Volume 21, Number 2 April 2017 – September 2017
Olympiad Corner
Below are the problems of the 2017 International Mathematical Olympiad (July 18-19, 2017) held in Brazil.
Problem 1. For each integer a0 >1, define the sequence a0, a1, a2, … by:
, 3 , int 1 otherwise a eger an is a if a a n n n n
for each n≥0. Determine all values of a0 for which there is a number A such that an=A for infinitely many values of n.
Problem 2. Let ℝ be the set of real
numbers. Determine all functions f : ℝ→ ℝ such that, for all real numbers x and y, f (f(x) f(y)) + f(x+y) = f(xy).
Problem 3. A hunter and an invisible
rabbit play a game in the Euclidean plane. The rabbit’s starting point, A0, and the hunter’s starting point, B0, are the same. After n−1 rounds of the game, the rabbit is at point An-1 and the hunter is at point Bn-1. In the nth round of the game, three things occur in order.
(continued on page 4)
Notes on IMO2017
Tat Wing Leung
International Mathematical Olympiad (IMO) 2017 was held in Rio De Janeiro, Brazil from 12 to 24 July, 2017. Members of Hong Kong Team are as follows.
Tat Wing Leung (Leader)
Tak Wing Ching (Deputy Leader) Man Yi Mandy Kwok, Shun Ming Samuel Lee, Yui Hin Arvin Leung, Cheuk Hin Alvin Tse, Jeff York Ye, Hoi Wai Yu (Contestants)
All contestants except Alvin Tse are entering universities during the academic year 2017-18. Thus we will have an essentially new team next year.
I went first to Brazil in July 13. Professor Shum Kar Ping, chairman of our Committee also went with me. He was to present the report of IMO2016. It was over quickly. Apparently members of the Advisory Board had nothing more to ask. Luckily it was done.
Upon arrival, I just had to follow the program closely and to attend Jury meetings. As claimed, I did experience the famous Brazilian Hospitality (this clause was copied from the program book) and I was quite happy in general.
As in these few years, in choosing the problems, first 4 easy problems, 1 from each of the four categories (Algebra, Combinatorics, Geometry and Number Theory) were selected. Then 4 medium problems, again 1 from each category was selected. Then members of the Jury (leaders) selected two easy problems of two categories, and the 2 medium problems from the two complementary categories were selected. It was claimed this scheme will help to produce a more balanced paper. But after a few years, I do think it is not necessarily true. First almost certain an easy geometry problem will be selected, thus all
medium but nice geometry problems will be discarded. It is also almost certain two combinatorics problems will be selected. The papers will then become more predictable. Anyway members still chose this scheme.
Our contestants arrived on July 16. During the opening ceremony, July 17, I had a chance to look at them (from far away). In the opening ceremony, the speech of Marcelo Viana, director of IMPA (Instituto de Mathematica Pura e Applicada) was particularly genuine and moving. He talked about the IMO training and selection in Brazil in these 38 years. (Certainly it was not an easy task to select a team of 6 from 18 million youngsters). Then he also talked about Maryam Mirzakhani, the Iranian Mathematician, who was a 1994 and 1995 IMO gold medalist, 2014 Fields’ medalist and passed away prematurely at age 40. Finally, he also talked about the upcoming International Congress of Mathematicians (ICM) 2018, to be held in Brazil.
The next two days (July 18 and 19) are contest days. The contestants had to sit for two 4.5 hours exam during the mornings. In the first half hours of the exams, there were Q&A times. In this year again they adapted a new scheme, namely they had 4 tables, 3 tables for problems 1, 2 and 3 (problems 4, 5 and 6 the next day), and so they were 4 queues. Clearly this is a more efficient scheme than before.
Again the next two days (July 20 and 21) were days of coordination, namely leaders and coordinators would decide the score of a particular problem. We followed the schedule to go to a particular table. We had a very capable deputy leader this year and so he knew well what our team members had done. So the process became relatively easy.
(continued on page 2)
Editors: 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/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 21, 2017.
For individual subscription for the next five issues for the 17-18 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
Mathematical Excalibur, Vol. 21, No. 2, Apr. 17 – Sep. 17 Page 2
Since the problems of IMO2017 is listed in this issue of Excalibur, I shall not reproduce them here nor to copy the proofs. I will only give a few comments of this year’s problems. First a few key words came to my mind. My first word is algorithm (or construction). Indeed the proposer has been trying hard to think of a new scenario that when you try to solve the problem, you need to invent a new algorithm to solve the problem. For example, problems 5 and 6 do not need to know a lot of higher math, but you do need to have some sense of ingenuity to think of a new scheme or method to solve a particular problem. In problem 3, you had to show an algorithm does not exist. The second word is induction, namely in these problems, small cases (cases with smaller parameters) were easy. So one might try to consider if the method of induction does work. It was not obvious. The third word is geometry. In this year, only 3 of us could solve the relatively easy geometry problem. Indeed this year’s geometry problem (P4), no new constructions are required, no new transformations (inversion, homothety, etc.) are needed. It is simply correct drawing and angle chasing. So I must admit that we have reverted back to our usual tradition. Now I will say a few more words on the individual problems and the performance of our team. Problem 1 is a number theory problem. Once a contestant tries a few cases and guess the correct answer (a0 ≣0 (mod 3)), then it is not too hard to prove a0 ≣1,2 (mod 3) do not work but a0≣0 (mod 3) works. Our team this year is relatively mature and relatively well trained. So all of them solved the problem and we have a perfect score.
Problem 2 is a functional equation, showing f(f(x)f(y))+f(x+y)=f(xy) for all real x and y will imply f(x)=0 or f(x) = ±(x−1). The most troublesome thing is the marking scheme. It is easy to get the first 3 points, but it is real hard to get an extra point, i.e., proving injectivity and onward. A leader secretly showed me the scores of problem 2 of his team, apparently he was dismayed by the performance. I was not sure. At the end I found their team scored 1 more point than us. For problem 3, I had (and still have) a very serious concern about it. Observe
only two contestants scored 7 points (a Russian and an Australian contestant), and also none of the USA team and the Chinese team (plus other teams) together scored any point at all. I suspect many contestants are like me and simply don’t know what exactly is going on. Indeed it is not quite sure what it means by “no matter how” and what exactly it means by a tracking device, I was told it is not like the “best strategy”. Indeed when you look at the solution, you get the idea such a strategy (or algorithm) does not exist. The solution is roughly as follows. Assume the rabbit moves in a straight line, and with luck (this term appears quite a few times in the solution) the tracking device also moves in a straight line. Because of this happening, the hunter can only move along a straight line (also with no justification but intuition) and follow the rabbit, and after finitely many steps, the distance between the rabbit and the hunter will only increase (easy to show by simple geometry). Thus there is no best strategy. I am still awaiting members to educate me on this problem.
Problem 4 was a relatively easy geometry exercise.
We did best in problem 5 among all teams, (our deputy leader reminded me about this point). Indeed altogether we scored 26 points. So essentially 4 of us solved the problem, while other teams scored at most 23 points. This shows our team does know something about problem solving. Indeed the problem is equivalent to say there are N(N+1) distinct integers randomly placed in a row, say, you can throw away N(N−1) of them, and among the remaining integers, the largest integer and the second largest integer will stick together, so are the third largest and the fourth largest integer will stick together, and so on. Not too hard?
For Problem 6, an ordered pair (x,y) of integers is a primitive point if gcd(x,y)=1. Now given a set of finitely many primitive points (xi, yi), 1≤i≤n, we need to find a homogeneous polynomial g(x,y) such that g(xi,yi)=1. If there is only one primitive point, then it is trivial, by Euclidean algorithm. The hard part is how to move on by induction. But it is not at all easy. At the end Shun Ming was awarded a gold medal (25 points), Mandy a silver (23 points), Jeff (18 points), Hoi Wai (17 points) and Cheuk Hin (17 points) all received Bronze medals. Yui Hin (11 points) managed to get a honorable
mention. Our rank is 26 among 111 countries/regions. Surely the result was not as good as last year nor as we had hoped for. Nevertheless there were certain things we can say. Indeed this was the 30th consecutive year that we sent teams to IMOs. No matter what, it is not an easy matter and it should be a date to remember. (Better still, we hosted the event in 1994 and 2016). Also Mandy Kwok was the second girls among all girl contestants. IMPA this year gave out 5 prizes to female contestants. Initially I thought Mandy should have a chance to get a prize. Later I found out the prizes were for the top female students who contribute the most to their respective team’s score. So I understand why she was not eligible for the prize. Nevertheless I must say we are very glad to see her improving very well in these few years. Finally we managed to get the highest score in Problem 5. I think this is an indication that our team is comparable with any other team. They really don’t have much special recipe we don’t envisage.
I hasten to say the cut scores of IMO this year cannot be said to be ideal. Indeed the cut scores for gold is 25, for silver 19, and bronze 16. One may say the easy problems (problems 1 and 4) were too easy and the four other problems too hard. The easy problem were too easy. Hence 14 points was not enough for a bronze and the hard problems too hard. Thus 25 points was good enough to get a gold. Really we expect a contestant to solve at least 2 problems (≥14 points) to get a bronze, at least 3 problems (≥21 points) a silver, and at least 4 problems (≥28 points) to get a gold. Some people expect a contestant should solve nearly at least 5 problems to get a gold. Really what is the point to set a problem so that only 2 out of 615 contestants can solve it?
Since we are trailing behind some other Asian countries this year, it was suggested that more money should be put into this activity. In my opinion the stakeholders (members of the Committee, the Academy and the Gifted Section of Education, but most important of all, past and present trainees) should sit together and sort out what exactly do we want, how much money/resource should be put into it and who will contribute what, etc. I suppose it is time to start thinking.
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 21, 2017.
Problem 501. Let x, y, s, m, n be
positive integers such that x+y=sm and x2+y2=sn. Determine the number of digits s300 has in base 10.
Problem 502. Let O be the center of
the circumcircle of acute ΔABC. Let P be a point on arc BC so that A, P are on opposite sides of side BC. Point K is on chord AP such that BK bisects ∠ABC and ∠AKB > 90°. The circle Ω passing through C, K, P intersect side AC at D. Line BD meets Ω at E and line PE meets side AB at F. Prove that ∠ABC = 2∠FCB.
Problem 503. Let S be a subset of
{1,2,…,2015} with 68 elements. Prove that S has three pairwise disjoint subsets A, B, C such that they have the same number of elements and the sums of the elements in A, B, C are the same.
Problem 504. Let p>3 be a prime
number. Prove that there are infinitely many positive integers n such that the sum of kn for k=1,2,…,p-1 is divisible by p3.
Problem 505. Determine (with proof)
the least positive real number r such that if z1, z2, z3 are complex numbers having absolute values less than 1 and sum 0, then |z1z2+z2z3+z3z1|2 + |z1z2z3|2 < r. *****************
Solutions
****************Problem 496. Let a,b,c,d be real
numbers such that a+sin b > c+sin d, b+sin a > d+sin c. Prove that a+b>c+d. Solution. Toshihiro SHIMIZU
(Kawasaki, Japan).
For x≥0, |sin x|≤ x. Let s = a − c and t = d −b. We have
s = a-c > sin d-sin b
= 2cos[(d+b)/2]sin[(d-b)/2] ≥ -2|sin (t/2)|
and t = d-b < sin a-sinc
= 2cos[
(
a+
c)/2]sin[(a-c)/2] ≤ 2|
sin(
s/2)|.If s ≥0, then t< 2|sin(s/2)|≤s. Similarly, if t≤0, then s > -2|sin(-t/2)| ≥ -2(-t/2) = t. Finally, if s < 0 < t, then –s < 2|sin(t/2)| ≤ t and t < 2|sin(s/2)| = |sin(−s/2)| ≤ −s, which leads to a contradiction.
Comment: The above solution avoided calculus as it used sin x ≤ x for 0≤x≤1, which followed by taking points A, B on a unit circle with center O such that ∠ AOB = 2x, then the length 2x of arc AB is greater than the length 2sin x of chord AB. Other commended solvers: Jason FONG and LW Solving Team (S.K.H. Lam Woo Memorial Secondary School).
Problem 497. Let there be three line
segments with lengths 1, 2, 3. Let the segment of length 3 be cut into n≥2 line segments. Prove that among these n+2 segments, there exist three of them that can be put to form a triangle where each side is one of the three segments.
Solution. William FUNG, Mark LAU (Pui Ching Middle School), LW Solving
Team (S.K.H. Lam Woo Memorial
Secondary School) and Toshihiro
SHIMIZU (Kawasaki, Japan).
Note line segments with lengths a≤b≤c form a triangle if and only if a+b>c. Let a1≤ a2 ≤⋯≤an be the lengths of such n segments with sum equals to 3. Assume there exists i such that ai>1. If 1<ai<2, then the segments with length 1,ai,2 forms a triangle since 1+ai>2. If 2≤ai, then the segments with length 1,2,ai forms a triangle since 1+2>ai. It remains to consider the case all ai≤1. Then i≥3. Assume no 3 of these segments form a triangle. Then a1+a2≤a3, a2+a3≤a4, …, an-2+an-1≤an, an+1≤2. Adding these and cancelling a3,…,an,1 on both sides, we have
3+a2 = (a1+a2+⋯+an)+a2 ≤ 2, which yields a2≤-1, a contradiction.
Problem 498. Determine all integers n>2
with the property that there exists one of the numbers 1,2,…,n+1 such that after its removal, the n numbers left can be arranged as a1,a2,…,an with no two of
|a1-a2|, |a2-a3|, …, |an−1-an|, |an-a1| being equal.
Solution. LW Solving Team (S.K.H. Lam Woo Memorial Secondary School), George SHEN and Toshihiro
SHIMIZU (Kawasaki, Japan).
Since no two of |a1-a2|, |a2-a3|, …, |an−1-an|, |an-a1| being equal and each is at most n, they must be 1,2,…,n in some order. So |a1-a2| + |a2-a3| + ⋯ + |an−1-an|+|an-a1|=1+2+⋯+n=n(n+1)/2. From a ≡ |a| (mod 2) and (a1-a2)+ (a2-a3)+⋯+(an−1-an)+(an-a1)=0, we see |a1-a2|+ |a2-a3| + ⋯ + |an−1-an| + |an-a1| is even. For n(n+1)/2 to be even, this implies n ≡ 0 or -1 (mod 4). In the case n=4k, remove k+1 and let a1=4k+1, a2=1, a3=4k, a4=2, …, a2k-1 =3k+2, a2k=k, a2k+1=3k+1, a2k+2=k+2, a2k+3=3k, a2k+4= k+3, …, a4k-1=2k+2 and a4k=2k+1.
In the case n=4k-1, remove 3k and let a1=4k, a2=1, a3=4k-1, a4=2, …, a2k-1 =3k+1, a2k=k, a2k+1=3k-1, a2k+2=k+1, a2k+3=3k-2, …, a4k-2= 2k-1, a4k-1=2k.
Problem 499. Let ABC be a triangle
with circumcenter O and incenter I. Let Γ be the escribed circle of Δ ABC meeting side BC at L. Let line AB meet Γ at M and line AC meet Γ at N. If the midpoint of line segment MN lies on the circumcircle of ΔABC, then prove that points O, I, L are collinear.
Solution. George SHEN.
O P A X Q C M N B J L I
Let P be the midpoint of MN. From AM=AN, we see AP⊥MN. So A,I,P are collinear. Let Q be on MN such that LQ⊥MN. Now ∠BMQ=∠CNQ and . cos cos 2 cos cos 2 cos cos NC MB LMB LNC NC LNC LMB MB LNQ NL LMQ ML NQ MQ
Mathematical Excalibur, Vol. 21, No. 2, Apr. 17 – Sep. 17 Page 4
This implies Δ BMQ, Δ CNQ are similar.
Let a=BC, b=CA, c=AB, s=(a+b+c)/2 = AM =AN and α =∠BAC.
We have
AP =AMcos(α/2) = s cos(α/2). By extended sine law, BC =a =2R sin α. From IP=BP=CP [see Math Excalibur, vol. 11, no. 2, page 1, Theorem in middle column−Ed.], we have
, 2 cos 2 2 180 sin 2BP BP BC a . ) 2 cos ( 2 ) ( 2 2 2 cos AI s a AI AP a IP a
Applying AI cos(α/2)=s-a and the last equation, we can get
. 2 sin 2 , 2 2 cos 2 2 2 a s c b a s s
Next MN = 2AMsin(α/2) = 2ssin(α/2) and (MQ+NQ) sin(α/2) = MB+NC. Using MQ/NQ=MB/NC, we get
MQ sin(α/2)=MB and
NQ sin(α/2)=NC,
which says ∠QBA = 90°=∠QCA. Then Q is on Γ and AQ is a diameter of Γ. Let line LQ meet the circumcircle Γ of ΔABC at X as labeled in the figure. Observe that APQX is a rectangle and AQ, XP are diameters of Γ intersecting at O. We claim LQ=AI (then LI∩AQ at O and so O,I,L are collinear).
Now BO=CO, BJ=CJ and ∠BAP = ∠CAP implies BP=CP. Hence, O, J, P are collinear. Next OJ⊥BC implies ∠LJP=90°=∠LQP. Then, J,P,Q,L are concyclic. Hence,
XL·XQ=XJ·XP
Let R be the circumradius of ΔABC. From , 2 cos , 2 sin 2 , 2 , 2 cot 2 s AP R IP R XP a XJ
We get XJ·XP=IP·AP. Then XL·XQ =IP·AP. Since XQ=AP, so XL=IP. Then QL=XQ-XL=AP-IP=AI. The conclusion follows.
Other commended solvers: LW Solving
Team (S.K.H. Lam Woo Memorial
Secondary School) and Toshihiro
SHIMIZU (Kawasaki, Japan).
Problem 500. Determine all positive
integers n such that there exist k≥2 positive rational numbers such that the sum and the product of these k numbers are both equal to n.
Solution. Mark LAU (Pui Ching Middle School), LW Solving Team (S.K.H. Lam Woo Memorial Secondary School) and
Toshihiro SHIMIZU (Kawasaki, Japan).
Observe that for a composite number n, there exist integer s,t≥2 such that n=st, the sequence s,t,1,1,…,1 (with st-s-t 1’s) has sum and product equals st=n.
For prime numbers n≥11, the sequence n/2,1/2,2,2,1,1,…,1 (with n-4-(n+1)/2 1’s) satisfies the condition by a simple checking.
For n=7, the sequence 9/2, 4/3, 7/6, satisfies the condition by a simple checking.
Next we claim the cases n=1,2,3,5 have no solution. Assume a1, a2,…,ak are positive rational numbers with sum and product equals to n. By the AM-GM inequality, we have . 1 1 k k k k a a n k a a k n
Then n≥kk/(k−1)>k. Since n>k≥2, cases n=1 or 2 are impossible.
Finally, for n=3 or 5, since 33/(3−1) =5.1… implies k=2, so only cases (n,k) = (3,2) and (5,2) remain. Now
(a1-a2)2 = (a1+a2)2-4a1a2 = n2-4n = -3 or 5, which have no rational solutions a1, a2. Therefore, the answers are all positive integers except 1,2,3,5.
Olympiad Corner
(Continued from page 1)
Problem 3 (Cont’d).
(i) The rabbit moves invisibly to a point An such that the distance between An-1 and An is exactly 1.
(ii) A tracking device reports a point Pn to the hunter. The only guarantee provided by the tracking device to the hunter is that the distance between Pn and An is at most 1.
(iii) The hunter moves visibly to a point Bn such that the distance between Bn-1 and Bn is exactly 1.
Is it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after 109 rounds she can ensure that the distance between her and the rabbit is at most 100?
Problem 4. Let R and S be different
points on a circle Ω such that RS is not a diameter. Let ℓ be the tangent line to Ω at R. Point T is such that S is the midpoint of the line segment RT. Point J is chosen on the shorter arc RS of Ω so that the circumcircle Γ of triangle JST intersects ℓ at two distinct points. Let A be the common point of Γ and ℓ that is closer to R. Line AJ meets Ω again at K. Prove that the line KT is tangent to Γ.
Problem 5. An integer N≥2 is given. A
collection of N(N+1) soccer players, no two of whom are of the same height, stand in a row. Sir Alex wants to remove N(N−1) players from this row leaving a new row of 2N players in which the following N conditions hold:
(1) no one stands between the two tallest players,
(2) no one stands between the third and fourth tallest players,
∶
(N) no one stands between the two shortest players.
Show that this is always possible.
Problem 6. An ordered pair (x,y) of
integers is a primitive point if the greatest common divisor of x and y is 1. Given a finite set S of primitive points, prove that there exist a positive integer n and integers a0, a1, … , an such that, for each (x,y) in S, we have:
. 1 1 1 2 2 2 1 1 0 n n n n n n n ax y ax y a xy ay x a
5o{µ,-fi:17Y\.S.
fo
Of
1117~~ &V\£.r,
~
2 ( ,
r'l!J.2
Problem 1. We have 4 cases. In case 1, where
an
=
-1 (mod 3), as k2 ¢ -1 (mod3), so an is not a square. Then an+I
=
an+
3 >an
and an+l is not a square. Thenan< an+i < .... So this case fails. In case 2, where an¢ -1 (mod 3) and an> 9, let t2
be the largest square integer less than an. Since an > 9, t ~ 3. The first few squares in the sequence an,an+3,an+6, ... will be (t+l)2,(t+2)2 or (t+3)2. Then there is indexm
>
nsuch that CLm ::; t + 3 < t2 < a,.. In case 3, where an
=
0 (mod 3), note by the definition ofthe sequence, a multiple of 3 is always followed by another multiple of 3. If
an
E {3, 6, 9},the sequence will eventually be 3, 6, 9, 3, 6, 9, .... If
an >
9, let j be an index such that a; is the least integer in { an+1, an+2, ... }. We must have a; $ 9. (Otherwise apply case 2 to a; will contradict a; is least.) So a; E {3, 6, 9}. In case 4, where an= l(mod 3), we claim there ism> n such that an= -l(mod 3). Note 4 is always followed by 2=
-l(mod3). So the claim is true for an=
4. If an=
7, the next few terms will be 10, 13, 16, 4, 2, ... and the claim is true. For an ~ 10, we take an index j>
n such that a; equals the minimum of an+1, an+2, an+a, ... , which by definition of the sequence consists of non-multiples of 3. Ha;= l(mod 3), then we have a; $ 9 by case 2 and minimality of a;. So we must have a;= -l(mod 3).It follows from these claims that if ao is a multiple of 3, then eventually the terms will be 3, 6, 9, 3, 6, 9, .... If ao
=
±1 (mod 3), the sequence will eventually be strictly increasing. So the answer is a0=
3, 6, 9, 12, ....Problem 2. Let (A) denote the functional equation. Note if f is a solution, then so is -f. So we may suppose f(O) ::; 0. For x =I-1, choosing y so that x+y
=
xy (i.e. y=
x/(x-1)), we get (B) J(f(x)J(x/(x -1))=
O. Then letting x=
0, we get (C) J(f(0)2)=
0. So /(0)2 is a root off. If /(0)=
0, then setting y=
0 in (A), we get f is the zero function. Iff(O) < 0, then we claim f(O)
=
-1, J(l)=
0 and the only root ofJ
is 1. Let a be any root off. Assume a# 1. Then setting x=
a in (B), we get f(O)=
0, contradiction. So a= l. Next, from (C), we get /(0)2 =a= l. As J(O)<
0, so /(0)=
-1.Setting y
=
1 in (A) and using /(1)=
0, we get f(x + 1)=
f(x)+
1. By induction, (D) f(x + n)=
f(x) + n for all real x and integer n is true. Next, we show f is injective. Assume J(b)=
J(c) for some b =f. c. By (D), we get (E) J(b + m + 1)=
J(c + m) + 1. Let m < -b. Then there exist real x, y such that x+
y=
b + m + 1 and xy=
c+
m. Fromb =I-c, we see x =f. 1 and y # l (so f(x) # 0, f(y) :f 0). From (A), using (E),(D) and (C), we get J(x)J(y)
=
0, contradiction.For any real t, plugging (x,y)
=
(t, -t) into (A), we use (D), (E) and f injective to get (F) f(t)J(-t)=
-t2 + 1. Plugging (x, y)=
(t, 1 - t) in (A), we use (D) and finjective to get f(t)f(l - t)
=
t(l - t). Finally, since /(1 - t)=
1+ /
(-t) by (E), we getf(t)f(l - t)
=
t(l - t) iff f(t)(l + f(-t))=
t - t2 iff f(t)+
f(t)f(-t)=
t - t2. Using (F),we get f(t)
=
t-1. Then it is easy to check f=
0, f(x)=
x -1 and f(x)=
1-x are the only solutions.Problem 3. If the answers were "yes", the hunter would have a strategy that would "work'', no matter how the rabbit moved or where the devic reported Pn. We will show the opposite: from the device reports, there is no strategy for the hunter that guarantees that the distance stays below 100 in 109 rounds.
So, let dn be the distance between the hunter and the rabbit after n rounds,. Of course, if dn 2:: 100 for any n < 1D9, the rabbit has won - it just needs to move straight away from the hunter, and the distance will be kept at or above 100 thereon.
We will now show that, while dn
<
100, whatever given strategy the hunter follows, the rabbit has a way of increasingd;
by at least 1/2 every 200 rounds. This way,d;
will reach 104 in less than 2 -1D4 · 200<
109 rounds, and the rabbit wins.Let the hunter be at Bn and the rabbit be at An. Suppose even that the rabbit reveals its position at this moment to the hunter. Let r be the line BnAn, and Yi and
Y2
be points which are 1 unit away from rand 200 units away from An, as in figure 1 below.The rabbit's plan is simply to choose one of the points Yi or Y2 and hop 200 rounds straight towards it. Since all hop stay within 1 distance unit from r, it is possible that all device reports stay on r. In this case, the hunter has no way of knowing whether the rabbit chose Yi or }'2.
If the hunter's strategy is to go 200 rounds straight to the right, he will end up at point H' in the figure. Indeed, after these 200 rounds he will always end up at a point to the left of H'. If his strategy took him to a point above r, he would end up even further from Yi_. No matter what strategy the hunter follows, he can never be sure his distance to the rabbit will be less than y
=
H'Yi=
H'}'2 after these 200 rounds.To estimate y2, we take Z as the midpoint of segment Y
1 Y2 , we take R' as a point 200 units to the right of An and define c
=
ZR' (note that H'R!=
dn). Then y2=
1 + (H' Z)2
= 1 +
(dn - c)2, where c=
200 - AnZ=
200 - v'2002 - 1 > 1/400. Nowe?- +
1 = 400c:. So y2=
d; -
2c:dn+
e?- +
1 =d;
+
c:(400 - 2dn). Since c: > 1/400 and we assumed dn < 100, this shows that y2 > ~+
! .
So, as we claimed, with this list ofdevice reorts, no matter what the hunter does, the rabbit may achieve
<i;+
200 > d! +!-The rabbit wins.
Yi r fl' Yi
F,,.z.
"TProblem 4. See figure 2 above. Inn and r, we have LKRS
=
LKTS=
LATS.Also, since RA is tangent to 0, we get LSKR
=
LSRA. So J::..AKI',...., !::,.SKR. ThenT Rf RK
=
AT/SR=
AT/ ST. The last relation, together with LATS=
LK R:r, yieldsProblem 5. Split the people into N groups by height: group G1 has the N
+
1 tallest ones, group G2 has the next N + 1 tallest and so on, up to group G N with the N + 1shortest people.
Now scan the original row from left to right, stopping as soon as you have scanned two people { consecutively or not) from the same group, say G •. Since we have N groups,
this must happen before or at the (N+l)-th person'ofthe row. Choose this pair of people, removing all the other people from the same group Gi and also all people that have been
scanned so far. The only people that could separate this pair's heights were in group Gi
(and they are gone); the only people that could separate this pair's positions were already scanned (and they are gone too).
We are now left with N - 1 groups (all except Ci)- Since each of them lost at most one person, each one has at least N unscanned people left in the row. Repeat the scanning process from left to right, choosing the next two people from the same group, removing this group and everyone scanned up to that point. Once again we end up with two people who are next to each other in the remaining row and whose heights cannot be separated by anyone else who remains (since the rest of their group is gone). After picking these 2 pairs, we still have N - 2 groups with at least N - 1 people each.
If we do the scanning process N times, it is easy to see we will end up with 2 people from each group, for a total of 2N people remaining. The height order is guaranteed by
the grouping. The scanning construction from left to right guarantees that each pair from a group stand next to each other in the final row. Done.
Problem 6. A homogeneous polynomial is a function of the form f(x,y)
=
aoxn+
a1xn-ly+ · · ·
+ an-1xyn-l + anyn. Observe that finding a homogeneous polynomial such that f(x,y)=
±1 is enough since /2(x,y)=
1. Label the primitive points (x1,y1 ) to( Xn, Yn). If any two of these points (Xi, Yi) and (xi, Yi) lie on the same line through the origin, then ((xi,Yi)
=
(-xi,-yi) because both points are primitive. We then have f (xi, Yi)=
±f (x., Yi)- So we can assume that no two of the lattice points are collinearwith the origin by ignoring the extra lattice points.
Consider h;,(x, y)
=
y.x - XiY and define 9i(x, y) be the product of the hi(x, y)'s forj =I= i. Then h;,(xi,Yi) = 0 if and only if j = i because there is only one lattice point on each line through the origin. Thus 9i(xi, Yi) = 0 for all j =/= i. Define ai = g.(x., Yi) and note that ai =/= 0. Also, note that 9i(x, y) is a degree n - 1 polynomial with the following two properties: (1). g.(xi, Yi) = 0 if j =/= i and (2} 9i(xi, Yi) = a;.
For any N ~ n - 1, there also exists a homogeneous polynomial of degree N with the same two properties. Specifically, let I.(x, y) be a degree 1 polynomial such that
Ii(Xi, Yi)= 1, which exists since (xi, Yi) is primitive. Then Ii(x, y)N-(n-l)gi(x, y) satisfies
both of the above peoprties and has degree N.
We .may reduce the problem to the claim: For any positive integer a, there is a
homogeneous polynomial fa(x, y), with integer coefficients, of degree at least 1, such that
!a(x,y)
=
1 (mod a) for all coprime (x,y).To see this claim solves the problem, take a to be the least common multiple of the numbers a; (1 ~ i ~ n). Take fa given by the claim, choose some power !a(x, y)k that
has degree at least n - 1, and subtract appropriate multiples of the 9i constructed above
to obtain the desired polynomials.
We prove the claim by factoring a. First, if a is a power of a prime (a= pk), then we may choose either fa(x,y)
=
(xP-l +yP-l)ef>(a) if pis odd or fa(x, y)=
(x2 +xy+y2)<f>(a) ifp
=
2. Suppose a is any positive integer. Let a=
q1 a2 · · · qk, where the qi are prime powers,pairwise coprime. Letfq, be the polynomial just constructed and let Fq, be the powers of these that have the same degree. Let c. =a/qi.Note that for coprime x, y, c.Fq, (x, y)
=
Ci(mod a). By Bezout's lemma, there is an integer linear combination of the c.'s that equal 1. Thus, there is a linear combination of the Fq, such that Fq.(x, y)
