Volume 17, Number 1 May-August 2012
IMO 2012 (Leader Perspective)
Tat-Wing Leung
Olympiad Corner
Below are the problems of the 2012 International Math Olympiad.
Problem 1. Given triangle ABC the point J is the centre of the excircle opposite the vertex A. This circle is tangent to the side BC at M, and to the lines AB and AC at K and L, respectively. The lines LM and BJ meet at F, and the lines KM and CJ meet at G. Let S be the point of intersection of the lines AF and BC, and let T be the point of intersection of the lines AG and BC. Prove that M is the midpoint of ST.
(The excircle of ABC opposite the vertex A is the circle that is tangent to the line segment BC, to the ray AB beyond B, and to the ray AC beyond C.)
Problem 2.Let n ≥ 3 be an integer, and let a2,a3,…,an be positive real numbers such that a2a3⋯an =1. Prove that
(1+a2)2(1+a3)3⋯(1+an)n > nn. Problem 3. The liar’s guessing game is a game played between two players A and B. The rules of the game depend on two positive integers k and n which are known to both players.
(continued on page 4) 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, 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 Septembert 20, 2012.
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, 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
As leader, I arrived Mar del Plata, Argentina (the IMO 2012 site) four days earlier than the team. Despite cold weather, jet lag and delay of luggage, I managed to get myself involved in choosing the problems for the contest. Once the “easy” pair was selected, the jury did not have much choice but to choose problems of possibly other topics for the “medium” and the “difficult” pairs. The two papers of the contest were then set. We had to decide the various official versions and the marking scheme of the contest. After that, I just had to wait for the contestants to finish the contest and get myself involved in the coordination to decide the points obtained by our team. Here I would like to discuss the problems. (Please see Olympiad Corner for the statements of the problems.)
Problem 1. Really problem 1 is quite easy, merely a lot of angle chasings and many angles of 90° (tangents) and similar triangles, etc, and no extra lines or segments needed to be constructed. First note that ∠AKJ =∠ALJ = 90°, hence A,K,L,J lie on the circle ω with diameter AJ. The idea is to show that F and G also lie on the same circle. Looking at angles around B, we see that 4∠MBJ +2∠ABC = 360°. Thus∠MBJ = 90°− ½∠ABC. Also,∠BMF=∠CML = ½∠ACB (as CM=CL). Then ∠LFJ =
∠MBJ −∠BMF = ½∠BAC.
Thus ∠LFJ =∠LAJ. Hence, F lies on ω. By the same token, so is G. Now AB and SB are symmetric with respect to the external bisector of ∠ABC, so is BK and BM. Now SM = SB+BM = AB+BK =AK. Similarly, TM=AL. So SM=TM. It is relatively easy to tackle the problem using coordinate geometry. For instance, we can let the excircle be the unit circle with J=(0,0), M=(0,1), BC is aligned so that B=(b,1) and C=(c,1). Coordinates of other points are then calculated to verify the required property. But one must be really careful if he tries to use coordinate method. It was somehow decided that if a contestant cannot get a full solution using coordinate method, then he will be “seriously penalized”!
Problem 2. As it turned out, this problem caused quite a bit of trouble and many students didn’t know how to tackle the problem at all. More sophisticated inequalities such as Muirhead do not work, since the expression is not “homogeneous”. The Japanese leader called the problem a disaster. There were trivial questions such as “why is there no a1?” A more subtle issue is how to isolate a2,a3,…,an. Clearly(1+a2)2 ≥ 22a2 by the AM-GM
inequality. But how about (1+a3)3?
Indeed the trick is to apply AM-GM inequality to get for k=2 to n−1,
1 1 1 1 1 1 ) 1 ( + + + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + + = + k k k k k k a a L . ) 1 ( ) 1 ( 1 1 1 1 1 k k k k k k k k a k k a k + + + + + = + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + ≥
By multiplying the inequalities, the constants cancelled out and we get the final inequality. That the inequality is strict is trivial using the conditions of AM equals GM. The above inequality can also be used as the inductive step of proving the equivalent inequality
(1+a2)2(1+a3)3⋯(1+a
n)n > nna2a3⋯an. w T S G F K M L J A B C
Mathematical Excalibur, Vol. 17, No. 1, May-Aug. 12 Page 2 Problem 3. Comparing with problem 6,
I really found this problem harder to approach! Nevertheless there were still 8 contestants who completely solved the problem. Among them three were from the US team. That was an amazing achievement!
We can deal with this combinatorial probabilistic problem as follows. Ask repeatedly if x is 2k. If A answers no k+1 times in a row, then the answer is
honest and x≠2k. Otherwise B stops
asking about 2k at the first time answer yes. He then asks, for each i=1,2,…,k, if the binary representation of x has a 0 in the i-th digit. Whatever the answer is, they are all inconsistent with a certain number y in the set {0,1,2,⋯, 2k−1}.
The answer yes to 2k is also
inconsistent with y. Hence x≠y. Otherwise the last k+1 answers are not honest and that is impossible. So we find y and it can be eliminated. Or we can eliminate corresponding numbers with nonzero digits at higher end. Notice we may need to do some re-indexing and asking more questions about the indices of the numbers subsequently. With these questions, we can reduce the size of the set that x lies until it lies in a set of size 2k.
Part 2 makes use of a function so that using the function, A can devise a strategy (to lie or not to lie, but lying not more than k times consecutively) so that no extra information will be provided to B and hence B cannot eliminate anything for sure. Due to limit of space, I cannot provide all details here.
It was decided that part 1 answered correctly alone was worth 3 points and part 2 alone worth 5 points. But altogether a problem is worth at most 7 points. So 3 + 5 = 7! At the end it really did not matter. After all, not too many students did the problem right.
The problem is noted to be related to the Lovasz Local Lemma. See N. Alon et al, The Probabilistic Methods, Wiley, 1992. In the book it seems that there is an example that deals with similar things. One may check how the lemma and the problem are related!
Problem 4. Despite being regarded as an easy problem, this problem is not at all easy. It is much more involved than expected. Also this problem eventually
caused more trouble because of the disputes about the marking scheme. First, by putting a=b=c=0, one gets f(0)=0. By
putting b=−a and c=0, one gets f(a) =
f(−a). More importantly, by putting
c=−(a+b) and solving f(a+b) = f(−(a+b)) as a quadratic equation of f(a) and f(b), one gets
( ) ( ) ( ) 2 ( ) ( ).
f a b+ = f a +f b ± f a f b
Putting a=b and c=−2a into the original equation, one gets f(2a)=0 or f(2a)=4f(a). Now the problem becomes getting all possible solutions from these two relations. Using the two conditions, one checks that there are four types of solution: (i) f1(x)≡0 , (ii) f2(x) = kx2 , (iii) 3 0, even ( ) , odd x f x k x ⎧ = ⎨ ⎩ and (iv) 4 0, 0 (mod 4) ( ) , 1(mod 4). 4 , 2(mod 4) x f x k x k x ≡ ⎧ ⎪ =⎨ ≡ ± ⎪ ≡ ⎩
The “k” in the solutions is essentially f(1). Indeed if f(1)=0, then f(2)=0, one then show by induction f(x)=0 for all x. (Or by showing f(x) is periodic of period 1.) Now if f(1)=k, using the condition f(2a)=0, one can show again by induction f(x) is k for x odd and is 0 for x even. Now if f(1)=k and f(2)=4k, then f(4)=0 or 16k. In the first case we get a function with period 4 and arrive at the solution f4(x). In the second case we get f2(x). (One needs to verify the details.) By checking the values of a, b and c mod 2 or 4, or other possible forms, one can check the solutions are indeed valid. Eventually if a contestant claimed that all the solutions are easy to check, but without checking, one point would be deducted. If a contestant says nothing about the solutions satisfy the functional equation and check nothing, then two points would be deducted!
Problem 5. The following solution was obtained by one of our team members.
Extend AX to meet the circumcircle of ABC at A’, likewise extend BX to meet the circle at B’. Now extend AB’ and BA’ to meet at H, which is exactly the orthocentre of ABX and it lies on the extension of DC.
Since BK2 = BC2 = BD·BA, we have ∆ABK~∆KBD, so ∠BKD =∠BAK = ∠BHD, which implies B, D, K, H concyclic. So ∠BKH =∠BDH=90°. This implies HK2 = BH2 −BK2 = BH2− BD·BA = BH2 − BA’·BH = HA’·HB.
Similarly HL2= HB’·HA. But HA’·HB
= HB’·HA. Hence HK=HL. Using similar arguments as above, we have ∠ALH = 90° ( =∠BKH.) Along with
HK=HL, we see ∆MKH≅∆MLH.
Therefore, MK=ML.
Problem 6. Clearing denominators of
1 2 1 2 ... 1, 3a 3a 3an n + + + =
one gets x1+2x2+⋯+nxn=3a, where
x1,x2,…,xn are non-negative integer powers of 3. Taking mod 2, one gets n(n+1)/2 ≡ 1(mod 2). This is the case only when n ≡ 1, 2 (mod 4). The hard part is to prove the converse also holds. The cases n=1 or 2 are easy. By trials, for n=5, (a1,…,a5)=(2,2,2,3,3) works. The official solution gave a systematic analysis of how to obtain solutions by using identities 1/2a=1/2a+1+1/2a+1 and
w/3a=u/3a+1+v/3a+1, where u+v=3w.
For n=4k+1≥5, one can arrive at the
solution a1=2=a3, a2=k+1, a4k = k+2 = a4k+1 and am = [m/4]+3 for 4≤m<4k.
Similarly, for n=4k
+
2≥6, one canarrive atthe solution a1 = 2, a2 = k+1, a3 = a4 = 3, a4k+1 = k+2 = a4k+2 and am = [(m−1)/4]+3 for 4<m≤4k. One can check these are indeed solutions by math induction on k. In the inductive steps of both cases, just notice a2, an−1, an are increased by 1 so to balance the new an+1, an+2, an+3, an+4 terms.
This reminds me of the 1978 USAMO problem: an integer n is called good if
we can write n=a1+a2+⋯+ak, where
a1,a2,…,ak are positive integers (not necessarily distinct) satisfying
. 1 1 1 1 2 1 = + + + k a a a L
Given 33 to 73 are good, prove that all integer greater than 33 are good. The idea there is to show if n is good, then 2n+8 and 2n+9 are good by dividing both sides of the above equation by 2
and adding the terms 1/4+1/4 and
1/3+1/6 respectively. H A' B' M L K D A C B X
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 September 20, 2012. Problem 396. Determine (with proof) all functions f : ℝ→ℝ such that for all real numbers x and y, we have
f (x2 + xy + f (y)) = (f(x))2 + xf(y) + y.
Problem 397. Suppose in some set of 133 distinct positive integers, there are at least 799 pairs of relatively prime integers. Prove that there exist a,b,c,d in the set such that gcd(a,b) = gcd(b,c) = gcd(c,d) = gcd(d,a) = 1.
Problem 398. Let k be positive integer and m an odd integer. Show that there exists a positive integer n for which the number nn−m is divisible by 2k. Problem 399. Let ABC be a triangle
for which ∠BAC=60°. Let P be the
point of intersection of the bisector of ∠ABC and the side AC. Let Q be the point of intersection of the bisector of ∠ACB and the side AB. Let r1 and r2 be the radii of the incircles of triangles ABC and APQ respectively. Determine the radius of the circumcircle of triangle APQ in terms of r1 and r2 with proof.
Problem 400. Determine (with proof) all the polynomials P(x) with real coefficients such that for every rational number r, the equation P(x) = r has a rational solution.
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Solutions
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Problem 391. Let S(x) denote the sum of the digits of the positive integer x in base 10. Determine whether there exist distinct positive integers a, b, c such that S(a+b)<5, S(b+c)<5, S(c+a)<5, but S(a+b+c)>50 or not.
Solution. AN-anduud Problem Solving Group (Ulaanbaatar, Mongolia), CHEUNG Ka Wai
(Munsang College (Hong Kong Island)), LI Jianhui (CNEC Christian College, F.5), LO Shing Fung (Carmel Alison Lam Foundation Secondary School), Andy LOO (St. Paul’s Co-educational College),
YUEN Wai Kiu
(
St. Francis’ CanossianCollege) and ZOLBAYAR Shagdar (9th
grader, Orchlon International School, Ulaanbaatar, Mongolia).
Yes, we can try a=5,555,554,445 and b=5,554,445,555 and c=4,445,555,555. Then S(a+b)=S(11,110,000,000)=4, S(b+c)=S(10,000,001,110)=4, S(c+a)=S(10,001,110,000)=4. Finally, S(a+b+c)=S(15,555,555,555)=51.
Other commended solvers:Alice WONG
(Diocesan Girls’ School), Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
Problem 392. Integers a0, a1, ⋯, an are all greater than or equal to −1 and are not all zeros. If
a0+2a1+22a2+⋯+2na n= 0, then prove that a0+a1+a2+⋯+an>0. Solution. AN-anduud Problem Solving
Group (Ulaanbaatar, Mongolia), Kevin LAU (St. Paul’s Co-educational College, S.3), Simon LEE (Carmel Alison Lam Foundation Secondary School), Harry NG Ho Man (La Salle College, Form 5), SHUM Tsz Hin (City University of Hong Kong), Alice WONG (Diocesan Girls’
School), ZOLBAYAR Shagdar (9th
grader, Orchlon International School, Ulaanbaatar, Mongolia),Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania). For all the conditions to hold, n ≠ 0. We will prove by mathematical induction. For n=1, if a0+2a1=0, then the conditions on a0 and a1 imply a0 is an even positive integer. So a0+a1 = a0/2 > 0. Suppose the case n=k is true. For the case n=k+1, the given equation implies a0 is even, hence a0 ≥ 0. So a0=2b, with b a nonnegative integer. Then dividing the equation by 2 on both sides, we get that (b+a1)+2a2+⋯+2kak+1 = 0. From the cases n=k and n=1 (in cases a2=⋯=ak+1=0), we get a0+a1+a2+⋯+an ≥ (b+a1)+a2+⋯+an > 0, ending the induction. Problem 393. Let p be a prime number and p ≡ 1 (mod 4). Prove that there exist integers x and y such that
x2 − py2 = − 1.
Solution. AN-anduud Problem
Solving Group (Ulaanbaatar, Mongolia), Kevin LAU (St. Paul’s Co-educational College, S.3), Simon LEE (Carmel Alison Lam Foundation Secondary School), Andy LOO (St. Paul’s Co-educational College), Corneliu MĂNESCU-AVRAM (Dept of Math, Transportation High School, Ploiesti, Romania), Alice WONG (Diocesan Girls’ School) and
ZOLBAYAR Shagdar (9th grader,
Orchlon International School, Ulaanbaatar, Mongolia).
Let (m,n) be the fundamental solution (i.e. the least positive integer solution) of the Pell’s equation x2 − py2 = 1 (see Math Excal., vol. 6, no. 3, p.1). Then
m2 − n2 ≡ m2 − pn2 = 1(mod 4). Then m is odd and n is even. Since
2 2 2 1 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = + ⋅ − n p m m
and (m−1)/2, (m+1)/2 are consecutive integers (hence relatively prime), either
uv n v m pu m 2 , 2 1 , 2 1= 2 + = 2 = − or m u m pv ,n 2uv 2 1 , 2 1= 2 + = 2 = −
for some positive integers u and v. In the former case, v2−pu2=1 with 0 < v ≤ v2 = (m+1)/2 < m and 0 < u = n/(2v) < n. This contradicts the minimality of (m,n). So the latter case must hold, i.e. u2−pv2 = −1.
Problem 394. Let O and H be the circumcenter and orthocenter of acute ΔABC. The bisector of ∠BAC meets the circumcircle Γ of ΔABC at D. Let E be the mirror image of D with respect to line BC. Let F be on Γ such that DF is a diameter. Let lines AE and FH meet at G. Let M be the midpoint of side BC. Prove that GM⊥AF.
Solution 1. AN-anduud Problem
Solving Group (Ulaanbaatar, Mongolia), Kevin LAU (St. Paul’s Co-educational College, S.3), MANOLOUDIS Apostolos (4° Lyk. Korydallos, Piraeus, Greece), Mihai STOENESCU (Bischwiller, France),
ZOLBAYAR Shagdar (9th grader,
Orchlon International School, Ulaanbaatar, Mongolia), Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
Mathematical Excalibur, Vol. 17, No. 1, May-Aug. 12 Page 4 O A B C F D X H E M G
As AD bisects ∠BAC, D is the
midpoint of arc BC. Hence, FD is the perpendicular bisector of BC. Thus, (1) FE || AH. Let line AH meet Γ again at X. Since
∠BCX=∠BAX=90°−∠ABC=∠BCH, H is the mirror image of X with respect
to BC. Therefore, ∠HED=∠XDE=
∠AFE. Thus, (2) AF || HE. By (1) and (2), AFEH is a parallelogram. Hence, G is the midpoint of AE. As M is also the midpoint of DE, we get GM || AD. Since DF is the diameter of Γ, AD⊥AF,
hence GM⊥AF.
Solution 2. Andy LOO (St. Paul’s
Co-educational College).
Place the figure on the complex plane and let the circumcircle of ΔABC be the unit circle centered at the origin. Denote the complex number representing each point by the respective lower-case letter. Without loss of generality we may assume a = 1 and that the points A, B and C lie on the circle in anticlockwise order. Let b =u2 and c=v2, where |u|=|v|=1. Then d=uv and hence f =−uv. Next, E is the mirror image of D with respect to BC means
e b d b c b c b − ⎛ − ⎞ = ⎜ ⎟ − ⎝ − ⎠ ,
giving e = u2 − uv + v2. By the Euler line theorem, h=a+b+c=1+u2+v2.Now G on lines AE and FH means
g a g a e a e a − = − − − and g f g f h f h f − − = − − .
Solving these simultaneously for G, we get g = (u2−uv+v2+1)/2. Also, m = (b+c)/2 = (u2+v2)/2.
To show GM⊥AF, it suffices to prove
that (m−g)/(f−a) is an imaginary
number. Indeed, 1 1 2 1 m g uv f a uv − = ⋅ − − + and 1 1 1 1 1 1 1 1 2 1 2 1 m g u v uv m g f a uv f a u v − ⋅ ⎛ − ⎞= ⋅ = ⋅ − = − − ⎜ − ⎟ + − ⎝ ⎠ + ⋅ as desired.
Other commended solvers: Simon LEE (Carmel Alison Lam Foundation Secondary School), and Alice WONG (Diocesan Girls’ School).
Problem 395. One frog is placed on every vertex of a 2n-sided regular polygon, where n is an integer at least 2. At a particular moment, each frog will jump to one of the two neighboring vertices (with more than one frog at a vertex allowed). Find all n such that there exists a jumping of these frogs so that after the moment, all lines connecting two frogs at different vertices do not pass through the center of the polygon.
Solution. Kevin LAU (St. Paul’s
Co-educational College, S.3), Simon LEE (Carmel Alison Lam Foundation Secondary School), LI Jianhui (CNEC Christian College, F.5) and Andy LOO (St. Paul’s Co-educational College).
If n ≡ 2 (mod 4), say n=4k+2, then label the 2n=8k+4 vertices from 1 to 8k+4 in clockwise direction. For j ≡ 1 or 2 (mod 4), let the frog at vertex j jump in the clockwise direction. For j ≡ 3 or 4 (mod 4), let the frog at vertex j jump in the counter-clockwise direction. After the jump, the frogs are at vertices 2, 6, …, 8k+2 and 3,7, …, 8k+3. No two of these vertex numbers have a difference of the form 2 (mod 4). So no line through two different vertices with frogs will go through the center.
If n ≢2 (mod 4), then assume there is such a jump. We may exclude the cases all frogs jump clockwise or all frogs jump counter-clockwise, which clearly do not work. Hence, in this jump, there is a frog, say at vertex i, jumps in the counter-clockwise direction, then the frog
at vertex i+m(n−2) (mod 2n) must jump
in the same direction as the frog at vertex i for m=1,2,….
If n is odd, then gcd(n−2,2n) = 1. So there are integers
a
and b such that a(n−2) + b(2n) = 1. For every integer q in [1,
2n], letting m = (q−i)a, we have i+m(n−2) ≡ q (mod 2n). Thismeans all frogs jump in the counter-clockwise direction, which does not work.If n is divisible by 4, then gcd(n−2,2n) = 2. So there are integers c and d such that
c(n−2)+d(2n)=2. Letting m=nc/2, we
have i+m(n−2)≡i+n (mod 2n). Then frogs at vertices i and i+n jump in the counter-clockwise direction and the line after the jump passes through the center, contradiction.
Therefore, the answer is n ≡ 2 (mod 4). Other commended solvers: Alice WONG (Diocesan Girls’ School).
Olympiad Corner
(continued from page 1) Problem 3. (Cont.) At the start of the game A chooses integers x and N with
1≤x≤N. Player A keeps x secret, and
truthfully tells N to B. Player B now tries to obtain information about x by asking player A questions as follows: each question consists of B specifying an arbitrary set S of positive integers (possibly one specified in some previous question), and asking A whether x belongs to S. Player B may ask as many such questions as he wishes. After each question, player A must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any
k
+
1
consecutive answers, at least one answer must be truthful. After B has asked as many questions as he wants, he must specify a set X of at most n positive integers. If x belongs to X, then B wins; otherwise, he loses. Prove that:1. If n ≥2k, then B can guarantee a win. 2. For all sufficiently large k, there exists an integer n ≥1.99k such that B cannot guarantee a win.
Problem 4. Find all functions f: Z → Z such that, for all integers a, b, c that satisfy a+b+c=0, the following equality holds:
f (a)2 + f (b)2 + f (c)2
= 2f (a) f (b) + 2f (b) f (c) + 2f (c) f (a). (Here Z denotes the set of integers.) Problem 5. Let ABC be a triangle with ∠BCA = 90°, and let D be the foot of the altitude from C. Let X be a point in the interior of the segment CD. Let K be the point on the segment AX such that BK=BC. Similarly, let L be the point on the segment BX such that AL=AC. Let M be the point of intersection of AL and BK. Show that MK=ML.
Problem 6. Find all positive integers n for which there exist non-negative integers a1,a2,…,an such that
1 2 1 2 1 1 1 1 2 ... ... 1. 2a 2a 2an 3a 3a 3an n + + + = + + + =
Q
~:I~:C~: ~ :AB~?=::~' :::::,III",~:'~ ~~'~~~O~8.
so LJAK "" L.JAL "" l' By LAKJ = LALJ = 90° the points I( and L lie on the droll) w
with diameter AJ.
Thetiiangle l( B M Is isosceles as B [(and B M are tangents to tbe excirole.Since B J Is. the bisector of LI(BM, we have LMBJ == 90° - ~ and LBMl( = ~. Likewise, LMGJ = 90" - ~ and LOML=~. Also LBMF' = LOML, therefore
..
Hence F lies On the circle w. ~By the angle computation, F and A are on the ~ame ~i4¢ 'of BO)
AnalogoUilly, G also lies Iln w. Since AJ js I). 9.!ameter of W, 'we obtaln LAF J= LAGJ ",. 90 .
The' lines AB and BO are BY1)lInetric with respect to. the eJ<temal bisector: BF. Becaus~.
AF:l BE' llild KM 1.. BF, thesegmeilts 8M and AI( are sYJllllletrlo with respect to BF, henCE) 8M = Art. By syrnrnetty T/vf = AI,. Since AI( and AL are equal as tangents to thll excirGie, it follows that
aM"" TM,
apd the proof IS complete... ' • • . . X2 X3 X l . f' h " I bl"
Solution. The 8ubstitutl<;m a2 = - , as = - , ... , an = -'-.-trans Qrms t e ongma pro ern
. Xl X2 Xn_i
into th60hlequ!iUty
(*)
for 011 !Ill, ... ,$,,-1 >' O. To prove this, we use the AM-OM in~q~o1ity fot each factor o[ the left-hand side as follows:
(Xl + (2)2 ($2 +x~)3 (Xa + (4)4 (2 ("t) +X3)3 (3 (~) + (4)4 2: 22xIX2 >" 39 (a)2 ~ .2 X3 2: 44 (lji)3 $4
Multiplyhlg these Inequalities together gives (*), with mequality sign.;:; instead of '>. Howl)ver [or the equality to occur it is necessary .that. WI == Xli, x~ = 2ar:it ; .. ; !l!1\_1 = (n~ 1 )t1, I~plylng. XI '" (n ~ l)!xI' ThiS is impossible smce Xi >-a and n ~ 3, 'rhl1tefore ~he metju!illty:is sttict
We 6ay that A is 4lc.o~istent with a mlmber iiI A = yes lind £ Ii 8, or if A = flO and i E S.
ObaerYe ·thaJ;.an a.ru;wer incQl!sistent with the target num.ber"X is a Ill).
a) :SUPP9si) tha:~Ben hi'S detl,lrmmed ll. set T of siZe m that c.ontainB:n, This is true Initially with m ;N and T == {I, 2, ... ,N}, For m >2'" wc show how Ben clin find a .numb~r yET
tl\~tis dlfferl\nt frol,li,$ •. By performing thi$ steP repM.tedly htl ca.n ted.n.dlj T to 'he of size 2~ ~ 11 Mid thu$\vin.;
. iSil1~e only t4e, size m:> 2-" of.T is ttll"vant, assUIil~ th.ll.t T = {O.1'''''J2!<.'''''1m~I}. Ben begins by lIljking r~peatedly whether", .is 2-". If A;ny !iJ'LSwets rIP k + 1 titiles In a rQW, .on" of, these answers;is truthful, and so x ,;, 2"'. Otherwi$e Ben stopsii&ldl;lg ~boi.lt 2'" lit the first answer yes. He then asks, for each i '= 1, ... I k, if the binary repr\!Seriti,i.tibll of x has. a 0 in
the i.th dIgit. Regardless of what the. k answers are, they ateilll incQnilis~f'ln~ with a certain number y E {a, i, ... , 2'" -1}. Tile precedin~ answer ye~ iJ,bo\lt 2k i& also inconsistent with y . He~1 ce y i' w. Otherwise the last k+ 1 answers \ITe nd~ truthful, which i~ iIDpQssible.
Either way, B, en fi,nds a numbetiu T that is di..ft. eteJit frdIj1'x, and the. Glahn is PT,oven. b) We prove that if 1 < A <. 2 and n = l(2 - ),),k+1J -1 then Ben Camlot lIuarllntee awill. To: oomplete the proof, then it suffices to take A such that 1.99 < ~ < 2 anti k large enough so
thl).t
n'" t(2 - ,\),-"+IJ -1 ~ L99k •
. Consider the following strategy for Amy. First she chooses N = n+ 1 and:j; E {I! 2, ... , n+ I}
atbltrarily.. Afhlr every answer of hers Amy determines, (or each i = 1, ~, ... ,n + 1, the rii.mib.er Ill; of cQnsecutfve anSwers she has given by tha~point that are incol).Sistent with i .. To decide on her' next answer, she. then uses the quantity
No1 matter whab BeJ'!'s ilElXt question is, Amy
choos~-the
!lll.Swer which minirnlzes,p.
:We claim that with this strategy <p will always stay less than ),"'+.1. Consequently no expo-pent 7ili in <p. will ~J(el," el(.c.eed h, .hence Amy will never give more than Ie tonsecuti;'e answers mcOnsiste.n~ wlthsoml;l j. In particular this applies to the target number "', so she ",iii never lie more th\ln kc times m a [OW, Thns, given the claim, Amy's strategy is legal. 1)lnoo. the strategy does. not depend (io.x·in anrway, Ben can make no deductions ahout x, and therefore he cannot gui;rantee e. Win.i
It remains. to show tha,t 4> < ,\k+1 at all times, Initio1ly ead> m; is 0, SO thi!, condition hulds luthe begi)li!.iilg due to l <. ),. < 2 and n = l(2 - ),).\k+1J -
1. Supposetpat if! < ,\~+1 at some point, a.nd Ben hBB just 8$ked if x E B J()r some set S. ACGotding as Amy answers y~s or no;the MW value, of ~ becomes
'h
i=L
1 +E
);IiIi+1 or <P2 =E,\m
l+! +I>
. I~S W; iES i~S
Since Amy chQoses, the ollWmininirnlzing <p, the new <P will equal min(qll' <P2)' Now we have
Because <p < >,Hl, the asstltllptions'\ < 2 and 11. = l(2 - ,\),\~+1
J
~ 1 lead toJIiin(<plt<P2)
-<
~C,\k+2 + (2 _A),k+1) <=; t.k+1,~ Solution. The Buhstitutioll a = b = c = 0 gives 3f(0)2 =6/(0)2, hellee
frO) =0. (1)
The substitution b = -a and c = 0 gives «(f(a) - f( _a»' = O. Hence f is I!ll ¢ven full9tibIit
fea) = fe-a) for all a E;~.
Now set b = a !\Ild 0 = -2a to obtain 2f(a)2 + f(2a)2 "" 2/(a)2 + 4.f(a)J(2a). Hence
f(2a) = 0 or f(2a) = 4f(a) for nil 4 E Z.
(2)
(3)
If f(l") = 0 for some I" 2: 1 then the substitution b = rand c = -a-I" giVes (1(<1+>')-j(a))' = O. So f Is periodic with period r, i. e.
f(a + t) = f(o.) for all a E Z.
In particular, if f(l) '" 0 then f is cohstant, ,thus f(a) = 0 for all a E Z. This functign clearly satisfies the functional equation. For lhe rest of'the analysis, we assume j(l) = k =/; O.
By (3) we have f(2} = b or f(2) = 4k. If f(2) = 0 then f is periodio o~ pe,riod 2, thus J(even) = 0 and f(ddd} = k. This function is a solution for every k. We postpone the verification; for the ~cqucl !lS$umc f(2) = 41> f O.
By (3) again, we have f(4) =0 or f(4) "" 16k. In the first case f is periodic at period 4, and
f(3) = f(-I) = f(l) = k, BO we haV\l j(4,,) = 0, f(4n+!l = f(4n+3) =-/0, and /(4n+2) = 4k
for all n E Z. This fundi';n is 0. solutio" too, wllich we justify later. For the rest of the analysis, we assume f(4) = 16k t 0.
We Shdw now that f(3) = 910. In order to do 60, we need two 6ubstitutions:
a = l,b = 2,e = -3 =* f(3)' - 1Okf(3) + ek' = 0 => /(3) E {k,9k},
a = 1, b = 3, c = -4 => N)' -34kf(3) + 225k' = 0 => f(3) E {llk,2&1;}.
Therefore j(3) = 9k, ~ clllimed. Now we prove inductively th~t the only remai!lip~ funptiQu is
f(x) = ka;~, x E Z. We prove<! thi!>for;2) = 0, 1,2,3,4. A&sume tha.t n ;::; 4 IUjd th/lt f(w.) 't= I;w·
holds for all integers :t E (0, n]. Then tlie substitutions a = n, b =; 1, ¢ = -II -1 lind q,= 1l-1,
b = 2, C'= -n - 1 lead reSpectively to
/(n+l) E {k(n+1)2.k(n-1)'} and f(n + 1) e {k(T\ + I)', k(tt- 3)2}.
Since ken - 1)~ f h(n - 3)2 fot " f 2. the oaiy possibility Is I(n +1) = ken + I)'. This
!'Qmpletes the induction, so I(x} '" kre' for all ;2) 2. O. The same expression is valid (or neg/ltive valu,," of x since f is ~ven. To verify that f(x) = kz' is actually a ~ol\ltion, we ne!X\ to ~heck the identity a4
+
b4+
(a +W
= 2a'b' + 2a2(il +W
+ 2b'(a+
~)2, which followsdirect\y byeXpimding both sides. .
Thetefore the only possible solutions of'lhe !\Inctlonal equa.tion are thfl constant functiQl'! b(x) =' 0 and the foliowing fundions:
I( )3 x -- { 0 k xeven xodd {
Ox == 0 (mod 4)
f4(:») = k x == 1 (mod 2)
4k x:.ii (mod 4)
for any non·zero integer k. The vcrifiCl)tion that tiJoy arc iI\d9cd sQlutiDn$ was daM Jot tho. first two. For fs note that if il + b + c = 0 then either Ii, li, 0 are llll eVll1!, in which ¢'ISe
f(a) = feb) == ftc) = 0, or ona of theni is even and the othet twu !lrll odd, 80 both Iildei3 of the eq\!ation equal-~k'. Fbr f. we use similar parity considerations and the ,sYWrl1~tty of the e<juatioll. which redu~ the verification to the f;riples (0, k, k), (4k, k, k), (Q,O,O), (0, 4i;!, 4k).
They all satisfy the equation. ' .
B
@
:5 .
"S.oIUtlon'LetO!betherellectiQnofOillthelineAB,andlet with c;eqt~ro A and B, passing throllgh L IlllQ J( respectively. Since"'I
AO' and = 41; AO be = the AL and cIr,cl~ BO'.= Ba = BK, both 41, and w~ pass ,through C and 0'. By t:BOA = 90·, AO is tangent til 41, at G, and BO Ii; tangent to 41, at C. Let KI f K be the second intersection of AX nodWi. and let, L, " L ,be th~ seCOlld intersection of BX alld 14,. . By the powers 9f X. with respect to
w.
alld WI,XK·XKI =XQ·XC'=XL·XL"
so th~ PQii\~ K" L, K, LI lie pn II cirole W3.
Tlw pflWer of A with respect to Wi gives
AL' = AC' = AK ·Ak\o
iqdi1:at;pg' thnt AL is tangent to 413 at L. Analollou~ly, BK is tangent to W3 at K. Hence M K
and ML ate the two tangents from M to W3 and tQe~efQre M J( = M L. . .
Solution. Such numbers ali 02,,' •• ,On exist if and only if n == 1 (mod 4) or 1\ == 2 (mod 4).
Let l:l=1 3!:~ = 1 witb ai, a2,,"
,°
0 nonnegative integers,. Then 1,xI +2'X2+" ·+n,xn = 3"With ~i",., Xn powero of 3 end-a ,2: O. The right-hand side is odd, and the left-hand side hss the sl\IIIe parity as 1 + 2+· .. + n. Hence the latter sum is odd, which implies n == 1,2 (mod 4).
Now we prove th~ converse.
Callfea,.lble !l~~g\leii~e bl , b" ...• bn If there are nonnegative integero ai, 02, •. ',' On such that
.2....+.2....+ ... +~=~+~+, .. +~=1. 241 ·2"3 2un. 3di 30l 30n
Let b. be a te(mpf a feasibTesequence b"b" ... ,bn with exponen~ '11,0., ... ,0" like ahove, and let u, v be nO,nliei\atlve integers with sum 3b •. Observe that
1 1 1
--' -'
-+--=-2d.+1 ll~Hl 2.' arid
It follows that the sequence bl, ... ,b.-I, II, v, bk+l," ., bn is fensible. The expooellts a. are the
same for the ljIlchanged te!1DB b" i t k; the new termB tt, v have <mJonellts al< + 1.
We state the c()t\ci'usIQn in reverse. If two terms u, v of a sequence are replaced by one . term ~ alln ~hl'i obtained sequence is feasible, then the original sequence is feasible too. Denote' by a" t!l<~ s'!,,\U¢Qce 1,!2, ... , n. TQ ,show that an is feasible {or n == 1, '2 (mod 4), we transform It by n-'f replneements {u, v} N
"t"
to the on'i'-term ~equence al. The la.lter is f"ilsibIe, with at '" 0, Note that it m alld 2m nre terms of a sequence then {m,2m} f-lt Tn, so2m can he, ignored if' necessary,
Let n ~ 16. W~ )?rQv~ that an can be red\lced to a,,_12 by 12 ol'eratlons. Write n:, l~k+ r
whete " ?: 1 and 0 51 r :s; 11. If 0 $. r $. (i then the last 12 terms of an can be partition!X\ into
'2 singletons {12k-6}, {12A;} and the followjng 5 pairs:
{12k ~ 6-i, Hlf9 -6+i},i = 1 .... ,5 - r; {12k - i, 12k+ j},i = 1, ... ,I'. (There i~ poly (ln~ kind of pairs if r E {DIU}.) One CiI" ignore 12k - ~ and 12k since a" oontains 6k -1! lind aI>, Furthermore the~ operations {12k ~ 6 - t; 1219-6+ i} t-t 8k _ 4 and
{12k -j, 12k + j} I-t 81; remove the 10 terms in the pairs Il)\d bring In 5 new tetms equal to Bk - 4 or 81,:. AU ()f these can be ignored too as 4k - '2 and 4k are still present io the sequence. lnde\1\l,41~ ~ n ~ 12j!l. equivalent to 8k?: i2 -r, which is true for r E {4.5}. And if r E {O,i, 2,3}tben n ~ 16implles k
a.
2. ~o SA: ~ 12 -" also hqlds. Thus an reduces til an-I.':he,case6:$ r S 11 is ,il.nalogous. ComMer the singletons {12k}. {12k +6} and the 5 pairs
{12k -i, 12k'+i},i:o 1, ... , n -r; {12k + 6 - j,12k+ tj
+
i},j = 1, ..• ,r - 6. Ignore the sil\gleloilS like-before, thell remove the pairS via operations {12k - i, 12k + i} t-t 8/0 and {12~ + 6 - i.12k+ 6 + j} t-t 810 + 4. The!i newly-appeared terms 8k lilld lik + 4 can be Ignored too sill~i} 410 + 2 :=; n -12 (t\lis follows from k;:: 1 andr i:: 6). We obe-ain an-1' a~ain.The problem reduces t02 $. n $15. In fact n e {2, 5, 6, 9,10,13, 14} by n a 1,2 (mod 4). The c!!SBS n =; 2,6; 10, 14 reduce to n = 1, 5, 9, 13 tespectively because the last even term of an can be ignol'ed. Fot n "" 5 apply f4,5} t-t 3. then {3,3} t-t 2, then ignore the 2 occurrel\CW of 2. For n '" lligilarc 6 firstl. then apply {5,7} >-+ 4, {4, S} t-t 4, {3,9} t-t 4. Now ignore the 3 occurrences of 4, then ignore 2. Finally n = 13 reduces tori = 10 by {Il, 13} t-t Ii and Ignoring 8 and 12. The proof is complete.
