British Mathematical Olympiad 1975Source: Mathematics in School, Vol. 4, No. 5 (Sep., 1975), p. 10Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30211427 .

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British Mathematical Olympiad 1975

The Examination Paper

24th March, 1975

Time allowed-3 hours

Do as much as you can. The earlier questions carry slightly fewer marks. Aim at answering whole questions. In any question, marks may be added for elegance and clarity or subtracted for obscure or poor presentation.

1. Given that x is a positive integer solve

[,] 1+ [-2 ]+ . . . + [3 '1 ] = 400 (where [z] means the integral part of z) and prove your solution is complete.

2. The first n prime numbers, 2,3,5,.. ,pn are partitioned into two disjoint sets A and B. The primes in A are al,a2, ... ,ah and the primes in B are bl,b2, ... ,bk where h+k = n.

he kg h h ai

Tmk

y The two products[ ai and i ibi are formed whereai

and Oi are any positive integers. If d divides the difference between these products prove

that either d = 1 or d>P . n"

3. Use the pigeonhole principle (i.e. if more than n objects are put into n pigeonholes then at least one pigeonhole must con- tain more than one object) to answer the following question.

A disc S is defined as the set of all points P in a plane such that IOPI < 1, where IOPI is the distance of P from O, a given point in the plane, called the centre of the disc.

Prove that if the disc S contains 7 points such that the distance from any of the 7 points to any other is greater than or equal to 1, then one of the 7 points is 0.

4. Three parallel lines AD, BE, CF are drawn through the vertices of triangle ABC meeting the opposite sides in D,E,F respectively.

The points P,Q,R divide AD, BE, CF, respectively in the same ratio k:1 and P,Q,R are collinear. Find the value of k.

5. For any positive integer m you are given that

where there are 2m+1 terms on the left hand side. Both these expressions are defined to be f(0). The function g(0) is defined by

g(o) =

1+[2] cos

2024m]cos 40+...+cos 2m0.

2 4

Given that there is no rational k for which a = kTr find the values of a for which

gL a) z

Lim

mf(a) gn~al

6. Prove that if n is a positive integer greater than 1 and x>y>l, then

x -1 ynl -1

x(xn-1 -1) ylyn-1 -1)

7. Prove that there is only one set of real numbers xl,x2, ... ,xn such that

(1-x1)2+(xl-x2)2+. .

.+(Xn_1 -xn)2+X n2-

1 n+1 "

10

8. The interior of a wine glass is a right circular cone. The glass is half filled with water and then slowly tilted so that the water starts and continues to spill from a point P on the rim.

What fraction of the whole conical interior is occupied by water when the horizontal plane of the water level bisects the generator of the cone furthest from P?

Some Selected Solutions 1. Call the left side f(x). As we increase x (given a positive integer) by unity, we have

f(x+1)-f(x) = [3 x3]+[3/(x3+1)] +. . .+ [3V((x+1)3-1)].

For each of the (x+1)3_X3 positive integers n such that x3 -<n < (x+1)3, [Vn] = x. Hence f(x+l)-f(x) = x {(x+1)3-x3}. f(2) = 1.7, f(3)-f(2) = 2.19 = 38, etc. f(5)-f(4) = 4(125-64) = 244. One can see that f(x) will soon pass 400 but it may be that

thoIre is no x for which f(x) = 400. However, 7+38+111+244 = 400 and since f(x) is increasing, x = 5 is the unique solution.

2. Call the products H and K respectively. Let d be any diviso, of (H-K). (d = 1) is a divisor. If d t 1 let - be the smallest prime divisor of d. dl(H-K) = YI(H-K). We show by contradiction that - cannot be one of the first n primes.

Assume y is one of the first n primes. Then yeA UB. Since A and B are disjoint, - is a member of only one of A and B. Say yeA. Then 3jH. 7yJH and yJ (H-K) yJ1fH-(H-K) } = K = eB. Similarly yeB = yeA. This contradicts the statement above and shows that the assumption yeA UB is false. So y is not one of the first n primes. Since the smallest prime divisor of d is not one of the first n primes, d> Pn-

4. If the parallel lines AD, BE, CF are parallel to a side of tri- angle ABC one of D, E, F is at infinity and any k will suffice. Otherwise one and only one of the parallel lines through the vertices will cut the opposite side internally. Without loss of generality, let this li ine be AD.

BE = DA xIBQ= k DA. CB BE = DA xCD, BO =k~x C CD i k~lCD k DA . CB

Similarly CR = DxDB k+1 DB

F

RD C B

Figure 1

CR-DP DP-BO 1 Since P lies on RO, CR-DPDP-B where DP = DA CD DB k+1D

Substitute for BQ, CR and DP k DA .CB 1 DA 1 k DA. CB

k+7' DB k+1 k+1DA k+l ' CD

CD BD

kCB-BD = CD-kCB = 2kCB = BD+CD = k = (based on the solution of R. J. Burns, Tonbridge School)

The results of the 1975 BMO were printed at the end of the NMC results on page 22 of the last issue. A circular giving details of the 1976 contest is enclosed with this issue. Entries should be sent to the Mathematical Association, 259 London Road, Leicester LE2 3BE through Regional Secretaries.

A t the International Mathematical Olympiad held in Bulgaria in July, the first 5countries (out of 17) were Hungary, E. Germany, USA, USSR and Britain. First prizes were awarded to J. J. Hitchcock (Kingston GS) and J. R. Rickard (City of London S), both of whom gained full marks, 0

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