,171)

Arithmetic:

distance-rate-time: 2003A-4(17,95), 2004A-15(30,119),2006B-5(53,169), 2007A-4(57,180), 2007B-2(61,189)

estimation: 2006B-4(53,169)

exponents: 2001-5(1,67), 2002A-3(8,78), 2004A-6(28,117)

factorials: 2001-5 (1,67)

factoring: 2001-5(1,67), 2003B-1(22,106)

modeling: 2001-7(2,68), 2002A-10(9,80), 2003A-2(17,95),2003A-4(17,95), 2003A-10(18,96), 2004A-1(28,117),2004A-2 (28, 117), 2004A-15 (30,119), 2004B-1 (33, 126),2005A-4 (38, 135), 2005A-19 (41,141), 2005B-1 (43, 145),2005B-3 (43, 145), 2005B-4 (43, 145), 2006A-1 (47, 157),2006A-5 (47, 157), 2006B-4 (53, 169), 2006B-5 (53, 169),2006B-6 (53, 169), 2006B-11 (54, 170), 2007A-1 (57, 180),

2007A-4 (57, 180), 2007A-5 (57, 180), 2007B-1 (61, 189),2007B-2 (61, 189), 2007B-4 (61, 189), 2007B-5 (61, 190)

order of operations: 2002A-3 (8,78), 2006A-3 (47, 157)

percentages: 2003A-3 (17, 95), 2004A-1 (28, 117), 2004A-9 (29, 118),2005B-4 (43, 145), 2006A-3 (47, 157), 2007A-1 (57, 180),

2007A-5 (57, 180)

V')


ratios: 2002A-10(9,80), 2003A-10(18,96), 2004A-9(29,118),2006B-6(53,169), 2006B-11(54,170), 2006B-17(55,173),2007B-4(61, 189)

Complex numbers:

complex conjugate: 2002A-24(12,84), 2004A-23(32,123),2004B-16(35,129), 2005B-22(45,152),

complex plane: 2004B-16(35,129)

2007A-18(59,184)

graphs of complex equations: 2004B-16 (35, 129)

modulus: 2002A-24(12,84), 2005B-22(45,152)

roots of complex numbers: 2002A-24(12,84), 2005B-22(45,152)

Counting:

combinations: 2001-14(4,70), 2003A-20(21,101),2003A-22(21,103), 2004A-13(29,119), 2004B-20(36,131),2005B-11(44,147), 2006A-24(51,167), 2006A-25(52,167),2006B-9 (54, 170), 2007A-25 (60, 187)

floor function: 2001-12(3,69), 2003A-18(20,101), 2005A-18(41,140)

inclusion-exclusion principle: 2001-12(3,69), 2005A-18(41,140)

multinomial coefficients: 2006A-24 (51, 167)

one-to-one correspondence: 2006A-25(52,167), 2007A-25(60,187)

partitions: 2006A-24 (51, 167)

permutations: 2001-16(4,71), 2003B-19(26,112), 2005A-23(42,142),2005A-25(42,143), 2006B-7(53,169), 2007A-16(59,184),2007B-16(63,193)

Functions:

absolute value functions: 2003B-24 (27, 114)

addition of functions: 2003A- 19 (21, 101)

Index of Problems 203

binary operations: 2003A-6(17,95), 2006A-2(47,157),2006B-2(53,169)

composition of functions: 2002A-19 (11, 82), 2003B-8 (23, 107),2005A-20(41,141), 2005A-24(42,142), 2007B-9(62,191)

domain: 2003A-25(21,104), 2004A-16(30,120), 2006A-18(50,161)

floor function: 2005A-23(42,142), 2006B-20(55,174),2006B-22 (56, 175)

functional equations: 2001-9(3,68), 2004A-17(30,120),2006A-18 (50, 161)

inverse functions: 2004B-13(34,128), 2007B-9(62,191)

linear functions: 2003B-9(24,108), 2004B-13(34,128),2007B-9(62, 191)

quadratic functions: 2001-13(4,70), 2003A-25(21,104)

range: 2003A-25 (21, 104)

Geometry, circles:

annulus area: 2004A-24(32,124), 2004B-10(34,127)

arc length: 2001-8(3,68), 2002A-7(9,79)

Central Angle Theorem: 2007A-8(58,181), 2007B-3(61,189)

circle area: 2001-17(5,71), 2002A-5(8,78), 2003A-11(18,96),2005B-18 (45, 150)

circumscribed about square: 2003A-11(18,96)

circumscribed about triangle: 2003A-11(18,96), 2007A-10(58,182)

common tangent lines: 2001-18(5,72), 2002A-18(10,82),2005A-16(40,139), 2006A-16(49,161), 2006A-19(50,162),2006B-15(55,173)

dimensional proportionality: 2002A-7(9,79), 2003A-11(18,96),2007A-10(58, 182)

inscribed angle: 2001-17(5,71), 2005B-18(45,150)

vii

--O


Power of a Point: 2006A-17 (50,16 1)

sector area: 2003A-15(20,98), 2003B-16(26,111), 2004A-24(32,124)

tangent circles: 2001-18(5,72), 2002A-5(8,78), 2004A-19(31,121),2005A-16(40, 139)

tangent lines: 2001-18(5,72), 2004A-18(30,120), 2004B-10(34,127),2005A-16(40,139), 2005B-14(44,149), 2006A-16(49,161),2006B-15(55,173)

tangents from a common point: 2004A-18 (30, 120),2004B-19(35,130), 2005B-14(44,149), 2006B-15(55,173)

Geometry, coordinate:

distance formula: 2001-20(5,73), 2004A-12(29,119),2004B-18(35,129), 2005A-25(42,143), 2005B-16(44,149),2006B-16(55,173), 2007B-25(64,199)

distance from point to line: 2007A-19 (59, 185)

ellipses: 2006B-21(56,175)

equations of circles: 2002A-18(10,82), 2002B-23(15,92),2003A-17(20,100), 2006A-21(51,163)

midpoint formula: 2001-20(5,73), 2002B-18(15,90),2005B-24 (46, 154)

perpendicular lines, slopes of: 2001-20 (5,73), 2005B-24 (46, 154),2007A-13 (58, 183)

reflections: 2004B-9(34,127)

rotations: 2004B-9(34,127)

symmetry: 2002B-18(15,90), 2003A-9(18,96), 2004B-9(34,127),2005A-25 (42, 143), 2006A-22 (51, 165)

three-dimensional coordinates: 2005A-25 (42, 143),2005B-16(44,149), 2007B-25(64,199)

vectors: 2001-20(5,73), 2001-23(6,75)

V'1


Geometry, polygons:

dimensional proportionality: 2006B-15(55,173)

hexagon: 2006B-15(55, 173), 2006B-16(55, 173)

miscellaneous polygons: 2007A-8 (58,18 1)

octagon: 2003B-15 (25,110), 2007A-20(59,185)

pentagon: 2001-10(3,69), 2002B-5(13,87), 2003B-10(24,108),2007B-7 (62, 190)

sum of angle measures: 2002B-5 (13, 87)

symmetry: 2003B-10(24,108), 2003B-15(25,110), 2005B-7(43,146),2006B-15(55,173), 2006B-16(55,173)

triangulation: 2001-10(3,69), 2003B-15(25,110), 2006B-15(55,173),2006B-16(55,173), 2007B-7(62,190)

Geometry, quadrilaterals:

dimensional proportionality: 2006B-13 (54,17 1)

parallelogram: 2007B-20(63,194)

product of diagonals: 2002B-24(15,93)

rectangle: 200313-3 (22, 106), 200313-4 (23, 106), 200313-14 (25, 110),

2003B-22(27,113), 2005A-3(38,135), 2006A-6(47,158),2006B-21 (56, 175)

rhombus: 2003B-22(27,113), 2005B-7(43,146), 2006B-13(54,171),2007B-19(63,193)

square: 2001-20(5,73), 2002A-8(9,79), 2002B-18(15,90),2003A-11(18,96), 2003A-14(19,98), 2004B-7(33,126),2005A-7(39,136), 2006A-6(47,158), 2006A-17(50,161),2007A-23 (60, 186), 2007B-7 (62, 190)

sum of angles: 2007B-11 (62, 191)

symmetry: 2002A-8(9,79), 2003A-14(19,98), 2005A-7(39,136),2006B-13 (54,171)

LY

E.e0

.-.

+=

'

V'1


trapezoid: 2002B-18(15,90)

triangulation: 2002A-8(9,79), 2006B-13(54,171), 2007B-20(63,194)

Geometry, solid:

box: 2003A-3(17,95), 2005A-22(42,142), 2007A-2(57,180)

cone: 2001-8(3,68), 2003B-13(24,108), 2004B-19(35,130)

construction from planar figure: 2001-8(3,68), 2001-15(4,70),2003A-13 (19, 97)

cube: 2003A-13(19,97), 2005A-17(41,140)

cylinder: 2004A-9(29,118), 2007B-19(63,193)

pyramid: 2005A-17(41,140)

sphere: 2003B-13(24,108), 2004A-22(31,122), 2004B-19(35,130),2005A-22(42,142), 2005B-16(44,149)

tangent spheres: 2004A-22(31,122), 2005B-16(44,149)

tetrahedron: 2001-15(4,70), 2007A-20(59,185)

Geometry, triangles:

Angle Bisector Theorem: 2002A-23 (12, 84)

angle complement and supplement: 2002A-4 (8,78)

common altitude or base: 2001-22 (6,74), 2003A-16 (20, 99),2005A-15 (40, 139)

congruent triangles: 2003A-17(20,100), 2006B-23(56,176)

dimensional proportionality: 2004B-14 (35, 128), 2007B-22 (64, 195)

equilateral triangles: 2003A-11(18,96), 2003A-14(19,98),2003A-16(20,99), 2004A-22(31,122), 2005B-24(46,154),2006B-13(54,171), 2006B-16(55,173), 2007B-7(62,190),2007B-14(63,192), 2007B-22(64,195)

Heron's formula: 2002A-23 (12, 84)


isosceles triangles: 2001-24 (7,75), 2005B-6 (43, 146),2007A-6 (57, 181), 200713-3 (61, 189)

Pythagorean Theorem: 2001-18(5,72), 2002B-20(15,90),2003B-5(23,106), 2004A-18(30,120), 2004A-19(31,121),2004B-6(33,126), 2004B-10(34,127), 2005A-3(38,135),2005A-7(39,136), 2005A-16(40,139), 2005B-6(43,146),2006A-16(49,161), 2006A-17(50,161), 2006B-15(55,173),2006B-23 (56, 176), 2007A-10 (58, 182)

right triangles: 2001-18(5,72), 2001-24(7,75), 2002A-18(10,82),2002A-22(11,83), 2002B-20(15,90), 2003A-17(20,100),2003B-5(23,106), 2003B-22(27,113), 2004A-8(29,117),2004A-18(30, 120), 2004A-19(31, 121), 2004B-6(33, 126),2004B-10(34,127), 2004B-14(35,128), 2005A-3(38,135),2005A-7(39,136), 2005A-15(40,139), 2005A-16(40,139),2005B-6(43,146), 2005B-14(44,149), 2006A-16(49,161),2006A-17(50,161), 2006B-13(54,171), 2006B-15(55,173),2006B-16(55,173), 2006B-23(56,176), 2007A-10(58,182),2007B-25(64,199)

similar triangles: 2001-22(6,74), 2002A-18(10,82),2002B-20(15,90), 2003A-17(20,100), 2003B-14(25,110),2004B-14(35,128), 2005A-15(40,139), 2006A-16(49,161),2006B-15(55,173)

sum of angles: 2001-24(7,75), 2007A-6(57,181)

symmetry: 2003A-16(20,99), 2005B-5(43,145), 2006B-23(56,176),2007B-22(64,195)

triangle area: 2001-22(6,74), 2002A-22(11,83), 2002A-23(12,84),2003A-15(20,98), 2003A-16(20,99), 2003B-16(26,111),2003B-22(27,113), 2004A-8(29,117), 2004A-24(32,124),2004B-14(35,128), 2005A-15(40,139), 2005B-18(45,150),2006B-23(56,176), 2007A-8(58,181), 2007A-10(58,182),2007B-14(63,192), 2007B-20(63,194), 2007B-22(64,195)

triangle inequality: 2003A-7(17,95), 2006B-10(54,170)

triangle perimeter: 2003A-7(17,95), 2006B-10(54,170),2007B-6 (62,190)

`''


Graphs:

graphical transformations: 2001-13 (4, 70), 2003A-19 (21, 101)

graphs of absolute value functions: 2003B-24 (27, 114),2005B-7 (43, 146)

graphs of inequalities: 2002B-25 (16, 93)

graphs of polynomial functions: 2002A-19 (11, 82), 2003B-20 (26, 112)

graphs of quadratic functions: 2001-13 (4, 70), 2003A-19 (21, 101),2003A-25 (21, 104)

intersections of graphs: 2004B-21 (36, 131)

miscellaneous graphs: 2002A-25 (12, 84), 2005A-20 (41, 141),2006A-11 (48, 159)

Logarithms:

change of base: 2002B-22 (15, 91), 2003A-24 (21, 104),2005A-23(42,142), 2005B-13(44,148)

conversion to exponential form: 2003B-17 (26, 111),2004B-17(35,129), 2005A-21(41,141), 2005B-13(44,148),2005B-17(45,150), 2005B-23(45,153), 2006A-21(51,163),2006B-20(55,174)

logarithm ofpower: 2003B-17 (26, 111), 2005A-23 (42, 142),2007A-23 (60, 186)

logarithm of product: 2002A-14 (10, 81), 2002B-22 (15, 91),2003B-17(26,111), 2004B-17(35,129), 2005B-17(45,150)

logarithm of quotient: 2002B-22(15,91), 2003A-24(21,104)

logarithmic equation: 2007A-23 (60, 186)

logarithmic inequality: 2007B-17(63,193)

Logic:

logic: 2001-6(2,67), 2002A-9(9,80), 2002B-8(13,87),2003A-12(18,96), 2004A-4(28,117), 2004A-7(29,117)

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Number theory:

divisibility: 2001-7 (2,68), 200213-12 (14, 88), 2003A-18 (20, 101),

2003B-12(24,108), 2003B-18(26,112), 2005B-19(45,151),2006A-9(48,158), 2006B-14(54,172), 2006B-19(55,174),2007A-22(60,186), 2007B-20(63,194), 2007B-24(64,196)

greatest common factor: 2006A-14(49,159)

integer divisors: 2001-7(2,68), 2001-21(6,74), 2002A-20(11,82),2002B-13(14,88), 2002B-21(15,91), 2003A-23(21,103),2004B-3(33,126), 2005A-8(39,136), 2005B-17(45,150),2005B-21(45,151), 2006A-8(48,158), 2006B-14(54,172),2006B-22(56, 175), 2007A-11(58,182), 2007A-14(59, 183),2007B-20(63,194), 2007B-23(64,195), 2007B-24(64,196)

modular arithmetic: 2002B-10(14,87), 2005B-15(44,149),2007A-22 (60, 186)

number bases: 2004A-25(32,124), 2005A-19(41,141),2007B-21(63,194)

number of divisors: 2003A-23(21,103), 2005B-21(45,151),2006A-8 (48, 158)

prime factorization: 2001-7 (2,68), 2001-21 (6,74), 2002A-20 (11, 82),2003A-23 (21, 103), 2004B-3 (33, 126), 2005A-8 (39, 136),

2005B-17(45,150), 2006B-14(54,172), 2006B-22(56,175),2007A-11 (58, 182), 2007B-20(63, 194)

Polynomials:

complex zeros: 2001-23(6,75), 2004A-23(32,123), 2007A-18(59,184)

determined by set of points: 2005A-24 (42, 142)

factorization: 2001-23 (6,75), 2003A-21 (21, 102), 2004B-23 (36, 132)

multiplicity of zeros: 2003A-21 (21, 102)

product or zeros: 2001-19 (5, 72), 2004B-17 (35, 129),2007A-21 (59, 186)

rational conjugate zeros: 2004B-23(36,132)

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rational zero theorem: 2004B-23(36,132)

sum of coefficients: 2002A-25 (12, 84), 2004A-23 (32, 123),2007A-18(59,184), 2007A-21(59,186)

sum of zeros: 2001-19(5,72), 2004A-23(32,123),2007A-21 (59, 186)

Probability, continuous:

proportional to area: 2001-17(5,71), 2002A-22(11,83),2002B-18(15,90), 2003B-25(27,115), 2004A-20(31,121)

proportional to length: 2003B-21(26,113), 2006B-20(55,174),2007B-13 (62, 192)

Probability, discrete:

combinations: 2003A-22(21,103), 2004B-20(36,131),2005B-11 (44, 147)

conditional probability: 2003B-19(26,112), 2005A-14(40,138)

defining equally likely cases: 2001-11 (3,69)

independent events: 2002A-16(10,81), 2002B-16(14,89),2003A-22(21,103), 2007A-12(58,182)

permutations: 2003B-19(26,112), 2005A-23(42,142)

Sequences:

arithmetic sequences: 2002B-5(13,87), 2002B-9(14,87),2002B-13(14,88), 2003B-6(23,107), 2004A-14(30,119),2004B-8(34,127), 2004B-24(36,132), 2005A-13(39,137),2006A-12(48,159), 2007A-7(57,181)

arithmetic series: 2002B-13(14,88), 2004B-8(34,127),2006A-12 (48, 159)

cyclic sequences: 2001-25(7,76), 2002A-21(11,83),2005B-10(44,147), 2007A-11(58, 182)

V'1


geometric sequences: 2002B-9(14,87), 2004A-14(30,119),2004A-21(31,122), 2004B-24(36,132), 2007B-15(63,192)

geometric series: 2004A-21(31,122), 2007B-15(63,192)

recursively defined sequences: 2001-25(7,76), 2002A-21(11,83),2004B-12(34,128), 2005B-10(44,147), 2006A-23(51,166),2006B-25(56,178), 2007B-16(63,193), 2007B-21(63,194)

Statistics:

mean: 2001-4(1,67), 2002A-15(10,81), 2004A-10(29,118),2004B-11(34,128), 2005A-5(38,135), 2005B-9(44,147),2007A-15 (59, 183)

median: 2001-4(1,67), 2002A-15(10,81), 2004A-10(29,118),2005B-9(44, 147), 2007A-15 (59, 183)

mode: 2002A-15 (10, 81)

range: 2002A-15 (10, 81)

Trigonometry:

Law of Cosines: 2002A-23 (12, 84), 2002B-23 (15, 92),

2003A- 14 (19,98)

Law of Sines: 2006A-22 (51, 165)

right triangle definitions: 2002A-23 (12, 84), 2003A-17 (20, 100),2004B-24(36,132), 2007B-19(63,193)

slope as tangent of angle: 2005B-24(46,154), 2006A-19(50,162),2006B-23(56,176)

trigonometric equations: 2003B-23(27,114), 2004B-24(36,132),2005B-24(46,154), 2006A-15(49,160), 2006B-24(56,177),2007A-17 (59, 184)

trigonometric graphs: 2003B-23(27,114), 2007A-24(60,187)

trigonometric identities: 2003A-17(20,100), 2004A-21(31,122),2004B-24(36,132), 2005B-24(46,154), 2006A-19(50,162),2006A-22(51,165), 2006B-24(56,177), 2007A-17(59,184)


trigonometric inequalities: 2003B-21 (26, 113)

values at special angles: 2003A-14 (19, 98), 2003B-21 (26, 113),

2004B-24(36,132), 2005B-24(46,154), 2006A-15(49,160),2006A-22(51,165), 2006B-23(56,176), 2006B-24(56,177)

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About the Editors

J. Douglas Faires received his BS in mathematics from Youngstown StateUniversity in 1963, and earned his PhD in mathematics from the University ofSouth Carolina in 1970. He has been a Professor at Youngstown State Univer-sity since 1980, and has been actively involved in the MAA for many years. Forexample, he was Chair of the Ohio Section in 1981-1982 and Governor from1997-2000. He is currently a member of the MAA's Undergraduate Researchand Undergraduate Student Activities Committees. He was the National Direc-tor of the AMC 10 Competition of the American Mathematics Competitionsuntil 2007.

Faires is a past President of Pi Mu Epsilon and he was a member of theirCouncil for many years. Faires has been the recipient of many awards andhonors. He was named the Outstanding College-University Teacher of Math-ematics by the Ohio Section of the MAA in 1996; he has also received fiveDistinguished Professorship awards from Youngstown State University, and anhonorary Doctorate of Science Degree in May 2006. Faires has authored andcoauthored numerous books including Numerical Analysis (now in its eighthedition!), Numerical Methods (third edition), PreCalculus (fourth edition), andFirst Steps for Math Olympians, a book in the MAA Problem Book Series.

David Wells received a PhD in mathematics from the University of Pittsburghin 1973. He taught at Ohio Dominican University for six years and has been atthe New Kensington campus of Penn State University since 1979. During hiscareer at Penn State he has received both the Teresa Cohen Service Award fromthe Department of Mathematics and the Excellence in Teaching Award fromthe New Kensington Campus. He has been a member of the MAA since 1974,holding several sectional and national positions during that time. He servedas Governor of the Allegheny Mountain Section from 1996 to 1999 and has

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recently been appointed as Chair of the MAA Committee on the AmericanMathematics Competitions. He has chaired the AMC 12 Contest Committeesince 2001. Dr. Wells received the Meritorious Service Award from the Al-legheny Mountain Section of the MAA in 2005. His non-mathematical inter-ests include hiking and songwriting.

the contest problem book ix 2001-2007

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