Mathematical Excalibur, Volume 14, Number 2

Volume 14, Number 2 May - September, 2009

Remarks on IMO 2009

Leung Tat-Wing

2009 IMO Hong Kong Team Leader

Olympiad Corner

The following were the problems of the first day of the 2008 Chinese Girls’ Math Olympiad.

Problem 1. (a) Determine if the set

{1,2,⋯,96} can be partitioned into 32 sets of equal size and equal sum. (b) Determine if the set {1,2,_{⋯,99} can} be partitioned into 33 sets of equal size and equal sum.

Problem 2. Let φ(x) = ax3_+bx2_{+cx+d be}

a polynomial with real coefficients. Given that φ(x) has three positive real roots and that φ(x) < 0, prove that 2b3 ₊

9a2_{d − 7abc ≤ 0.}

Problem 3. Determine the least real

number a greater than 1 such that for any point P in the interior of square ABCD, the area ratio between some two of the triangles PAB, PBC, PCD, PDA lies in the interval [1/a, a].

Problem 4. Equilateral triangles ABQ,

BCR, CDS, DAP are erected outside the

(convex) quadrilateral ABCD. Let X, Y,

Z, W be the midpoints of the segments PQ, QR, RS, SP respectively. Determine

the maximum value of . BD AC YW XZ + +

Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK

高 子 眉 (KO Tsz-Mei)

梁 達 榮 (LEUNG Tat-Wing)

李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST

吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU

Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU

Acknowledgment: Thanks to Elina Chiu, Dept. of Math.,

HKUST for general assistance.

On-line:

http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is October 3, 2009.

For individual subscription for the next five issues for the 09-10 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin LI, Dept. of Mathematics, The Hong Kong University of Science and Technology, Clear Water Bay, Kowloon, Hong Kong

Fax: (852) 2358 1643 Email: [email protected]

© Department of Mathematics, The Hong Kong University of Science and Technology.

The 50th _{International Mathematical}

Olympiad (IMO) was held in Bremen, Germany from 10th _{to 22}nd _{July 2009. I}

arrived Bremen amid stormy and chilly (16°C) weather. Our other team members arrived three days later. The team eventually obtained 1 gold, 2 silver and 2 bronze medals, ranked (unofficially) 29 out of 104 countries/regions. This was the first time more than 100 countries participated. Our team, though not among the strongest teams, did reasonably well. But here I mainly want to give some remarks about this year’s IMO, before I forget.

First, the problems of the contest:

Problem 1. Let n be a positive integer

and let a1, a2, …, ak (k ≥ 2) be distinct integers in the set {1,2,…,n} such that n divides ai(ai+1−1) for i=1,…, k−1. Prove that n does not divide ak(a1−1).

This nice and easy number theory problem was the only number theory problem in the contest. Indeed it is not easy to find a sequence satisfying the required conditions, especially when k is close to n, or n is prime. Since adding the condition n divides ak(a1−1) should

be impossible, it was natural to prove the statement by contradiction. Clearly 2 ≤ k ≤ n, and we have a1 ≡ a1a2 (mod n),

a2 ≡ a2a3 (mod n), …, ak−1 ≡ ak−1ak (mod

n). The extra condition ak ≡ aka1 (mod

n) would in fact “complete the circle”.

Now a1 ≡ a1a2 (mod n). Using the

second condition, we get a1 ≡ a1a2 ≡

a1a2a3 (mod n) and so on, until we get a1

≡ a1a2⋯ak (mod n). However, in a circle every point is a starting point. So starting from a2, using the second

condition we have a2 ≡ a2a3 (mod n). By

the third condition, we then have a2 ≡

a2a3a4 (mod n). As now the circle is

complete, we eventually have a2 ≡

a2a3⋯aka1 (mod n). Arguing in this

manner we eventually have a1 ≡ a2 ≡ ⋯

≡ ak (mod n), which is of course a contradiction!

Problem 2. Let ABC be a triangle with

circumcenter O. The points P and Q are interior points of the sides CA and AB, respectively. Let K, L and M be midpoints of the segments BP, CQ and

PQ, respectively, and let Γ be the circle

passing through K, L and M. Suppose that PQ is tangent to the circle Γ. Prove that OP=OQ.

The nice geometry problem was supposed to be a medium problem, but it turned out it was easier than what the jury had thought. The trick was to understand the relations involved. A very nice solution provided by one of our members went as follows.

As KM||BQ (midpoint theorem), we have

∠

AQP =

∠

QMK. Since PQ is

tangent to Γ, we have

∠

QMK =

∠

MLK (angle of alternate segment).

Therefore,

∠

AQP =

∠

MLK. By the

same argument, we have

∠

APQ =

∠

MKL. Hence, ΔAPQ ~ ΔMKL. Therefore, 2 . 2 AP MK MK BQ AQ= ML = ML =CP

This implies AP·PC = AQ·QB. But by considering the power of P with respect to the circle ABC, we have

AP·PC = (R+OP)(R−OP)

= R2 _{− OP}2_,

where R is the radius of the circumcircle of ΔABC. γ M L K O A C B P Q

Mathematical Excalibur, Vol. 14, No. 2, May-Sep. 09 Page 2 Likewise,

AQ·QB = (R+OQ)(R−OQ)

= R2_−OQ2_.

These force OP2 _{= OQ}2_{, or OP = OQ,}

done!

Problem 3. Suppose that s1, s2, s3, …

is a strictly increasing sequence of positive integers such that the subsequences 1

,

2

, ,...

3 s s s

s s s

and 11

,

21

,

31

,...

s s s

s

₊

s

₊

s

₊

are both arithmetic progressions. Prove that the sequence s1, s2, s3, … is

itself an arithmetic progression. This was one of the two hard problems (3 and 6). Fortunately, it turned out that it was still within reach.

One trouble is of course the notation. Of course,

1 s

s

stands for the

1

th

s

term of the si sequence and so on. Starting from an arithmetic progression (AP) with common difference d, then it is easy to check that both

1

,

2

,

3

,...

s s s

s s s

and 11

,

21

,

31

,...

s s s

s s

_{+ +}

s

₊

are APs with common difference d2_.

The question is essentially proving the “converse”. So the first step is to prove that the common differences of the two APs i s

s

and ₁ i s

s

₊ are in fact the same, say s. It is not too hard to prove and is intuitively clear, for two lines of different slopes will eventually meet and cross each other, violating the condition of strictly increasing sequence. The next step is the show the difference between two consecutive terms of si is indeed

s

,

(thus s is a square). One can achieve this end by the method of descent, or max/min principle, etc.

Problem 4. Let ABC be a triangle with

AB = AC. The angle bisectors of

∠

CAB and

∠

ABC meet the sides BC

and CA at D and E, respectively. Let K be the incenter of triangle ADC. Suppose that

∠

BEK = 45_{°. Find all}

possible values of

∠

CAB.

This problem was also relatively easy. It is interesting to observe that an isosceles triangle can be the starting point of an IMO problem. With geometric software such as Sketchpad, one can easily see that

∠

CAB should

be 60° or 90°. To prove the statement of the problem, one may either use synthetic method or coordinate method. One advantage of using the coordinate method is after showing the possible values of

∠

CAB, one can go back to show these

values do work by suitable substitutions. Some contestants lost marks either because they missed some values of

∠

CAB or forgot to check the two

possible cases do work.

Problem 5. Determine all functions f

from the set of positive integers to the set of positive integers such that, for all positive integers a and b, there exists a non-degenerate triangle with sides of lengths a, f(b) and f(b+f(a)−1). (A triangle is non-degenerate if its vertices are not collinear.)

The Jury worried if the word triangle may be allowed to be degenerate in some places. But I supposed all our secondary school students would consider only non-degenerate triangles. This was a nice problem in functional inequality (triangle inequality). One proves the problem by establishing several basic properties of f. Indeed the first step is to prove f(1)=1, which is not entirely easy. Then one proceeds to show that f is injective and/or f(f(x)) = x, etc, and finally shows that the only possible function is the identity function f(x) = x for all x.

Problem 6. Let a1, a2, …, an be distinct positive integers and let M be a set of

n−1 positive integers not containing s=a1+a2+⋯+an. A grasshopper is to jump along the real axis, starting from the point

O and making n jumps to the right with

lengths a1, a2, …, an in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in M.

It turned out that this problem was one of the most difficult problems in IMO history. Only three of the 564 contestants received full scores. (Perhaps it was second to problem 3 posed in IMO 2007, for which only 2 contestants received full scores.)

When I first read the solution provided by the Problem Committee, I felt I was reading a paper of analysis. Without reading the solution, of course I would say we could try to prove the problem by induction, as the cases of small n were easy. The trouble was how to establish the

induction step. Later the Russians provided a solution by induction, by separating the problem into sub-cases min M < an or min M ≥ an, and then applying the principle-hole principle, etc judiciously to solve the problem. Terry Tao said (jokingly) that the six problems were easy. But in his blog, he admitted that he had spent sometime reading the problem and he even wrote an article about it (I have not seen the article.)

The two hard problems (3 and 6) were more combinatorial and/or algebraic in nature. I had a feeling that this year the Jury has been trying to avoid hard number theory problems, which were essentially corollaries of deep theorems (for example, IMO 2003 problem 6 by the Chebotarev density theorem or IMO 2008 problem 3 by a theorem of H. Iwaniec) or hard geometry problem using sophisticated geometric techniques (like IMO 2008 problem 6).

The Germans ran the program vigorously (obstinately). They had an organization (Bildung und Begabung) that looked after the entire event. They had also prepared a very detailed shortlist problem set and afterwards prepared very detailed marking schemes for each problem. The coordinators were very professional and they studied the problems well. Thus, there were not too many arguments about how many points should be awarded for each problem.

Three of the problems (namely 1, 2 and 4) were relatively easy, problems 3 and 5 were not too hard, so although problem 6 was hard, contestants still scored relatively high points. This explained why the cut-off scores were not low, 14 for bronze, 24 for silver and 32 for gold.

It might seem that we still didn’t do the hard problems too well. But after I discussed with my team members, I found that they indeed had the potential and aptitude to do the hard problems. What may still be lacking are perhaps more sophisticated skills and/or stronger will to tackle such problems.

Problem Corner

We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li,

Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for sending

solutions is October 3, 2009.

Problem 326. Prove that 345 +456is the product of two integers, each at least 102009_.

Problem 327. Eight pieces are placed

on a chessboard so that each row and each column contains exactly one piece. Prove that there is an even number of pieces on the black squares of the board.

(Source: 1989 USSR Math Olympiad)

Problem 328. (Due to Tuan Le,

Fairmont High School, Anaheim, Ca., USA) Let a,b,c > 0. Prove that

2 2 3 3 2 2 3 3 2 2 3 3 a c a c c b c b b a b a + + + + + + + + _. ) )( )( ( ) ( ) ( 6 a c c b b a c b a ca bc ab + + + + + + + ≥

Problem 329. Let C(n,k) denote the

binomial coefficient with value

n!/(k!(n−k)!). Determine all positive

integers n such that for all k = 1, 2, _⋯,

n−1, we have C(2n,2k) is divisible by C(n,k).

Problem 330. In ΔABC, AB = AC = 1

and ∠ BAC = 90°. Let D be the midpoint of side BC. Let E be a point inside segment CD and F be a point inside segment BD. Let M be the point of intersection of the circumcircles of ΔADE and ΔABF, other than A. Let N be the point of intersection of the circumcircle of ΔACE and line AF, other than A. Let P be the point of intersection of the circumcircle of ΔAMN and line AD, other than A. Determine the length of segment AP with proof.

(Source: 2003 Chinese IMO team test) *****************

Solutions

****************

Problem 321. Let AA’, BB’ and CC’ be

three non-coplanar chords of a sphere and let them all pass through a common point

P inside the sphere. There is a (unique)

sphere S1 passing through A, B, C, P and a

(unique) sphere S2 passing through A’, B’,

C’, P.

If S1 and S2 are externally tangent at P,

then prove that AA’=BB’=CC’.

Solution. NGUYEN Van Thien (Luong The Vinh High School, Dong Nai, Vietnam) and Jim Robert STUDMAN (Hanford, Washington, USA).

Consider the intersection of the 3 spheres with the plane through A, A’, B, B’ and P.

B A A' B' P M N

Let MN be the common external tangent through P to the circle through A, B, P and the circle through A’, B’ P as shown above. We have∠ABP = ∠APM = ∠A’PN = ∠A’B’P = ∠A’B’B = ∠BAA’ = ∠BAP.

Hence, AP=BP. Similarly, A’P = B’P. So

AA’ = AP+A’P = BP+B’P = BB’.

Similarly, BB’ = CC’.

Other commended solvers: CHUNG

Ping Ngai (La Salle College, Form 6) and LAM Cho Ho (CUHK Math Year 1). Problem 322. (Due to Cao Minh Quang,

Nguyen Binh Khiem High School, Vinh Long, Vietnam) Let a, b, c be positive real

numbers satisfying the condition a+b+c = 3. Prove that . 2 ) 1 ( ) 1 ( ) 1 ( 2 2 2 ≥ + + + + + + + + + + + ca a c a c bc c b c b ab b a b a

Solution. CHUNG Ping Ngai (La Salle College, Form 6), NGUYEN Van Thien (Luong The Vinh High School, Dong Nai, Vietnam) and the proposer independently. Observe that 2( 1) _. ab b a ab a ab b a b a + + − = + + + _(*)

Applying the AM-GM inequality twice, we have . 9 1 3 3 3 3 2 2 + + ≤ = ≤ + + b a ab b a ab ab b a ab By (*), we have . 9 1 8 9 1 ) 1 ( 2 _{− −} = + + − ≥ + + + a b a b a ab b a b a

Adding two other similar inequalities and using a+b+c = 3 on the right, we get the desired inequality.

Other commended solvers: LAM Cho

Ho (CUHK Math Year 1), Manh Dung NGUYEN (Special High School for

Gifted Students, HUS, Vietnam),

Paolo PERFETTI (Math Dept,

Università degli studi di Tor Vergata Roma, via della ricerca scientifica, Roma, Italy), Stefan STOJCHEVSKI (Yahya Kemal College, Skopje, Macedonia), Jim Robert STUDMAN (Hanford, Washington, USA) and

Dimitar TRENEVSKI (Yahya Kemal

College, Skopje, Macedonia).

Problem 323. Prove that there are

infinitely many positive integers n such that 2n_{+2 is divisible by n.}

Solution. CHUNG Ping Ngai (La Salle College, Form 6), LAM Cho Ho (CUHK Math Year 1) and WONG Ka

Fai (Wah Yan College Kowloon, Form

4).

We will prove the stronger statement that there are infinitely many positive even integers n such that 2n_{+2 is divisible by}

n and also that 2n_{+1 is divisible by n−1.} Call such n a good number. Note n = 2 is good. Next, it suffices to prove that if n is good, then the larger integer m = 2n₊₂ is also good.

Suppose n is good. Since n is even and m = 2n_{+2 is twice an odd integer, so m = nj} for some odd integer j. Also, the odd integer m−1 = 2n_{+1 = (n−1)k for some} odd integer k. Using the factorization

ai_{+1 = (a+1)(a}i−1_−ai−2_{+⋯+1) for positive}

odd integer i, we see that 2m_{+2 = 2(2}(n−1)k₊₁₎ = 2(2n−1_{+1) (2}(n−1)(k−1)_−⋯+1) is divisible by 2(2n−1_{+1) = m and} 2m_{+1 = 2}nj_{+1 = (2}n₊₁₎₍₂n(j−1)_−⋯+1) is divisible by 2n_{+1= m−1. Therefore, m} is also good.

Problem 324. ADPE is a convex

quadrilateral such that

∠

ADP =

∠

AEP. Extend side AD beyond D to

a point B and extend side AE beyond E to a point C so that

∠

DPB =

∠

EPC.

Let O1 be the circumcenter of ΔADE

and let O2 be the circumcenter of

ΔABC.

If the circumcircles of ΔADE and

Mathematical Excalibur, Vol. 14, No. 2, May-Sep. 09 Page 4 then prove that line O1O2 bisects line

segment AP.

Solution. Jim Robert STUDMAN (Hanford, Washington, USA).

Let the circumcircle of ΔADE and the circumcircle of ΔABC intersect at A and Q.

Observe that line O1O2 bisects chord

AQ and O1O2⊥AQ. Hence, line O1O2

bisects line segment AP will follow if we can show that O1O2 || PQ, or

equivalently thatPQ⊥AQ.

O₂ C A B O₁ D P E Q M N

Let points M and N be the feet of perpendiculars from P to lines AB and

AC respectively. Since ∠ANP = 90° =

∠AMP, points A, N, P, M lie on a circle Г with AP as diameter. We claim that

∠MQN =∠MAN. This would imply Q is also on circle Г, and we would

have PQ⊥AQ as desired.

Since we are given

∠

ADP =

∠

AEP,

we get

∠

BDP =

∠

CEP. This

combines with the given fact

∠

DPB =

∠

EPC imply _{∆DPB and ∆EPC are}

similar, which yields DB/EC =

DP/EP=DM/EN.

Since A,E,D,Q are concyclic, we have

∠

BDQ =180°−

∠

ADQ

=180°−

∠

AEQ =

∠

CEQ.

This and

∠

DBQ=∠ABQ =∠ACQ =

∠

ECQ imply _{∆DQB and ∆EQC are}

similar. So we have QD/QE=DB/EC. Combining with the equation at the end of the last paragraph, we get

QD/QE=DM/EN.

Using _{∆DQB and ∆EQC are similar,} we get

∠

MDQ =

∠

BDQ =

∠

CEQ

=

∠

NEQ. These imply _{∆MDQ and}

∆NEQ are similar. Then

∠

MQD =

∠

NQE.

Finally, for the claim, we now have

∠

MQN =

∠

MQD +

∠

DQN =

∠

NQE +

∠

DQN =

∠

DQE

=

∠

DAE =

∠

MAN.

Comments: Some solvers used a bit of

homothety to simplify the proof.

Other commended solvers: CHUNG

Ping Ngai (La Salle College, Form 6), LAM Cho Ho (CUHK Math Year 1), NG Ngai Fung (STFA Leung Kau Kui

College, Form 7).

Problem 325. On a plane, n distinct lines

are drawn. A point on the plane is called a

k-point if and only if there are exactly k of

the n lines passing through the point. Let

k2, k3, …, kn be the numbers of 2-points, 3-points, …, n-points on the plane, respectively.

Determine the number of regions the n lines divided the plane into in terms of n,

k2, k3, …, kn.

(Source: 1998 Jiangsu Province Math

Competition)

Solution. LAM Cho Ho (CUHK Math Year 1).

Take a circle of radius r so that all intersection points of the n lines are inside the circle and none of the n lines is tangent to the circle. Now each line intersects the circle at two points. These 2n points on the circle are the vertices of a convex 2n-gon (call it M) as we go around the circle, say clockwise. Let the n lines partition the interior of M into P3 triangles,

P4 quadrilaterals, ⋯, Pj j-gons, ⋯. These polygonal regions are all convex since the angles of these regions, which were formed by intersecting at least two lines, are all less than 180°. By convexity, no two sides of any polygonal region are parts of the same line. So we have Pj = 0 for j>3n.

Consider the sum of all the angles of these regions partitioning M. On one hand, it is 180°(P3+2P4+3P5+⋯) by counting region

by region. On the other hand, it also equals 360°(k2+k3+⋯+kn)+(2n−2)180° by counting all the angles around each vertices of the regions. Cancelling 180°, we get

P3+2P4+3P5+⋯=2(k2+k3+⋯+kn)+(2n−2). Next, consider the total number of all the edges of these regions partitioned M (with each of the edges inside M counted twice). On one hand, it is 3P3+4P4+5P5+⋯ by

counting region by region. On the other hand, it is also (4k2+6k3+⋯2nkn)+4n by counting the number of edges around the k-points and around the vertices of

M. The 4n term is due to the 2n edges

of M and each vertex of M (being not a k-point) issues exactly one edge into the interior of M. So we have

3P3+4P4+5P5+⋯=4k2+6k3+⋯2nkn+4n. Subtracting the last two displayed equations, we can obtain

. 1 ) 1 ( 23 2 5 4 3+P+P+ =k + k + n− k +n+ P L n

Finally, the number of regions these n lines divided the plane into is the limit case r tends to infinity. Hence, it is exactly k2+2k3+⋯+(n−1)kn+n+1.

Other commended solvers: CHUNG

Ping Ngai (La Salle College, Form 6)

and YUNG Fai.

Remarks on IMO 2009

(continued from page 2) As I found out from the stronger teams (Chinese, Japanese, Korean, or Thai, etc.), they were obviously more heavily or vigorously trained. For instance, a Thai boy/girl had to go through more like 10 tests to be selected as a team member.

Another thing I learned from the meeting was several countries were interested to host the event (South-East Asia countries and Asia-Minor countries). In fact, one country is going to host three international competitions of various subjects in a row for three years. Apparently they think hosting these events is good for gifted education.

The first IMO was held in Romania in 1959. Throughout these 51 years, only one year IMO was not held (1980). To commemorate the fiftieth anniversary of IMO in 2009, six notable mathematicians related to IMO (B. Bollabas, T. Gowers, L. Lovasz, S. Smirnov, T. Tao and J. C. Yoccoz) were invited to talk to the contestants. Of course, Yoccoz, Gowers and Tao were Fields medalists. The afternoon of celebration then became a series of (rather) heavy lectures (not bad). They described the effects of IMOs on them and other things. The effect of IMO on the contestants is to be seen later, of course!

1_ ·The ~wer is DQ for

p3rt

(a) an~yes for

pru:t

(b). _ : . >

(~)' ~ince

1 + 2 + •••

+

96 ~ 48·97 is' Dot divisible by

32~

-

we camiot.partitiOD the the

~et

. {1.2, ••

~$}

inio'32

s.~tS

'of

~Ua1

m. .' \:-:-..

(b) Sincel+2+···+99=50·99 •. ~h of the 33 subsets nltiSf" have-sum 150. We

partitiOD.~

Dumbers.#I the

se~

{1.2 •••• /b} mto.·

• 0" ... • • •

common difference 1: .

. .

~ {1,50}. {3,49}, {5.48}, ••• ,.f33,34}, {2,66}, {4;6S}; ... ;{32,51}.

. It is then

easy

to see

that

{1,50,9~}. {3,49,98};{5,4.8,97},: •• , {33,34,83}, {2,66,82}, {4 .. 65,81}, ... {32, 51, 67}:

is a partition:satisfyirig the conditions of the problem. _. .

. Remarlc·.Ill

general, the se~{1,2, ••• ,3n}

ca:n

-be beparti-tioned

. .

into· n sets'Qf equal'size and equal ~ if ~d

only

if n is odd. . . ..

2. So~ution . Denote by Ai,~,X:J the-~ts of ;(x). -By Vie~'~ relation, we have

b . c · J

Xt +X2+X:J =.,.--'-, ~X2 +X:lX:J +X:Jx,. =-, x,.~X:J

=--.

a a a

Since .rP<O)

=

d < 0 ~d .ll~X:J > 0 , _ it foliows that a

>

0 . Dividing both sides of the desired iriequaIlty by -~3 give~

- \. .

2(!)1

+9d

-i.!!..E..so

a a a a . '

or

-2(Xt + ~ + ~)3 -9Xt~11 + 7(4. + Xl + X:J)(Xt~ + ~X:J + x,Xt) ~ 0 . Ex.panding the te~ on the left-hand side of the last inequality _ .and simplifying gives

or

-_ X:X:l

+.llX;+X;X:J +~~+X:xJ +x,x;

s2(X:

+~

+.r:), .

(1) . One can .... eStablish (1) by noting that

(Ai """7r2X~2-X:)=(I:t ~~)l(Xt +Xz» 2

x;~ +~xi

$X:

+'~.

_ Addin~ the last inequality with its cyclic 'analogous forms -yields ( 1).

3. The

answe~

is the golden ratio r/J :::: 1 ..:

-15 .

'-- 2

. We first show that there must always be ~o triangles whose ratio

lies

~

the

interval

[;,~

J.. .

By scaling, we may assume without loss of generality that

ABeD has area 2. Then [PAB]+[PCD]=.!. [ApeD}::::l, and likewise

. - ' . _ . _ 2

[PBC]+[PDA1~1. Let [PAR] =x.

so.

that [PCD]=I-x; by symmetry we may assume .without loss of generality that x ~ 1-x, or

eq~.val~ntly,·

that

~~~.

Likewise, let

[P~C]=

y and [FDA]: . 1- y, and again we may assume y ~ 1- y. Finally, we may. also assume w.ithout

loSs

of generality that x

s y.

. ' . /

. We know. that

.!.

s

x < 1;' we now divide into case!; based on the

. ·2

v~lueofx.

-' _J

. I - . . I

.-Case 1: X~:...-.lnthlscase, l-x~t-->O, so

. r P '

rP

1

-=-s

_"_'=_l_=!:::

r/J-l-:x

l-.!.

,,-1

1:

r/J

rP

by the well-known identity that r/J -'1

=.!..

Since x

~

1-x

rP .

-~I~-,andso x -1" - E x

[1 ]

-,rP

as desired.

.

I-x

rP

I-x r/J.

1 1 - .' .

Case 2: x~-. Now, since -<xsy<l, weconc1udethat

rP . rP : ~ desired. . y 1 l~-<-=r/J x .1

r/J-,

We now show that no smaller value of

a

wOdcs. '

We first

note"tha~

for

~y x'~d

y strictly 'between 0 and 1, we

, '

can find a point P such that [PAB]=x, [PCD]~I-x"[PBCJ=y.

, . , .. x

, "-[PDA]

=

1-y : just ,take a POint,~ w~ch is distance

'Ji

away from ,

~~de AB'and,~~e ~ a~ay

from

side~C.,

r

,,'

We now set x=- and consider the

limitm8

case as

y-;

app~aches

O.

By

illspecti~

we

see

~t'

the smallest triangle

area.

rati~

greater than

lis

[PDA]

=(1:"

y);,

~hich

ap,proaches',;

iIi.

the

, ,[PCD]" ' , "

lirillt as y

~

0 ':'

Likewis~.

the

iarieSt

w'

ratio'less

tlwi"

1 is

, "

IPCD]

=

1 •

which'approach~ ..!..

hi

the.linii~g

case. Hence

[PD~] (1-y); ,,; , , '

, any a that satisfies

th~ ,co~ti~ns

of the

proble~

must, satisfy either:'

, ,

, ' , 1 1 '

(l-y);:5;;a or" ' > :eitherway,amustbeatl~(1-y);.

, (1-y); a : ' ,

T~

the linriting

,~

as

y

~

0, we ,see that' a

=,

is

tJi(f'

, minimal,

value ora

that wor~. ,

" " . l+J3, '.

4.,Solution L The answer i s - - . '

, , , , ' 2

We consider the CO?Jfiguration shown above. (Our proofs can be

adjusted slightly for other Coidigurations.) LetPl,

QI,

Rl, SI be.the

: mi~intS ,of sewnents DA. AB. 'Be, CA, respectively. It is well

known that ,PlQIRISl is a

parallelogram

(since ~G ,,'BDfl ~SI 804

, ,

,GRs II ACII

SIP.), Let M ~d Nbe th~ midpointS of segmen~ DP and

DS,

reSPectively.

Note

that '

~ bS1 :::S.N =DN=WN and ' Note also·that ,

LP"Ds"

'= 'LPDS -1200

=

₁₈₀0 - LDNW ..,.120° .

" =

60° - LDNW

=

L.WNS1• "'.'-_. ---,-,' By SAS. (WN=D.~,L.P,.DS.:=L.WNS" andNS. =DS.,), we

c~nclude that triangles WNS. and .fDS. are ,congruent, to each'other: Likewise. we can show that triangles ,~MW .and ~DS are 'congrUent to

each

~th~r.

It

follows that 'l'tSr =WS1'=l;W;'

that

is,-tri~gle

WS1!t

~s' equi1at~ In exactly the same wily, we can show that ,'yQ.~ is also equilateral.

~t

U

and

V be the' midpoints 'of segme~tS Sl~and G~.

" BD" ..

Th~ in parallelogram ~Ql~S., UV =~g = - . By the 1riangl~

. 2 " ,

'i~equa1ity. we have

. YW

:5;;'"

+

VU'+U:V,=

~.fj

+

~Ql

+

~J~J3

Sl~.Jj+BD+M.J3 BD+AC.Jj

'2 ' , 2

Similarly. we can show that

xz

~

,AC

+

BD.J3 .'

2 Adding the last two i~equalities yields

'1+../3 ' XZ+YW 1+../3

.xz+YW~-2-·(AC+BD) or AC+'BD ~~. ,

"