Mathematical Excalibur, Volume 19, Number 2

Volume 19, Number 2 September 2014 – October 2014

Olympiad Corner

Below are the problems of the 2014 Bulgarian National Math Olympiad on May 17-18, 2014.

Problem 1. (Teodosi Vitanov, Emil

Kolev) Given is a circle k and a point A

outside it. The segment BC is a diameter of k. Find the locus of the orthocenter of ȋ ABC, when BC is changing.

Problem 2. (Nikolay Beluhov) Consider a rectangular n×m table where n • 2 and m • 2 are positive integers. Each cell is colored in one of the four colors: white, green, red or blue. Call such a coloring interesting if any 2_{×2 square contains every color} exactly once. Find the number of interesting colorings.

Problem 3. (Alexander Ivanov) A real nonzero number is assigned to every point in space. It is known that for any tetrahedron IJ the number written in the incenter equals the product of the four numbers written in the vertices of IJ. Prove that all numbers equal 1. Problem 4. (Peter Boyvalenkov) Find all prime numbers p and q such that

p2 _{| q}3 _{+ 1 and q}2 _{| p}6 _{í 1.} (continued on page 4)

IMO2014 and Beyond (II)

Leung Tat-Wing

To discuss the IMO2014 problems, let’s proceed from the easier problems to the harder problems.

Problem 1. Let a0 < a1< a2 < ֥ be an infinite sequence of positive integers. Prove that there exists a unique integer

n • 1 such that 0 1 1 ... . n n n a a a a a n      d

This problem is nice and easy. It gave us no problem. All of us got full scores in this problem. Nevertheless the problem is not entirely trivial, and indeed about 100 contestants scored nothing in this problem! First notice the middle term is not an arithmetical mean. Really during the question and answer period, some contestants did ask why the sequence doesn’t start at index 1. Moreover the problem is not exactly an algebra problem, as it involves a strictly

increasing sequence of integers. Try

small cases, say n = 1. Then we need a1 < a0+a1 sure, but not necessarily a0+a1 ”

a2, why is that so? For n = 2, then we need a2 < (a0+a1+a2)/2, or a2< a0+a1, not necessarily true, but say when compared with the case of n = 1, if it is false, then

a0+a1 ” a2 is true and we have an n satisfying the inequality! And the other side a0+a1+a2 ” 2a3, why true again? If it is false, look at the left hand side for the case of n = 3. After several attempts, we really see what is going on. Indeed the inequality is equivalent to nan < a0+a1+֥+an ” nan+1. The left hand

inequality corresponds to (a0+a1+֥+an )

í nan >0, while the right hand inequality

corresponds to (a0+a1+֥+an)ínan+1” 0,

same as (a0+a1+֥+an+1)í(n+1)an+1 ” 0.

Alas, if we define dn=(a0+a1+֥+an ) í nan, then we just have to show there

exists a unique n such that dn > 0 • dn+1!

The proof is then complete if we can see (prove) dn is a strictly decreasing

sequence of integers. Not too bad. Using induction, or other measures on the expression (a0+a1+֥+an)/n, our

team members managed to solve the problem.

Problem 4. Points P and Q lie on side

BC of acute-angled triangle ABC such

that _{ӘPAB = ӘBCA and ӘCAQ =} ӘABC. Points M and N lie on lines AP and AQ, respectively, such that P is the midpoint of AM, and Q is the midpoint of AN. Prove that lines BM and CN intersect on the circumcircle of triangle

ABC.

This is the easiest problem in the competition, yet about 30 contestants did not get anything from it. Altogether more than 10 solutions were received, using synthetic geometry, coordinate geometry, complex numbers and the like. Some of us did it by coordinate geometry, setting the foot of A be (0,0), and coordinates A(0,a), B(b,0) and

C(c,0). Then get everything out of it via

complicated calculations. But indeed if we can draw the picture properly, and do the angle tracings correctly, the problem is really not hard at all.

Indeed suppose BM and NC meet at S. Let ӘABC=ӘCAQ=ȕ and ӘACB= ӘBAP = Ȗ, then ȋABP~ȋCAQ. Hence

.

BP BP AQ QN

PM PA QC QC

Also, ӘNQC=ӘBQA=ӘAPC=ӘBPM. The last two statements imply ȋBPM ~

ȋNQC, hence ӘBMP=ӘNCQ. Then

we also have ȋBPM ~ȋBSC!

Finally, we have _{ӘCSB =ӘMPB = ȕ+Ȗ} =180_{°íӘABC. So ӘCSB+ӘBAC=180°} and we are done.

Editors: ㇄ 䤍 ㅆ(CHEUNG Pak-Hong), Munsang College, HK 浧 ⷟ 䦘 (KO Tsz-Mei)

㬐 拣 㰽 (LEUNG Tat-Wing)

㧝 ⋴ 影 (LI Kin-Yin), Dept. of Math., HKUST ⛂ 晰 㽱 (NG Keng-Po Roger), ITC, HKPU

Artist: 㯙 䱏 喀 (YEUNG Sau-Ying Camille), MFA, CU

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

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 November 20, 2014.

For individual subscription for the next five issues for the 14-15 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin LI, Math Dept., Hong Kong Univ. 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

J E S N M B C A Q P

Mathematical Excalibur, Vol. 19, No. 2, Sep. 14 – Oct. 14 Page 2 Problem 2. Let n • 2 be an integer.

Consider a n_{×n chessboard consisting} of n2 _{unit squares. A configuration of n} rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer k such that, for each peaceful configuration of n rooks, there is a k_×k square which does not contain a rook on any of its k2 _{unit squares.}

All of us managed to give (basically) the correct answer ( ª n 1 º ) and knew essentially how to tackle the question. There were gaps here and there and few points eventually deducted, but in my opinion, not really serious mistakes. Here n rooks are placed in a n×n board so that they are not attacking each other, and this time we ask for the largest possible gap (square with no rook). Of course the k2 squares should be congruent to others and the “gap” square should be in one piece. Indeed several candidates had the same concern. This is really a classical chess board problem and I am not at all sure if the question was asked before somewhere.

First, given a n×n board with n rooks non-attacking (peaceful configuration). Suppose l is such that l2 _{< n, then we} can find a l×l square with no rook in it. Indeed there is a rook in the first

column, consider the l consecutive rows starting with the row where the

particular rook is placed. Now remove the first n í l2 _{columns of this piece} (hence at least one rook is removed). The remaining l×l2 _{piece can be} decomposed into l l×l pieces of squares, but contain at most lí1 rooks, hence we have an empty l×l square. Now we want to construct a peaceful configuration with largest possible square of size ª n1º × ª n1º. Most of us see what the configuration should look like. We first let n be of the form l2_{. Label the square with row}

_i

and column j as (i,j), with 0 ” i ” lí1 and 0 ” j ” lí1. The rooks are then placed on the positions (il+j,jl+i), 0 ”

i,j ” lí1. One can easily check that any l×l square contains a rook.

Now comes where the most common gap lies. If n < l2_{, we need to produce a} peaceful configuration with no rook in any l×l square. The idea is of course to remove columns and rows from the previous construction. Only when (say)

the top row and the leftmost column removed, two rooks may be removed, we have to put a rook back to an appropriate position (naturally where it should be) to return to a peaceful configuration! (A 9_{×9 peaceful configuration with 2×2} squares as largest possible empty squares.)

ġ

Ź

ġ

Ź

ġ

Ź

ġ

Ź

ġ

Ź

ġ Ź

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Źġ

Problem 5. For each positive integer n, the Bank of Cape Town issues coins of denomination 1/n.Given a finite collection of such coins (of not necessarily different denominations) with total value at most 99+½, prove that it is possible to split the collection into 100 or fewer groups, such that each group has total value at most 1. I am happy to see how our students handled this problem. In short, they used various grouping and induction techniques and tricks, and changed the problem to a format they can handle, thus solved the problems. Even though our arguments were sometimes rather unclear and convoluted, thus some points deducted because of gaps and other things, four of us essentially solved the problem. Indeed the main idea of solving the problem is by “merging” or “cleaning” the set of coins. Clearly if the process can still be completed after merging the coins, it can be done before merging!

Indeed the problem can be generalized as follows. Given coins of total value at most

Ní½, they can be split into N groups each

of value at most 1. The problem then can be completed by the following steps. (i) Two coins of values 1/(2k) may be merged into a coin of value 2_{×1/(2k)=1/2,} thus for every even number

m

,

we may assume there is at most one coin of value 1/m.

(ii) For every odd number m, there are at most mí1 coins of such value, otherwise they can be merged to form a coin of value 1 first.

(iii) Coins of value 1 must form a group of itself. Thus if there are d coins of value 1 in a group of N coins, we might as well consider a group of Níd coins of values less than 1.

(iv) Now consider coins of values 1/(2kí1) and 1/(2k), with k=1,2,…,N. We first place them into N groups according different values of k. In each group, the total value is at most

1 1 1 1 (2 2) 1 1. 2 1 2 2 1 2 k k k k k  u      

The total value of all N groups is at most Ní½. By taking average, there exists a group of total value at most

1 1 1

( ) 1 .

2 2

N

N   N

(v) All the remaining coins are of values less than 1/(2N). We may put them one by one into each group, as long as the value of each group does not exceed 1í1/(2N) and we are done! The problem is meant to be a number theory problem, but is really more like a combinatorial problem. Our members managed to give different proofs to this problem and it is very nice. But indeed it is natural to consider coins of larger values (greedy method) first then consider coins of small values (a lot of them).

Problem 3. Convex quadrilateral

ABCD has ӘABC = ӘCDA = 90°.

Point H is the foot of the perpendicular from A to BD. Points S and T lie on sides AB and AD, respectively, such that H lies inside triangle SCT and ӘCHS í ӘCSB = 90°, ӘTHC í ӘDTC = 90°. Prove that line BD is tangent to the circumcircle of triangle TSH.

In these few years, problems of this kind appear rather frequently. Proving a certain line is tangent to a certain (hidden) circle, or two (hidden) circles will touch each other, or the like, are generally not too easy. Still one should be able to handle them by first finding out some related geometric properties, and then obtain final results still by using only basic geometric properties and techniques.

Let us look at this problem. It is not easy to draw an accurate and nice picture, let alone proving it.

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 November 20, 2014. Problem 451. Let P be an n-sided convex polygon on a plane and n>3. Prove that there exists a circle passing through three consecutive vertices of P such that every point of P is inside or on the circle.

Problem 452. Find the least positive real number r such that for all triangles with sides a,b,c, if a • (b+c)/3, then

c(a+bíc) ” r ((a+b+c)2_{+2c(a+cíb)).} Problem 453. Prove that there exist infinitely many pairs of relatively prime positive integers a,b with a>b such that b2_{í5 is divisible by a and}

a2_{í5 is divisible by b.}

Problem 454. Let ī1, ī2 be two circles with centers O1, O2 respectively. Let P be a point of intersection of ī1 and ī2. Let line AB be an external common tangent to ī1, ī2 with A on ī1, B on ī2 and A, B, P on the same side of line

O1O2. There is a point C on segment

O1O2 such that lines AC and BP are perpendicular. Prove that _ӘAPC=90°. Problem 455. Let a1, a2, a3, … be a permutation of the positive integers. Prove that there exist infinitely many positive integer n such that the greatest common divisor of an and an+1 is at

most 3n/4.

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

Solutions

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

Problem 446. If real numbers a and b satisfy 3a₊₁₃b₌₁₇a _{and 5}a₊₇b₌₁₁b_{, then}

prove that a < b.

Solution. Kaustav CHATTERJEE (MCKV Institute of Engineering College, India), Ioan Viorel CODREANU (Secondary School Satulung, Maramures, Romania), KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S4), Elaine LAM (Tsuen Wan Secondary School),

Corneliu MĂNESCU-AVRAM (Transportation

High school, Ploieúti, Romania), NGUYÊN Viêt Hoàng (Hà Nôi, Viêt Nam), PANG Lok Wing, YAN Yin Wang (United Christian College (Kowloon East), Teaching Staff) and Simon YAU.

If a • b, then 3a₊₁₃a _{• 3}a₊₁₃b₌₁₇a_{. (*)}

Since 3/17<13/17<1, the function f(x) =(3/17)x_+(13/17)x _{is strictly decreasing.} By (*), f(a) • 1> f(1). So a < 1. Next, 5b₊₇b _{” 5}a₊₇b _{= 11}b_{. (**) Since} 5/11 < 7/11 < 1, the function g(x) = (5/11)x + (7/11)x _{is strictly decreasing. By (**),} g(b) ” 1 < g(1). So b>1>a, contradiction. Other commended solvers: Math Activity

Center (Carmel Alison Lam Foundation Secondary School), Nicuúor ZLOTA (“Traian Vuia” Technical College, Focúani, Romania), Titu ZVONARU (Comăneúti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).

Problem 447. For real numbers x, y, z, find all possible values of sin(x+y) + sin(y+z) + sin(z+x) if . ) sin( sin sin sin ) cos( cos cos cos z y x z y x z y x z y x        

Solution. KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S4), Corneliu MĂNESCU-AVRAM (Transportation High school, Ploieúti, Romania), YAN Yin Wang (United Christian College (Kowloon East), Teaching Staff), Titu ZVONARU (Comăneúti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania). Let S=x+y+z. Cross multiply and transfer all terms to one side. We get

0 = sin S cos x í cos S sin x + sin S cos y í cos S sin y + sin S cos z í cos S sin z = sin(Síx) + sin(Síy) + sin(Síz) = sin(y+z) + sin(z+x) + sin(x+y).

Other commended solvers: Kaustav

CHATTERJEE (MCKV Institute of

Engineering College, India), Ioan Viorel CODREANU (Secondary School Satulung, Maramures, Romania) and Math Activity Center (Carmel Alison Lam Foundation Secondary School).

Problem 448. Prove that if s,t,u,v are integers such that s2_í2t2_+5u2_í3v2_=2tv, then s = t = u = v = 0.

Solution. Ioan Viorel CODREANU (Secondary School Satulung, Maramures, Romania), KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S4), Corneliu MĂNESCU-AVRAM (Transportation High school, Ploieúti, Romania), Math Activity Center (Carmel Alison Lam Foundation Secondary School), NGUYÊN Viêt Hoàng (Hà Nôi, Viêt Nam), YAN Yin Wang (United Christian College (Kowloon East), Teaching Staff), Titu ZVONARU (Comăneúti, Romania) and

Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania). Assume s,t,u,v are not all zeros. By cancelling all common factors of s,t,u,v, we may assume they are relatively prime. We can rewrite the equation as 2(s2_+5u2_{) = (2t+v)}2_+5v2_{. (†)} For 0 ” x, y ” 4, we have 2x2 _Ły2 _{(mod 5)} if and only if x Ł y Ł 0 (mod 5). (‡) So s2_+5u2 _{Ł 2t+v Ł 0 (mod 5), which} implies s = 5m and 2t+v = 5n for some integers m,n. Substituting these into (†), we get 2(5m2_+u2_)=5n2_+v2_{. By (‡), u, v} are divisible by 5. Then s,t,u,v are divisible by 5, contradicting they are relatively prime. So s,t,u,v are all zeros.

Other commended solvers: Kaustav

CHATTERJEE (MCKV Institute of Engineering College, India),

Problem 449. Determine the smallest positive integer k such that no matter how {1,2,3,…,k} are partitioned into two sets, one of the two sets must contain two distinct elements m, n such that mn is divisible by m+n.

Solution. Titu ZVONARU

(Comăneúti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania). Call distinct positive integers m,n a

good pair if mn is divisible by m+n.

Collect all good pairs with m,n”40. We will try to separate m,n first. Let A ={1,2,3,5,8, 10, 12, 13, 14, 18, 19, 21, 22, 23, 30, 31,32,33,34} and B = {4, 6, 7, 9, 11, 15, 16, 17, 20, 24, 25, 26, 27, 28, 29, 35, 36, 37, 38, 39}. Each of A and B do not contain any good pair. For 1 ” k ” 39, we can remove integers greater than k from A and B to get 2 disjoint subsets of {1, 2, …, k} with no good pair in each subset.

For k=40, put 6, 12, 24, 40, 10, 15 and 30 around a circle. Notice any two consecutive terms in this circle is a good pair. No matter how we divide {1,2,…,40} into 2 disjoint subsets, one of the subsets will contain at least 4 of 7 numbers in the circle. So there will be a good pair in that subset. Therefore, 40 is the desired least integer.

Other commended solvers: NGUYÊN

Viêt Hoàng (Hà Nôi, Viêt Nam). Problem 450. (Proposed by Michel

Mathematical Excalibur, Vol. 19, No. 2, Sep. 14 – Oct. 14 Page 4 with no right angle and O be its

circumcenter. For i = 1,2,3, let the reflection of Ai with respect to O be Ai'

and the reflection of O with respect to line Ai+1Ai+2 be Oi (subscripts are to be

taken modulo 3). Prove that the circumcenters of the triangles OOiAi'

(i = 1,2,3) are collinear. O A₃ A₁ A₂ A'₁ H M₁ O₁ J₁ I₁

Solution. KWOK Man Yi (Baptist

Lui Ming Choi Secondary School, S4). Notice that O1 is the reflection of O with respect to the midpoint M1 of A2A3. By the nine point circle theorem (see

Math Excalibur, vol.3, no 1, p,1), AH, OM1 are parallel and their lengths are 2:1. Now A1O=OA1ƍ. So, in ȋA1A1ƍH,

M1 is the midpoint of A1ƍH, i.e. H is the reflection of A1ƍ with respect to M1. Let I1 be the circumcenter of ȋOO1A1ƍ. Then I1 lies on the perpendicular bisector A2A3 of OO1. Reflect I1 with respect to M1 to J1. Then J1 also lies on

A2A3. With respect to M1, J1 is the circumcenter of the reflection of ȋ

OO1A1ƍ, i.e. ȋOO1H. So, J1 also lies on the perpendicular bisector of OH. Define I2, I3, J2, J3 similarly. As J2, J3 also lie on the perpendicular bisector of

OH by a similar proof, J1, J2, J3 are collinear. Then by Menelaus’ theorem,

. 1 2 3 3 1 1 2 2 3 3 1 1 2 _{˜ ˜ } A J J A A J J A A J J A As A3I1/I1A2=A2J1/J1A3 (due to I1, J1 are reflection of the midpoint of A2A3) and similarly for I2, J2, I3, J3, we have

. 1 1 3 3 2 3 2 2 1 2 1 1 3 _{˜ ˜ } A I I A A I I A A I I A

By the converse of Menelaus’ Theorem,

I1, I2, I3 are collinear as desired.

Other commended solvers: Andrea

FANCHINI (Cantú, Italy), Corneliu Mănescu-Avram (Transportation High school, Ploieúti, Romania), NGUYÊN Viêt Hoàng (Hà Nôi, Viêt Nam), Samiron SADHUKHAN (Kendriya Vidyalaya, Barrackpore, Kolkata, India), Titu

ZVONARU (Comăneúti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).

Olympiad Corner

(Continued from page 1)

Problem 5. (Nikolay Nikolov) Find all functions f : _ч+ _{ĺ ъ+ such that}

f(xy) = f(x+y)(f(x)+f(y)) for any x,y _Ӈч+_. Problem 6. (Nikolay Beluhov) The quadrilateral ABCD is inscribed in the circle k. The lines AC and BD meet in E and the lines AD and BC meet in F. Show that the line through the incenters of ǻABE and ǻABF and the line through the incenters of ǻCDE and ǻCDF meet on k.

IMO2014 and Beyond (II)

(Continued from page 2)

First, let the line passing through C and is perpendicular to SC meets AB at Q. Then ӘSQC=90°íӘBSC=180°íӘSHC. So C, H,

S, Q are concyclic. Moreover SQ is a

diameter of this circle, thus the circumcenter K of SHC lies on AB. Likewise, circumcenter L of the circle

CHT lies on AD. To show the circumcircle

of the triangle SHT is tangent to BD, it suffices to show the perpendicular bisectors of HS and HT meet at AH. But the two perpendicular bisectors coincide with the angle bisectors of AKH and ALH, thus by the bisector theorem, it suffices to show AK/KH=AL/LH. Let M be the midpoint of CH, then B,C,M,K are concyclic, L,C,M,D are concyclic. By the sine law, AK/AL= sin_{ӘALK / sinӘAKL =} (DM/CL)/(BM/CK) = CK/CL = KH/LH. Problem 6. A set of lines in the plane is in

general position if no two are parallel and

no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite areas; we call these its finite regions. Prove that for all sufficiently large n, in any set of n lines in general position it is possible to color at least ¥n of the lines blue in such a way that none of its finite regions has a completely blue boundary.

Notes: Results with ¥n replaced by c¥n

will be awarded points depending on the value of the constant c.

I have to admit that I don’t like this

problem at all. Indeed it was meant to be an “open end” problem, that students may produce different results with different degrees of difficulty. But when I first saw the problem, I thought we should give an algorithm, say a greedy algorithm, or other heuristic that gives good pattern (with as many blue colored lines as possible), and then analyze the pattern and give an estimate. Not so. (I guess I have become kind of intuitionist.) I doubt if there was any algorithmic solution anyway. Indeed in the official solution, a best possible solution is assumed, surely it exists, but we were not told how to get there.

Let me reproduce a part of the proof as follows. Given a set of n lines colored blue and red, and the lines colored blue is as large as possible (maximality argument), so that every finite region still has at least one boundary line colored red. Assume k lines are colored blue. Call a vertex which is the intersection of two blue lines blue as well, so there are kC2 blue vertices. Now take any red line l, using the maximality argument, there exists at least one region with this red line l as the only red side, (for if all regions have two or more red lines, surely we can change one more red line to blue). In this region there is at least one blue vertex v since any finite region has at least three lines. We then associate the blue vertex with the red line. Now finally every blue vertex v belongs to four regions, (some may be unbounded), hence it may be associated with at most four red lines. Therefore the total number of red lines is at most 4kC2=2k(kí1).

On the other hand, there are ník red lines, thus, ník ” 2k(kí1). Solving for

n, we get n ” 2k2_{ík ” 2k}2_{. Hence, k •} ª n/2 º and we get an estimate on the number of blue lines!

l

IV II

III I

v

By putting some weights on the blue vertices, or by refining local analysis, one may get the stronger result k•¥n.

Problem 1. (Teodosi Vitanov,_ Emil

!(~lev)

Given ls,a. ·cirgle k and a point A

outside of it. The segment BC is a diameter of k. Find the loc::us of the ofthocenter

of

6.ABC, when BC is changing.

First solution. Let 0 be the center of k and

w is the circle with diameter AO. Let AA1 and

BB1 be the altitudes of D..ABC and H is the

'orthocenter. Then the power of H with respect·, 1

tow is equal to

A.H·.

HA1 and the power of H

with respect to k is equal to BH · HB1. Since

AH · HA1 =·· BH · HB1 it fqllows that H lies

on the radical axis l of k and

w.

Cdnversely, it is

easy to see that each point of l is an orthocenter

of some triangle from given type. · ).

Second solution. Let ABnk =

B',

AC'nk

=

C' B

and BC'

n

B' C

=

H. Then H is the orthocenter

of D..ABC and H lies on the polar l of point A

with respect to k ( Wl)y?). Conversely, it is easy

to see that each point of l is an otthocenter of

some triangle from given type.

A

G

Problem 2. (Nikola,y Beluhov) Consider a rectangular n x m table where n;:: 2

and m ;:: 2 are positive integerS. Each cell is colored in one of the four colors:

white, green, red or blue. Call such a coloring interesting if any 2 x 2 square

contains every color exactly once. Find the number of interesting colorings.

Solution. Number the ~lumns of the table by .1, 2, .... , m and its rows by·

1, 2, ... , nand denote the cell in the i-th column and j~th row by (i,j).

. Consider an interesting coloring S. We show first th.at eith.er aily row in S

contains only two colors or any column in S contains only two colors.

Suppose S dpes .not contain .a rectangle 1 X 3 (with longer horizontal side)

containing 'three distinct colors. It follows that in any row of S we have two

alternating colors, as rieeded. . . . .

Let the cells (u, v),. (u+ 1, v) aild (u+2,v) b.e ¢<W¢i:~il..in three distinct colors.

·a, b and G· We obtain that (

u

+

1,

v,+

1) is !;ol?~ed in

t.he

f6urth color d, ( u,

v

+

1)

-inc, (u.+

2,

v+

1) -in a, (;i4-1,v +2)-in b, (u,·v+2)-in a, (u+2, tJ+ 1)-in

c, and so on. By--analogy w_e obtain that (u+l, v-1) is in color d, (u, v -1)-in

c, ( u

+

2, v -'- 1) - in a and so on. We conchide that column u contains only calm's.

a and c, column u

+

1 contains only colors,b and d, and column tt

+

2 contains

only colors a and c. It follows ti()W that !),rty•column having the same parity as u

contains only colors a and c;, a.il.d any coiurim ·having the parity of u

+

1 contains

only colors b and d.

Hence, either any row in S contains only two colors or any c,olumn in S

. contains only two colors.

The number of interesting colorings such that in any collllllil of odd number

two of the colors alternate and in any column of even parity the other two colo'rs

alternate is equal to

m

>< 2"":. The corresponding colorings when the rows arfil

colored in two colors equal

(i)

x 2n. The colorings in which the color of the cell

( i, j) depends only on the parity. of i and j are counted in both cases. Therefore,

the number of interesting cc;>lorings equa~ 6 x (2m+ 2n)-24.

· Problem 3. (Alexander Ivanov) A real nonzero number is assigned to every

point in t.he space. It is known that tor any tetrahedron r the number written in

the incenter equals the product of the four number~ written in the vertices of r.

Prove that all numbers equal 1.

~O.J~~i~~·

'-.

q~:~~~?~J·

t'Yg_

a.r:ki.~~~:?· poil}t.~,.

;X.

apd,.}j ..

·~d

let x and .11 be the

cottespondifig h'uli:lbers. dhoose pomts I and J on the hne XY such that XY

=

·yi = IJ. Let X' on the line XY be.such that XI= JX'. Consider the plane

>.perpendicUlar to IJ and passing through the 'midpoint of IJ. Let ABGX and

ABCX' be two equal regular triangular pyramids with base ABC in the plane

>.

having incenters I and J. Slnce.·nAl?,~'lianx = nr and nAnBnanxr = nJ we h ave th a t

nx

=

n"t.ni n • .,_;--· .. · .... ;-. . .. ·<: · <: •... .

J .

Move the plane>. towards point I and consider the spheres Sr and SJ with

~-e~ter~Iand J,·r,~sp~~~~Vf3!r t~,at:,a~e ta,p.~en,.t·to,>.= .P~t· A1lhf)rX{ beregulat;

tri-; a.ng-u1a.r

pyriirii:idtMtJ:i

oiisifA1mi

c

i

·hi

xran.tl

ins.PMr'Ei

S'J:

TheteguW

tri!ing\.Ilar

pyramid with ba.cie A1B1G1 and insphere Sr has vertex X1. It follows from the

b th t nxt.ni · . ·

a 9Ye .. a p.x1.,,~ ,· n , ,P,ttX•: . . .

:: ;,! ·:· -: . .... J ·~~·:(·· ;f:.·.:: ·:: (:'·;:· · .. ;· /'. . . •, ·' .· .:· ., .

Wheii

>.

inoves towa.tds.'lfthe radius df the sphere

Si

tends to zero and

:6A1B~ C1 tends to triangle which is the base' 6f regular· triangular pyramid with

vertex X' and inscribed sphere with center J and radius I J;

We conclude that X1. tends to I. Continuity arguments show that .. ail inner

points on the segillent XI (with point X) are assigned with the same number.

Thus, x ,;, y and all numbers are equal. It follows from x4 _{== x and x =/=}

_b

_that

X= 1..

Probleni 4 .. (Peter Bovval~nkpv) Fi~d all::prime:,)J,tliiJ,~~rs. p ·and·q such that

p?-_141~·+

t¥ci.

~2IP.6·~ 1. . .. . . .·. . ..

Solution. If p

=

3, then qfl36 _{- 1}

₌

₇₂₈

=

₂3_{.7.11 and therefore q ~ 2 which}

gives a solution. Let p

f.

3. Since (q

+

1, q2

- q

+

1) = 1 ·or 3, we have p2lq

+

1 or

p2lq2- q

+

1, which implies that p

<

q. If p

+

1

=

q, then p

=

2 and q

=

3 which

is another solution. In the sequel we assume that q ;::: p

+

2.

Since q2_lp6_-1

₌

_(p-1)(p+l)(p2_-p+1)(p2_{+p+1) and (q,p-1)}

₌

_(q,p+1)

₌

1, we have q2j(p2- p+ l)(p2 +p+ 1). Moreover, we have (p2 -p+ l,p2

+p+

1) =

(p2 _{+ p + 1, 2p) = 1 and therefore q}2_lp2

- p + 1 or q2jp2 + p + 1. However,

p2 - p + 1

<

p2

<

q2 and (p + 2)2 ::;; q2 ::;; p2 + p + 1, i.e. 3p + 3

<

0, a

,'f:rciblem 5.

(Nik~lay

Nikolov)Find all functiohs f: Q.+ - t

a+ such that _

.. ::::,:

;•, · f(xy) = f(x+v)(f(x)

+J(y)),

foi' ariy~,y E

i]+.

':·~.,.··.

(=+:) g(x

+

y)g(:v)f!(Y)

=

g(:t.y)(g(c)+ g(y)).

. ' 1 ' - . . ' . ': : . . - ·. '

Let f(l)

=-

c

>

0 i,e. g(l)

=

-.'From (:f) we get g(x

+

1)

=

cg(x)

+

1 ·and

c . . . lienee g(~ = 2, g(3)

=

~c

+

1, ~(4)

= 2¢

+

Q

+

1₁ 'g(5) ::: 2c3

+

c?

+

c

+

1, g( 6) = 2c

+

c3

+.

c?

+

c

+

L On

th§

gth~ h~nct ~!"lttfng a; = 2,

v

= 3 in (:f) lead us to .,·:r,· g(Q)g(~)fl(~) = p(t5)(

₉

(~)

-r

.9(3)),. which implies ... . 1 Therefore c

=

1 or -. . . . 2 . .. ,' .·: . .

If c::: 1 then g(x +I)= g(x) + 1. By induction g(n) = n for any n EN a:rld

moteover g(x +n)

=

g(x)

+n

for any'~ 6 Q't-and

n

EN.

By setting y = n in ( *) we obta!ri ~- . . .

· (g(x)

+n)g(~)n:;, g(ft~)'(g(;)

+

n)l

i.e. g(nx}:= ng(x).

S~ttiiigx_=.pfij ancfn'~

q,

w~erep,'q

EN we obtain g(x) = x,

for any x

E

Q+, i.e. f(x)

=

1/x, fur

any :v E. Q+. :, · .

. 1 ' 1 .. ·.. . . . ' .· . If 'c ':"

2-

then g(:t + 1) ~

₂

g(x)

+

1. Hen~t') . . ·.·. ': L ·• g(a;)22 .· :/:;, ··+- · . ' .. g(ri) = 2 . and g(x

+

n) ~ 2;;:: . 2n , , x E Q , n EN; Setting y.

=

n in (:f) we obtain 2g(x +n)g(x);., g(nx)(g(x)

+

2),

and it's

suffi.ci~ntto

.see that these equations imply g(x) = 2, for a.n_y' X E Q+,

J.e.

f

=

1/2.

. . . 1

f

1,

;,J!inally! the ?~;Y,:~?~ti.~ns sa~i~fyingth~.¢i~en ,Eiquali~ are

f

=

2

.and . ==· ;·

·~·~·;k:·:},:{:.;~.'.l;-'; ·, ·~-,~f'.:·': .. !;,::.:.·. :<.··:·'· ~... 11 · : . ,·\. ·• ~ ·:>·· .... :: '. . ~. ~ : . .

Remark: ·Fn fact cill ju:nctions f : JR+ - t lR satisfying the given equality are

f

=

0, ·

f

=

~'and}==i

t\ · ·.

Problem

6:

{NikolayB~tuho~) Til~ qtiadril~teral

AB01J

is

inscribed in the Circle

k. The lines AG and BD meetin E and the lines AD and BO meet' in F. Show .

that. the line through th~ mc~rit~rs of

·LiAEE

arid h.ABP a,r;d the line through

the incenters of .6.GDE and .6.CDF meet on /c.

Sohttfo'ti;

'te{ie~·Ij;

Je,

J1

beth~

incenters of .6.ABE, LABF, LODE, .6.CDF,

. resP,~gtiv:~l,r,.;J:,et

P,

=; 4-Ie

n

Bit, Q

=·A!!

n

BI.,, Up= CJe

n

DJJ; a.ncl V =

OJt f!DJe be the: ex~~nte~ of~AB?.9pposit~· to A, the excenter of 61-BD

opposit!l to B, the mcenter of fiACD, anl:l the mcenter of .6.BOD, respectively .

. : :·: .::.':.:' ~} .. ·.; : •.':, .: {: ·. ·.·~ ':·:· :::· ·~ . ;.;:..:... '': .: .. ; .... \ :: . ' ... ·... ' . . . : ., ' . . '

~

.

W~ h~ye ~A,P

1? ;::::

~ A~;;;, <tAQ$; ~he.r~fore, ABPQ is cyclic. Analogously,

GDUV is cyclic.·. . - . ·

··:.···~.'..; ·;·::.·.·:';:: .. :.• ::'::.;--. ~-· :::'· ... : :' · . : " , . : . ..--...· ' : : .

Wehave <tA..:f'B =

!

.Af77=

<tUCV, <tAQ!3 7 tAB= <tUDV; <l:APQ =

... : . ...-....: . .: . ·. .

-

..--...

<t.ABQ :::=

!

AD= <tUOD,

an~

-t.BQP = <l:BJ:\.P

=

!

BC= -t.V DC. Therefore,

the figur{l_s ABPQ and UVOD are ·simil!ll' and identically oriented. Let T be

their. ~~nfer.ofsi!Ullituqe, {f:e:i letT beth~ tiJted pojnt of the unique similitude

which i:naps· A, J3, P, Q onto

U,

V, O, D, respectively.) .. ·

\i/b

have LTAQ

'IV<iiun;':th~t~fbre',

L.rA.u

~-

.6.TQD and -t.ATQ

=

.

. ~ ·.·'· .

<l:(AU, QD)

=

!

AD= <l:Ait,Q; It foll()w:s that T lies pn the circumcircle of

LAI,Q. Analogously, T lies on the circumcircle of b.I{fe.P, LGJeV, a.nd f:?,DJ"U.

::. . ' . . ·:.._ ~ I . . . ; :· . , : . . . ·. . . , . ,...__

·.·we.

have, them, <XATB =· <XATie

+

<Xfe.TB

=

<XAL;?$,

+

·-t.APB =

!

AB

·~ .. ,.-:*-..: .• ~ :: ~··::.?-.. . ·. ' . : . ; ' .. •' . . ' : . . ' ' . '

+!AB~AB, sh6wing that Tlies on k. p

·.Let T' E IJI;' so

~hat

I1A · I1Q = _{Ifle . I1T}1 _{= I}

1B . I1P. Then T' is a

pomt other than Ie whiCh lies on the circumcircles of both .6.A1 Q and 6BI

P·

therefore, T1

=

T and T E IeiJ, e " '