solutions - canadian mathematical society308/ solutions solution 2, by ivko dimitri c, slightly modi...

SOLUTIONS /307

SOLUTIONSNo problem is ever permanently closed. The editor is always pleased to consider forpublication new solutions or new insights on past problems.

Statements of the problems in this section originally appear in 2016: 42(7), p. 313–317.

4161. Proposed by Peter Y. Woo.

A high-school math teacher discovered some geometry problems while sliding arug under his feet, over a floor with square tiles of length 1 unit. He chose x andy axes along two edges of some arbitrary tile. Today, he moved the square rugABCD of length between 1 and 2 units, so that A is on (0, 1) and C is on (c, 0)for some c > 2. Then surprise! He noticed that the edge BC goes through thepoint (2, 1). Can you find ∠BAE?

We received 34 submissions, out of which 33 were complete and correct. We presenttwo of the most elegant solutions.

Solution 1, by Kee-Wai Lau, slightly extended by the editor.

Let F be the point (2, 0) and let θ = ∠BAE = ∠FEC. Then AB = 2 cos θ,BE = 2 sin θ and

EC = BC −BE = AB −BE = 2(cos θ − sin θ).

Thus

1 = EF = EC cos θ = 2 cos θ(cos θ − sin θ),

or

cos(2θ)− sin(2θ) = 0,

which implies

tan(2θ) = 1

and hence θ = π8 .

Copyright c© Canadian Mathematical Society, 2017

308/ SOLUTIONS

Solution 2, by Ivko Dimitric, slightly modified by the editor.

Let the vertical line x = 1 intersect the lines AE and AB at points P and Qrespectively, and let F be the point (2, 0). Then the right triangles APQ andEFC are congruent, having two congruent angles ∠QAP = ∠CEF and adjacentlegs each of unit length. Hence AQ = EC, and thus BQ = BE, which impliesthat QEB is a right isosceles triangle. Therefore

∠AQE = π − ∠BQE =3π

4.

Since PQ halves AE at a right angle, the triangle AQE is also isosceles, and hence

∠BAE = ∠QAE =π

8.

4162. Proposed by George Apostolopoulos.

Let ABC be a triangle such that ∠B = 2∠C. We extend the side BC by a segmentCD equal to 1

3BC. Prove that

Area(ABC) =1

4|BC|2 · cot

θ

2,

where θ = ∠BAD.

There were 19 correct solutions. We present five different solutions. The firsttwo solutions are adapted from those of Daniel Dan, George Florin Serban, TituZvonaru and the proposer.

B C D

A

E

Solution 1.

Let a, b, c, d be the respective lengths of BC, CA, AB, AD, and suppose thatthe bisector of angle ABC intersects AC at E. Then, from the similarity oftriangles AEB and ABC and the equality BE = EC = ab(a+ c)−1, we have thatAC : AB = BC : BE, whence

b2 = c(a+ c). (1)

Crux Mathematicorum, Vol. 43(7), September 2017

SOLUTIONS /309

Applying Stewart’s Theorem to triangle ABD and cevian AC (or alternatively theCosine Law to triangles ABC and ACD and eliminating the cosine of the angleat C), we find that

ad2 +(a

3

)c2 =

Åb2 +

a2

3

ãÅ4a

3

ã.

This, along with (1), leads to 9d2 = 4a2 + 12b2 − 3c2 = (3c+ 2a)2, whence

d =1

3(3c+ 2a). (2)

Applying the Cosine Law to triangle ABD and using (2) yields that

16a2 = 9(c2 + d2 − 2cd cos θ) = (18c2 + 12ac+ 4a2)− 6c(3c+ 2a) cos θ,

so that6(3c2 + 2ac) cos θ = 6(3c2 + 2ac− 2a2),

whence

cos θ = 1− 2a2

3c2 + 2ac= 1− 2a2

3cd, (3)

and

cd =a2

3 sin2 θ2

.

Therefore

[ABC] =3

4[ABD] =

3

8cd sin θ =

1

4a2 cot2

θ

2.

Solution 2.

We begin by deriving an expression for cot θ/2. As in Solution 1, we can use (1)and (2) to derive (3), from which

1− cos θ =2a2

3c2 + 2ac, 1 + cos θ =

6c2 + 4ac− 2a2

3c3 + 2ac,

and

cot2θ

2=

1 + cos θ

1− cos θ=

(c+ a)(3c− a)

a2. (4)

From Heron’s area formula, (1) and (4), we obtain that

16[ABC]2 = (a+ c+ b)(a+ c− b)(b− a+ c)(b+ a− c)= [(a+ c)2 − b2][b2 − (c− a)2] = [a(a+ c)][a(3c− a)]

= a4 cot2θ

2.

whence [ABC] = a2

4 cot θ2 as desired.


310/ SOLUTIONS

Solution 3, by C.R. Pranesachar.

Using the notation of Solution 1 and the fact that 3d = 2a + 3c, we note thatthe triangle ABD has sides c, d and 4a/3 and semiperimeter a+ c. Applying theformula

tanA

2=

(s− b)(s− c)s(s− a)

=(s− b)(s− c)

[ABC]

for an arbitrary triangle ABC to the triangle ABD and angle θ, we obtain

cotθ

2=

[ABD]

a2/3=

4[ABC]

a2

as desired.

Solution 4, by Leonard Giugiuc.

Place the triangle in the Cartesian plane with A at (0, 1) and C at (k, 0) withk = cotC > 0. We find that the coordinates of A, B, C, D are given by

A(0, 1), B

Å1− k2

2k, 0

ã, C(k, 0), D

Å9k2 − 1

6k, 0

ã,

and that a = (3k2 − 1)/(2k) = 2[ABC]. Since x = cot 12∠BDA is the positive

solution of the equationx2 − 1

2x=

9k2 − 1

6k,

we have x = 3k and tan 12∠BDA = 1/(3k). Observe that θ + B + ∠BDA = π so

that

cotθ

2= tan

ÅC +

∠BDA2

ã=

4k

3k2 − 1=

2

a.

Therefore1

4a2 cot

θ

2=a

2= [ABC].

Solution 5, by Andrea Fanchini.

Let U and V be the respective feet of the perpendiculars from C and D to theside AB. Since UV : BU = CD : BC = 1 : 3, then 3(AU −AV ) = 3UV = BU .

B C D

A

U

V


SOLUTIONS /311

(To cater to the various configurations, the lengths along the vector−−→AB are signed,

a contingency that can be accommodated using barycentric coordinates.) SinceDV : CU = BD : BC = 4 : 3, then

cot θ =AV

DV=

3AV

4CU=

1

4

Å3AU −BU

CU

ã=

1

4(3 cotA− cotB).

Let S = 2[ABC], twice the area of the triangle, and define Sφ = S cotφ for anyangle φ. In particular

SA =S cosA

sinA= bc cosA =

1

2(b2 + c2 − a2),

with analogous expressions for SB and SC . It follows that

SB + SC = a2 and 4Sθ = 3SA − SB .

Since

2 cotB cotC = 2 cot 2C cotC = cot2 C − 1,

we have

2SBSC = S2C − S2,

so that

a2SB = SB(SB + SC) = (SB + SC)2 + (2SBSC − S2C)− 3SBSC

= a4 − S2 − 3SBSC = a4 + 3(S2 − SBSc)− 4S2.

From the identity cotA cotB + cotB cotC + cotC cotA = 1, we obtain

a2SB = a4 + 3SA(SB + SC)− 4S2 = a4 + 3a2SA − 4S2,

which can be rearranged to yield

3SA − SB4

=4S2 − a4

4a2.

We turn to the desired equality, which can be recast as

S θ2

=2S2

a2.

Since, for 0 < φ < π, there is a one-one relation between Sφ = (S2φ/2−S

2)/(2Sφ/2)and Sφ/2, the equality is equivalent to

3SA − SB4

= Sθ =(2S2/a2)2 − S2

4S2/a2=

4S2 − a4

4a2,

which has already been established.


312/ SOLUTIONS

Editor’s Comments. The solutions revealed several interesting relations, in par-ticular expressions for the cotangent in terms of the elements of triangle ABD:

cotθ

2=

2 sinC sin 2c

sin 3C=

√2b2 + c2 − a2

a.

Vaclav Konecny pointed out that if angle C is less than 45◦, then drawing thecircle with centre A through B along with the radius to its other intersection withBC gives the diagram for the trisection of the exterior angle at A with a markedstraightedge and compasses.

Miguel Amengual Covas observed that the condition ∠B = 2∠C figured in earlierproblems published in Crux and references to those can be found in the articleRecurring Crux Configurations: No. 7 by Chris Fisher in Volume 38(6) of June2012, p. 238–240.

4163. Proposed by Leonard Giugiuc.

Let a, b be real numbers with 0 < a < b and consider a positive sequence xn suchthat

limn→∞

Åaxn +

b

xn

ã= 2√ab.

Find limn→∞ xn or show that it does not exist.

We received 17 submissions, all of which were correct. We present two differentsolutions.

Solution 1, by Adnan Ali.

Define the sequence {yn} as yn = axn +b

xn. Then limn→∞ yn = 2

√ab. Solving

ax2n − ynxn + b = 0,

we have

xn =1

2a

Äyn ±

√y2n − 4ab

ä.

Sincelimn→∞

√y2n − 4ab = 0,

it follows that

limn→∞

xn =1

2alimn→∞

yn =2√ab

2a=

…b

a.

Solution 2, by Michel Bataille.

From Åaxn +

b

xn

ã2−Åaxn −

b

xn

ã2= 4ab


SOLUTIONS /313

and the hypothesis, we deduce that

limn→∞

Åaxn −

b

xn

ã2= 0, so lim

n→∞

Åaxn −

b

xn

ã= 0.

Since

2axn =

Åaxn +

b

xn

ã+

Åaxn −

b

xn

ã,

it follows that limn→∞ 2axn = 2√ab, and so limn→∞ xn =

…b

a.

Editor’s comment. Both Roy Barbara and Oliver Geupel pointed out that theassumption a < b is superfluous.

4164. Proposed by G. Di Bona, A. Fiorentino, A. Moscariello and G. G. N.Angilella.

In an election, N voters are to elect k representatives. Each voter must indicateexactly m distinct preferences, with m ≤ k < N . Every voter is a candidatethemselves, and all candidates have a distinct age. The candidates are then rankedaccording to the number of votes received, and the k candidates who receive thelargest number of votes are elected. In case of degeneracies, the eldest candidateis elected.

What is the minimum number of votes that a candidate should receive, in orderto be sure to get elected?

There were five submissions, but only that of the proposer was complete and correct.The faulty solutions tended to argue from extreme situations without making a solidconnection to the general one, however intuitively appealing this might be. Twosolvers gave answers that depended on the status of particular individuals, ratherthan a value V that applied to any of the candidates. We present the solution ofthe proposer.

The minimum number of votes that a candidate should receive in order to ensureelection is

V =

õNm

k + 1

û+ 1.

First, we show that a candidate receiving at least V votes will be elected. Letv1, v2, . . . , vk+1 be the number of votes received by the top k+ 1 candidates, withv1 ≥ v2 ≥ · · · ≥ vk+1. Since v1 + v2 + · · ·+ vk+1 ≤ Nm, the total number of votescast, the arithmetic mean M of these numbers must satisfy

vk+1 ≤M ≤Nm

k + 1< V.

Hence, any candidate with at least V votes must be among the top k candidatesand so be elected.


314/ SOLUTIONS

On the other hand, we construct a situation in which a candidate with V − 1votes fails to get elected. Suppose only k + 1 candidates receive votes, and wenumber them 1, 2, . . . , k + 1. Create a sequence with Nm terms by repeatingthe base sequence {1, 2, . . . , k + 1} as many times as needed to fill the Nm slots.Partition this sequence into N subsequences of m terms, and let the ith voter votefor the candidates in the ith subsequence. Each candidate will receive at leastbNmk+1c = V − 1 votes and the candidate numbered k+ 1 will receive exactly V − 1votes. Thus, this candidate may fail to get elected.

Note that the role of the age rule is to break a tie. Consider the situation wherethe N voters are arranged in a circle and each votes for the m candidates sittingimmediately to the right. Then each candidate receives the same number m ofvotes, and we need to decide on the winners by seniority.

4165. Proposed by Daniel Sitaru.

Prove that for all real numbers x1, x2, x3 and x4, we have,

|x1 + x2 + x3 + x4|+ 2(|x1|+ |x2|+ |x3|+ |x4|) ≥ 6 6

∏1≤i<j≤4

|xi + xj |.

We received six submissions, all of which were correct. We present the solutionwith generalization by Michel Bataille.

We prove the stronger result that for any complex numbers x1, x2, x3 and x4, wehave

|x1 + x2 + x3 + x4|+ 2(|x1|+ |x2|+ |x3|+ |x4|) ≥∑

1≤i<j≤4|xi + xj |. (1)

The proposed inequality then follows from (1) by the AM-GM Inequality. To prove(1), we will make use of Hlawka’s inequality which states that

|a+ b+ c|+ |a|+ |b|+ |c| ≥ |a+ b|+ |b+ c|+ |c+ a| (2)

for all complex numbers a, b, c. (See e.g., problem 2482 [1999 : 430 ; 2000 : 506].)Setting a = x1, b = x2 and c = x3 + x4, then from (2) we have

|x1+x2+x3+x4|+|x1|+|x2|+|x3+x4| ≥ |x1+x2|+|x2+x3+x4|+|x1+x3+x4|. (3)

Applying (2) again, we obtain

|x2 + x3 + x4| ≥ |x2 + x3|+ |x3 + x4|+ |x2 + x4| − |x2| − |x3| − |x4| (4)

and

|x1 + x3 + x4| ≥ |x1 + x3|+ |x3 + x4|+ |x1 + x4| − |x1| − |x3| − |x4| (5)

Adding (4) and (5) and denoting the right side of (3) by R, then we have

R ≥ |x3 + x4| − |x1| − |x2| − 2|x3| − 2|x4|+∑

1≤i<j≤4|xi + xj |. (6)


SOLUTIONS /315

From (3) and (6), we deduce that

|x1 + x2 + x3 + x4|+ |x1|+ |x2|+ |x3 + x4|

≥ |x3 + x4| − |x1| − |x2| − 2|x3| − 2|x4|+∑

1≤i<j≤4|xi + xj |

from which (1) follows immediately.

4166. Proposed by Mihaela Berindeanu.

Show that for all real numbers x, y and z, we have:

24x−y + 24y−z + 24z−x ≥ 2x+2y + 2y+2z + 2z+2x.

We received 14 solutions. We present three solutions.


Let 2x = a, 2y = b and 2z = c. Then the proposed inequality becomes

a4

b+b4

c+c4

a≥ ab2 + bc2 + ca2,

which is nothing but a consequence of the Cauchy-Schwarz Inequality:Ça4

b+b4

c+c4

a

å(bc2 + ca2 + ab2) ≥ (a2c+ b2a+ c2b)2 = (bc2 + ca2 + ab2)2.

Equality holds if and only if a = b = c or equivalently x = y = z.

Solution 2, by Salem Malikic.

Introducing substitution 2x = a, 2y = b and 2z = c our inequality becomesequivalent to

a4

b+b4

c+c4

a≥ ab2 + bc2 + ca2.

The last one follows after adding up the following obvious inequalities

a4

b+ bc2 ≥ 2a2c,

b4

c+ ca2 ≥ 2b2a,

c4

a+ ab2 ≥ 2c2b.

In order to achieve equality one must have

a4

b= bc2,

b4

c= ca2,

c4

a= ab2.

That implies a2 = bc, b2 = ca and c2 = ab, i.e.

(a− b)2 + (b− c)2 + (c− a)2 = 0,

so a = b = c, i.e. x = y = z.


316/ SOLUTIONS

Solution 3, by Michel Bataille.

Multiplying both sides by the positive real number 2−(x+y+z), we obtain

23x−2y−z + 23y−2z−x + 23z−2x−y ≥ 2y−z + 2z−x + 2x−y

or, setting a = 2y−z, b = 2z−x, c = 2x−y,

ac3 + ba3 + cb3 ≥ a+ b+ c. (1)

Thus, it suffices to prove (1) for positive reals a, b, c such that abc = 1. Sinceac3 = c(abc) cb = c · cb and similarly ba3 = a · ac , cb

3 = b · ba , (1) rewrites as X ≥ 1where

X =a

a+ b+ c· f( ca

)+

b

a+ b+ c· f(ab

)+

c

a+ b+ c· fÅb

c

ãwhere the function f is defined by f(t) = 1

t .

Now, since f is convex on the interval (0,∞), Jensen’s inequality yields

X ≥ fÅ

a

a+ b+ c· ca

+b

a+ b+ c· ab

+c

a+ b+ c· bc

ã= f(1) = 1

and therefore (1) holds.

4167. Proposed by Dao Thanh Oai and Leonard Giugiuc.

Consider triangle ABC and let D be any point in the plane. Let points A′, B′, C ′

be reflections of points A,B,C inD, respectively. Construct the 3 triangles AB′C1,CA′B1 and BC ′A1 outwardly as the given diagram indicates:

Show that A1B1C1 is an equilateral triangle.

We received 13 submissions, all correct. We feature two solutions; the first istypical of the ten that used complex coordinates.


SOLUTIONS /317

Solution 1, by Somasundaram Muralidharan.

We must assume that the triangles AB′C1, CA′B1, and AB′C1 are equilateral

with the same orientation as ∆ABC, but there is no loss of generality in assum-ing that D is at the origin. Let a, b, c be the complex numbers representing thevertices A,B,C respectively. Then A′, B′, C ′ are represented by the complex num-bers −a,−b,−c respectively. The vertex B1 is obtained by rotating the segmentCA′ about vertex C anticlockwise through 60◦. Hence, if the complex numberrepresenting B1 is b1, then

b1 = c+ ((−a)− c)e iπ3 = c− (a+ c)eiπ3 .

Similarly the complex numbers representing C1 and A1, namely c1 and a1, aregiven by

c1 = a− (a+ b)eiπ3 , a1 = b− (b+ c)e

iπ3 .

Now,

−−−→C1A1 = (b− (b+ c)e

iπ3 )− (a− (a+ b)e

iπ3 )

= (b− a) + (a− c)e iπ3

= b− ce iπ3 − a(1− e iπ3 )

= b− ce iπ3 − ae− iπ3

Similarly,−−−→C1B1 = −a+ be

iπ3 + ce−

iπ3 .

Because

ei2π3 = −1

2+ i

√3

2= −Ç

1

2− i√

3

2

å= −e− iπ3 ,

we have −−−→C1B1 =

−−−→C1A1e

iπ3 .

Thus−−−→C1B1 is obtained by rotating

−−−→C1A1 anticlockwise by 60◦ and the triangle

A1B1C1 is equilateral.

Solution 2, by Peter Y. Woo, with small corrections supplied by the editor.

Because the hexagonAC ′BA′CB′ is symmetric aboutD, the quadrilateralsBC ′B′C

and ABA′B′ are parallelograms. Compare the vector−−−→B1A1, which is the sum of

vectors−−−→B1A

′,−−→A′B, and

−−→BA1, to the vector

−−−→B1C1, which is the sum of vectors

−−→B1C,

−−→CB′, and

−−−→B′C1:

−−−→B1A

′ equals−−→B1C rotated 60◦ anticlockwise,−−→

A′B equals−−−→B′C1 rotated 60◦ anticlockwise,

−−→BA1 equals

−−→CB′ rotated 60◦ anticlockwise.


318/ SOLUTIONS

Because vector addition is commutative, the sum of the first vector of each ofthe three pairs equals the sum of the last three vectors rotated 60◦ anticlockwise.

Consequently−−−→A1B1 is the vector

−−−→C1B1 rotated 60◦ anticlockwise. It follows that

A1B1C1 is an equilateral triangle.

4168. Proposed by Leonard Giugiuc and Daniel Sitaru.

Let a, b, c be positive real numbers such that a2 + b2 + c2 = 18 and abc = 4. Provethat

6 ≤ a+ b+ c ≤ 2

»2√

6 + 4 +√

6− 2.

When does equality hold?

We received 15 solutions, all of which were correct. We present the one by JosephDiMuro.

We’ll use Lagrange multipliers to find the extreme values of f(a, b, c) = a+ b+ c,subject to the constraints g1(a, b, c) = a2 + b2 + c2 = 18 and g2(a, b, c) = abc = 4.

Note first that the extreme values must occur at points where5f = λ15g1+λ25g2for some real numbers λ1, λ2, unless they occur at points where the vectors 5g1and 5g2 are linearly dependent.

Since 5f = (1, 1, 1, ]), 5g1 = (2a, 2b, 2c), and 5g2 = (bc, ca, ab), 5g1 and 5g2are linearly dependent only if a = b = c. But the constraints a2 + b2 + c2 = 18and abc = 4 cannot be satisfied if a = b = c since the system 3a2 = 18 and a3 = 4has no solutions.

Hence it suffices to look for points that satisfy 5f = λ1 5 g1 + λ2 5 g2. That is,λ1(2a, 2b, 2c) + λ2(bc, ca, ab) = (1, 1, 1), or

2aλ1 + bcλ2 = 1 (1)

2bλ1 + acλ2 = 1 (2)

2cλ1 + abλ2 = 1 (3)

From (1)-(2) we obtain

2(a− b)λ1 + c(b− a)λ2 = 0 ⇐⇒ 2(a− b)λ1 = c(a− b)λ2,

which is true if and only if either a = b or 2λ1 = cλ2. Similarly, from (2)-(3) and(3)-(1) we deduce that either b = c or 2λ1 = aλ2; and either c = a or 2λ1 = bλ2.

Recall that a = b = c is impossible. On the other hand, if a, b, and c are alldistinct, then 2λ1 = cλ2 = bλ2 = aλ2 would yield λ1 = λ2 = 0 so 5f = 0, acontradiction. Hence, exactly two of a, b, and c must be equal.

Without loss of generality, assume a 6= b = c. Then the original constraints become

a2 + 2b2 = 18 and ab2 = 4,


SOLUTIONS /319

which yields

a2 +8

a= 18,

a3 − 18a+ 8 = 0,

(a− 4)(a2 + 4a− 2) = 0.

Hence, a = 4 or a = −2±√

6. Since a > 0, we obtain a = 4 or√

6− 2.

If a = 4, then b = c = 1, and if a =√

6− 2, then b = c =√

2√

6 + 4. These pointsthen yield the extreme values of

f(4, 1, 1) = 6 and f(√

6− 2,

»2√

6 + 4,

»2√

6 + 4) = 2

»2√

6 + 4 +√

6− 2.

Since

2

»2√

6 + 4 +√

6− 2 ≈ 6.4157 > 6,

the results follow. Note that the maximum is attained at the 3 points obtained bypermuting the coordinates of (4, 1, 1) and the minimum is attained at the 3 pointsobtained by permuting the coordinates of

(√

6− 2,

»2√

6 + 4,

»2√

6 + 4).

4169. Proposed by Michel Bataille.

Let a, b, c be positive real numbers. Prove that(a

b

a+ b+ b

…c

b+ c+ c

…a

c+ a

)Çb

…a+ b

b+ c

…b+ c

c+ a

…c+ a

a

å≤ (a+b+c)2.

There were 14 correct solutions submitted. Eight of these solvers independentlygave the solution presented here.

Using the Cauchy-Schwarz and arithmetic-harmonic means inequalities, we obtainthe following two inequalities:

a

b

a+ b+ b

…c

b+ c+ c

…a

c+ a=√a

ab

a+ b+√b

bc

b+ c+√c

…ca

c+ a

=√a+ b+ c

ab

a+ b+

bc

b+ c+

ca

c+ a

≤√a+ b+ c

…a+ b

4+b+ c

4+c+ a

4

=1√2

(a+ b+ c)


320/ SOLUTIONS

and

b

…a+ b

b+ c

…b+ c

c+ a

…c+ a

a=√b√a+ b+

√c√c+ b+

√a√c+ a

≤√a+ b+ c

»(a+ b) + (b+ c) + (c+ a)

=√

2(a+ b+ c).

Multiplying these inequalities gives the desired result.

Editor’s Comment. Nguyen Ngoc Tu used the concavity of the function√x and

reduced the problem to establishing that

ab(a+ b)−1 + bc(b+ c)−1 + ca(c+ a)−1 ≤ (a+ b+ c)/2.

4170. Proposed by Miguel Ochoa Sanchez and Leonard Giugiuc.

Let ABCD be a circumscribed quadrilateral (that is, a quadrilateral for whichan incircle can be constructed) and let P be the intersection point of AC andBD. Let ha, hb, hc and hd denote the distances from P to AB,BC,CD and DA,respectively. Prove that

AB · CDAD ·BC

=hb + hdha + hc

.

We received nine submissions, all correct, and will feature two of them. The firstis typical of those that reduced the problem to a fairly recent result; the second isprovided for those readers who prefer to see the details.


Let θ be the angle between the diagonals AC and BD. Then observe that we havethe following relations:

AB · ha = PA · PB sin θ,BC · hb = PB · PC sin θ,CD · hc = PC · PD sin θ,DA · hd = PD · PA sin θ.

It follows thatAB · CDAD ·BC

=hb · hdha · hc

.

The desired result is thereby reduced to proving

hb · hdha · hc

=hb + hdha + hc

.

But this is an immediate consequence of a known theorem, namely, a convexquadrilateral ABCD has an incircle if and only if the altitudes ha, hb, hc, hd fromthe intersection point of the diagonals to each of the four sides satisfy

1

ha+

1

hc=

1

hb+

1

hd.


SOLUTIONS /321

Proofs of the theorem can be found in [2], for example. Moreover, this result wasProblem 5 on the 2015 Indian National Mathematics Olympiad.

Solution 2 is a composite of similar solutions from Oliver Geupel and John G.Heuver.

Let A′, B′, C ′, and D′ be the points where the lines AB, BC, CD, and DA,respectively, touch the incircle Γ, and denote the tangent lengths by a = AA′ =AD′, b = BB′ = BA′, c = CC ′ = CB′, and d = DD′ = DA′. A degenerateversion of Brianchon’s theorem tells us that the chords A′C ′ and B′D′ also passthrough P . Here is an alternative proof of this claim: Let Q be the intersectionpoint of the lines AC and A′C ′.

The tangents AB and CD to Γ are symmetric with respect to the perpendicularbisector of the chord A′C ′. Hence ∠QA′A = ∠DC ′Q = 180◦ − ∠QC ′C. By theSine Law applied to triangles AQA′ and CQC ′ we obtain

AQ

AA′=

sin∠QA′Asin∠AQA′

=sin∠QC ′Csin∠CQC ′

=CQ

CC ′;

that is, AQ : CQ = a : c. Similarly, if R is the intersection point of AC and B′D′,we have AR : CR = a : c. It follows that Q = R so that the lines AC, A′C ′, andB′D′ are concurrent at point Q = R. Similarly, the lines BD, A′C ′ and B′D′ areconcurrent at Q = R. Thus, the lines AC, BD, A′C ′, and B′D′ are concurrent atP = Q = R. We deduce, therefore, that

AP

PC=a

cand

BP

PD=b

d.

It follows that [ABP ][BCP ] = AP

PC = ac , where we have used square brackets for the area

of a triangle. Analogous results involving the other relevant triangles give us

[ABP ]

ab=

[BCP ]

bc=

[CDP ]

cd=

[DAP ]

da. (1)


322/ SOLUTIONS

Note that 2[ABP ] = (a+ b)ha, so that

ha =2[ABP ]

a+ b=

2kab

a+ b, (2)

where we have set k equal to the common ratio in (1). Similar formulas hold forthe remaining triangles. We finally obtain

AB · CDAD ·BC

=(a+ b)(c+ d)

(d+ a)(b+ c)

=

1(b+c) ·

1(d+a)

1(a+b) ·

1(c+d)

=bcb+c + da

d+aaba+b + cd

c+d

=hb + hdha + hc

,

which completes the proof.

Editor’s comments. Note that equation (2) (together with the corresponding equa-tions for hb, hc, hd) lead immediately to a proof of the theorem mentioned in thefirst solution. Other proofs of the theorem can be found in [1], where Josefssontraces the result back to a 1995 problem in the Russian Journal Kvant [3]. Hereports that it also appeared in [4], and as a problem in the 4th stage of the 48thGerman Mathematical Olympiad (2009).

References

[1] Martin Josefsson, Similar Metric Characterizations of Tangential and Extan-gential Quadrilaterals, Forum Geometricorum, 12 (2012) 63–77.

[2] N. Minculete, Characterizations of a Tangential Quadrilateral, Forum Geo-metricorum, 9 (2009) 113–118.

[3] I. Vaynshtejn, N. Vasilyev and V. Senderov, Problem M1495, Kvant (in Rus-sian) no. 6 (1995) 27–28.

[4] A. Zaslavsky, Problem M1887, Kvant (in Russian) no. 3 (2004) p. 19.


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