Volume 16, Number 1 May 2011
Harmonic Series (II)
Leung Tat-Wing
Olympiad Corner
Below are the problems of the 2011 Asia Pacific Math Olympiad, which was held in March 2011.
Problem 1. Let a, b, c be positive
integers. Prove that it is impossible to have all of the three numbers a2+b+c,
b2+c+a, c2+a+b to be perfect squares. Problem 2.
Five points A
1, A2, A3, A4,A5 lie on a plane in such a way that no
three among them lie on a same straight line. Determine the maximum possible value that the minimum value for the angles ∠AiAjAk can take where i, j, k
are distinct integers between 1 and 5.
Problem 3. Let ABC be an acute
triangle with ∠BAC=30°. The internal and external angle bisectors of ∠ABC meet the line AC at B1 and B2,
respectively, and the internal and external angle bisectors of ∠ACB meet the line AB at C1 and C2, respectively.
Suppose that the circles with diameters B1B2 and C1C2 meet inside the triangle
ABC at point P. Prove that ∠BPC = 90°.
(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 June 25, 2011.
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 usual, for integers a, b, n (with n > 0), we write a≡b (mod n) to mean a−b is divisible by n. If b≠0 and n are relatively prime (i.e. they have no common prime divisor), then 0, b, 2b, …, (n−1)b are distinct (mod n) because for 0 ≤ s < r<n, rb ≡ sb (mod n) implies (r−s)b = kn. Since b, n have no common prime divisor, this means b divides k. Then 0 < (k/b)n = r−s < n, contradicting b ≤ k. Hence, there is a unique r among 1, …, n−1 such that rb≡ 1 (mod n). We will denote this r as b−1 or 1/b (mod n). Further, we can
extend (mod n) to fractions by defining a/b ≡ ab−1 (mod n). We can easily
check that the usual properties of fractions holds in mod n arithmetic. Next, we will introduce Wolstenholme’s theorem, which is an important relation concerning harmonic series.
Theorem (Wolstenholme): For a prime
p ≥ 5, 2 1 1 ( 1) 1 ... 0 (mod ). 2 1 H p p p − = + + + ≡ −
(More precisely, for a prime p ≥ 5, if
1 1 ( 1) 1 ... , 2 1 a H p p b − = + + + = − then p2 | a.) Example We have 1 1 1 7381 (10) 1 ... 2 3 10 2520 H = + + + + = and 112 | 7381. First proof We have
1 1 2 1 1 ) 1 ( − + + + = − p p H L
∑
−∑
= − = − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + =( 1)/2 1 2 / ) 1 ( 1 . ) ( 1 1 1 p n p n n p n p n p n So we need to prove ( 1) / 2 1 1 0 (mod ). ( ) p n p n p n − = ≡ −∑
Now( 1) / 2 ( 1) / 2 2 1 1 1 1 (mod ). ( ) p p n n p n p n n − − = = ≡ − −∑
∑
Since every 1/n2 is congruent to exactly
one of the numbers 12, 22, … , [(p−1)/2]2
(mod p) and 1/n2 are all distinct for n
=1,2,…,(p −1)/2, we have when p ≥ 5, ). (mod 0 24 ) 1 ( 1 2 2 / ) 1 ( 1 2 / ) 1 ( 1 2 2 p p p k n p n p k ≡ − = ≡
∑
−∑
= − =Wolstenholme’s theorem follows.
Second proof (using polynomials mod
p) We use a theorem of Lagrange, which says if f (x) = c0 + c1x + ⋯ + cnxn is a
polynomial of degree n, with integer coefficients, and if f (x) ≡ 0 (mod p) has more than n solutions, where p is prime, then every coefficient of f (x) is divisible by p. The proof is not hard. It can be done basically by induction and the division algorithm mod p. The statement is false if p is not prime. For instance, x2 − 1 ≡ 0 (mod 8) has 4
solutions. Here is the other proof. From Fermat’s Little theorem, xp−1 ≡ 1
(mod p) has 1, 2, …, p −1 as solutions. Thus xp−1−1 ≡ (x−1)(x−2) ⋯ (x−p+1) (mod p). Let (x−1)(x−2) ⋯ (x−p+1) = xp−1 − s 1xp−2 + ⋯ − sp−2x + sp−1. (*) By Wilson’s theorem, sp−1 = (p−1)! ≡ −1 (mod p). Thus 0 ≡ s1xp−2 + ⋯ − sp−2x (mod p).
The formula is true for every integer x. By Lagrange’s theorem, p divides each of s1, s2, …, sp−2. Putting x = p in (*), we
get (p−1)!=pp−1−s
1pp−2 + ⋯ − sp−2p+ sp−1.
Canceling out (p−1)! and dividing both sides by p, we get
0 = pp−2 − s
1 pp−3 + ⋯+sp−3 p − sp−2.
As p ≥ 5, each of the terms is congruent to 0 (mod p2). Hence, we have s
p−2≡ 0 (mod p2). Finally, 2 1 1 ( 1)!(1 ... ) ( 1)! . 2 1 p a s p p p b − = − + + + − = −
Mathematical Excalibur, Vol. 16, No. 1, May. 11 Page 2 Using Wolstenholme’s theorem and
setting x = kp in (*), we get (kp−1)(kp−2)⋯(kp−p+1) = (kp)p−1−s 1(kp)p−2+⋯ +sp−3(kp)2−sp−2kp+sp−1 ≡sp−3(kp)2 − sp−2kp + sp−1 ≡ (p −1)! (mod p 3).
Upon dividing by (p−1)!, we have
3 1 1 (mod ), 1, 2,... . 1 kp p k p − ⎛ ⎞ ≡ = ⎜ − ⎟ ⎝ ⎠
This result may in fact be taken as the statement of Wolstenholme’s theorem. Here are a few further remarks. Wolstenholme’s theorem on the congruence of harmonic series is related to the Bernoulli numbers
B
n.
For instance, we have2 3 4 3 1 1 1 ( ) (mod ) 1 3 p kp k k p B p p − − ⎛ ⎞≡ − − ⎜ − ⎟ ⎝ ⎠ ,
which is usually called Glaisher’s congruence. These numbers are related to Fermat’s Last Theorem. It is known that for any prime p ≥ 5,
3 1 1 (mod ). 1 kp p p − ⎛ ⎞≡ ⎜ − ⎟ ⎝ ⎠
Are there primes satisfying
4 2 1 1 (mod )? 1 p p p − ⎛ ⎞ ≡ ⎜ − ⎟ ⎝ ⎠
These primes are called Wolstenholme primes. (So far, we only know 16843 and 2124679 are such primes). In another direction, one can ask if there exist composite numbers n such that
3 1 1 (mod )? 1 kn n n − ⎛ ⎞ ≡ ⎜ − ⎟ ⎝ ⎠
All these are very classical questions.
Example 10 (APMO 2006): Let p ≥ 5 be a prime and let r be the number of ways of placing p checkers on a p×p checkerboard so that not all checkers are in the same row (but they may all be in the same column). Show that r is divisible by p5.
Solution Observe that
. 1 )! 1 ( )) 1 ( ( ) 1 ( 2 2 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − − − − = − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = p p p p p p p p r L
Hence it suffices to show that
(p2−1)(p2−2)⋯(p2−(p−1)) − (p−1)! ≡ 0 (mod p4) (1) Now let f(x) = (x−1)(x−2)⋯(x−(p−1)) = xp−1 + s 1xp−2 + ⋯ + sp−2x + sp−1. (2)
Thus the first congruence relation is the same as f(p2) − (p−1)! ≡ 0 (mod p4).
Therefore it suffices to show that sp−2 p2≡
0 (mod p4) or s
p−2≡ 0 (mod p2), which is
exactly Wolstenholme’s theorem.
Example 11 (Putnam 1996): Let p be a prime number greater than 3 and k =[2p/3]. Show that 2 ... 0 (mod ) 1 2 p p p p k ⎛ ⎞ ⎛ ⎞+ + +⎛ ⎞≡ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ For example, ). 7 (mod 0 98 4 7 3 7 2 7 1 7 2 ≡ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ Solution Recall ( 1)...( 1) . 1 2 ... p p p p i i i ⎛ ⎞= − − + ⎜ ⎟ ⋅ ⋅ ⎝ ⎠ This is a multiple of p if 1 ≤ i ≤ p−1. Modulo p, the right side after divided by p is congruent to . 1 ) 1 ( 2 1 )) 1 ( ( ) 1 ( 1 i i i = − i− ⋅ ⋅ − − − L L
Hence, to prove the congruence, it suffices to show ). (mod 0 1 ) 1 ( 3 1 2 1 1 1 p k k ≡ − + − + − − L
Now observe that
1 1 1
(mod ).
2i 2i p i p
− ≡ +
−
This allows us to replace the sum by
1 1
1 ... 0 (mod ),
2 p 1 p
+ + + ≡
−
which is Wolstenholme’s theorem. We can also give a more detailed proof as follow. Let n n H 1 3 1 2 1 1 ) ( = + + +L+ and 1 1 1 ( ) 1 ... ( 1) . 2 n P n n − = − + + −
Then the problem is reduced to showing that for any p > 3, p divides the numerator
of P([2p/3]). First we note that p divides the numerator of H(p−1) because 2H(p−1) ) 1 1 1 ( ) 2 1 2 1 ( ) 1 1 1 ( + − + + − + + − + = p p p L ). (mod 0 1 ) 2 ( 2 1 p p p p p p p ≡ − + + − + − = L
Next we have two cases.
Case 1 (p = 3n+1) Then [2p/3] = 2n. So we must show p divides the numerator of P(2n). Now (3 ) (2 ) 1 1 1 1 1 2(1 ... ) ( ... ) 2 2 2 1 2 2 3 1 1 1 1 1 (1 ... ) ( ... ) 2 2 1 2 2 3 1 1 1 1 (1 ) (2 ) ... ( ) 1 2 ... . 1 2( 2) ( ) H n P n n n n n n n n n p p n p n p p p p p n p n − = + + + + + + + + + = + + + + + + + + + = + + + + + + − − − = + + + − − −
So p divides the numerators of both H(3n) and H(3n) − P(2n), hence also the numerator of P(2n).
Case 2 (p = 3n+2) Then [2p/3] = 2n+1. So we must show p divides the numerator of P(2n+1). Now H n(3 + −1) P n(2 +1) 1 1 1 1 1 2(1 ... ) ( ... ) 2 2n 2 2 2 2n n 3 1n = + + + + + + + + + + 1 1 1 1 1 (1 ... ) ( ... ) 2 n 2 2 2 2n n 3 1n = + + + + + + + + + + 1 1 1 1 (1 ) (2 ) ... ( ) 1 2 p p n p n = + + + + + + − − − ... . 1 2( 2) ( ) p p p p p n p n = + + + − − −
So, p divides the numerator of H(3n+1)−P(2n+1), and hence P(2n+1).
Example 12: Let p ≥ 5 be a prime, show that if 1 1 1 ... , 2 a p b + + + = then p4| ap −b. (continued on page 4)
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 June 25, 2011.
Problem 371. Let a1, a2, a3, … be a
sequence of nonnegative rational numbers such that am+an= amn for all
positive integers m, n. Prove that there exist two terms that are equal.
Problem 372. (Proposed by Terence ZHU) For all a,b,c > 0 and abc = 1, prove that ) 1 ( ) 1 ( 1 ) 1 ( ) 1 ( 1 + + + + + + + abab bb bcbc a a . 4 3 ) 1 ( ) 1 ( 1 ≥ + + + + ca ca c c
Problem 373. Let x and y be the sums of positive integers x1,x2,…,x99 and
y1,y2,…,y99 respectively. Prove that
there exists a 50 element subset S of {1,2,…,99} such that the sum of all xn
with n in S is at least x/2 and the sum of all yn with n in S is at least y/2. Problem 374. O is the circumcenter of acute ∆ABC and T is the circumcenter of ∆AOC. Let M be the midpoint of side AC. On sides AB and BC, there are points D and E respectively such that ∠BDM=∠BEM=∠ABC. Prove that BT⊥DE.
Problem 375. Find (with proof) all odd integers n > 1 such that if a, b are divisors of n and are relatively prime, then a+b−1 is also a divisor of n.
*****************
Solutions
****************Problem 366. Let n be a positive integer in base 10. For i=1,2,…,9, let a(i) be the number of digits of n that equal i. Prove that
1
10
9
3
2
a(1) a(2) a(8) a(9)≤ n
+
L
and determine all equality cases.
Solution. LAU Chun Ting (St. Paul’s
Co-educational College, Form 2).
Let f(n)=2a(1)3a(2)⋯9a(8)10a(9). If n is a
number with one digit, then f (n) = n+1. Suppose all numbers A with k digits satisfy the given inequality f (A) ≤ A+1. For any (k+1) digit number, it is of the form 10A+B, where A is a k digit number and 0 ≤ B ≤ 9. We have
f(10A+B) = (B+1) f(A) ≤ (B+1)(A+1) = (B+1)A+B+1 ≤ 10A+B+1. Equality holds if and only if f (A) = A+1 and B = 9. By induction, the inequality holds for all positive integers n and equality holds if and only if all but the leftmost digits of n are 9’s.
Other commended solvers: CHAN Long
Tin (Diocesan Boys’ School), LEE Tak
Wing (Carmel Alison Lam Foundation Secondary School), Gordon MAN Siu
Hang (CCC Ming Yin College) and
YUNG Fai.
Problem 367. For n = 1,2,3,…, let xn and
yn be positive real numbers such that 2 1 2 + + = n+ n n x x x and . 1 2 2 + + = n+ n n y y y
If x1, x2, y1, y2 are all greater than 1, then
prove that there exists a positive integer N such that for all n > N, we have xn > yn.
Solution. LAU Chun Ting (St. Paul’s
Co-educational College, Form 2) and
Gordon MAN Siu Hang (CCC Ming Yin
College).
Since x1, x2, y1, y2 are all greater than 1, by
induction, we can get xn+1> xn2 > 1 and
yn+1> 1+yn > n for n ≥ 2. Then xn+2 =
xn+xn+12 > xn+12 > xn4 and yn+2 = yn2+yn+1 =
yn2+yn−12+yn < 3yn2 < yn3 for all n ≥ 4.
Hence, log xn+2 > 4 log xn and log yn+2 < 3
log yn. So for n ≥ 4, . log log 3 4 log log 2 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ > + + n n n n y x y x (*)
As 4/3 > 1, by taking logarithm, we can solve for a positive integer k satisfying the inequality . 1 log log , log log min 3 4 5 5 4 4 > ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ y x y x k
Let N = 2k+3. If n > N, then either n = 2m+4 or n = 2m+5 for some integer m ≥ k. Applying (*) m times, we have
. 1 log log , log log min 3 4 log log 5 5 4 4 > ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ > y x y x y x m n n This implies xn > yn.
Other commended solvers: LEE Tak
Wing (Carmel Alison Lam Foundation Secondary School) and NGUYEN
Van Thien (Luong The Vinh High
School, Dong Nai, Vietnam).
Problem 368. Let C be a circle, A1,
A2, …, An be distinct points inside C
and B1, B2, …, Bn be distinct points on
C such that no two of the segments A1B1, A2B2,…, AnBn intersect. A
grasshopper can jump from Ar to As if
the line segment ArAs does not intersect
any line segment AtBt (t≠r,s). Prove
that after a certain number of jumps, the grasshopper can jump from any Au
to any Av.
Solution. William PENG.
The cases n = 1 or 2 are clear. Suppose n ≥ 3. By reordering the pairs Ai, Bi, we
may suppose the convex hull of A1,
A2, …, An is the polygonal region M with
vertices A1, A2, …, Ak (k≤n). For
1≤m≤k, if every AmBm intersects M only
at Am, then the n-th case follows by
removing two pairs of Am, Bm separately
and applying case n −1.
M
AI Ax A1 Ak AJ Ay Am Bm Dm Bx Dx By BI BJ DyOtherwise, there exists a segment AmBm
intersecting M at more than 1 point. Let it intersect the perimeter of M again at Dm.
Since AiBi’s do not intersect, so AjDj’s
(being subsets of AiBi’s) do not intersect.
In particular, Dm is not a vertex of M.
Now AmDm divides the perimeter of M
into two parts. Moving from Am to Dm
clockwise on the perimeter of M, there are points Ax,Dx such that there is no
Dw between them. As Dx is not a vertex,
there is a vertex AI between Ax and Dx.
Then AIBI only intersect M at AI. Also,
moving from Am to Dm anti-clockwise
on the perimeter of M, there is AJ such
that AJBJ only intersects M at AJ. Then
AIBI and AJBJ do not intersect any
diagonal of M with endpoints different from AI and AJ.
Mathematical Excalibur, Vol. 16, No. 1, May. 11 Page 4 Removing AI, BI and applying case n−1,
the grasshopper can jump between any two of the points A1,…,AI−1, AI+1,…,An.
Also, removing AJ, BJ and applying
case n−1, the grasshopper can jump between any two of the points A1,…,AJ−1, AJ+1,…,An. Using these two
cases, we see the grasshopper can jump from any Au to any Av via At (t≠I,J).
Other commended solvers: T. h. G.
Problem 369. ABC is a triangle with BC > CA > AB. D is a point on side BC and E is a point on ray BA beyond A so that BD=BE=CA. Let P be a point on side AC such that E, B, D, P are concyclic. Let Q be the intersection point of ray BP and the circumcircle of ΔABC different from B. Prove that AQ+CQ=BP.
Solution. CHAN Long Tin (Diocesan
Boys’ School), Giorgos KALANTZIS (Demenica’s Public High School, Patras, Greece) and LAU Chun Ting (St. Paul’s Co-educational College, Form 2). C B A E D P Q
Since A,B,C,Q are concyclic and E,P,D,B are concyclic, we have
∠AQC = 180°−∠ABC =∠EPD and
∠PED=∠PBD =∠QAC. Hence, ΔAQC and ΔEPD are similar. So we have AQ/AC=PE/DE and CQ/AC = PD/DE. Cross-multiplying and adding these two equations, we get (AQ+CQ)×DE = (PE+PD)×AC. (*) For cyclic quadrilateral EPDB, by the Ptolemy theorem, we have
BP×DE = PE×BD+PD×BE = (PE+PD)×AC (**) Comparing (*) and (**), we have AQ+CQ=BP.
Other commended solvers: LEE Tak
Wing (Carmel Alison Lam Foundation Secondary School).
Problem 370. On the coordinate plane, at every lattice point (x,y) (these are points where x, y are integers), there is a light. At time t = 0, exactly one light is turned on. For n = 1, 2, 3, …, at time t = n, every light at a lattice point is turned on if it is at a distance 2005 from a light that was on at time t = n − 1. Prove that every light at a lattice point will eventually be turned on at some time.
Solution. LAU Chun Ting (St. Paul’s
Co-educational College, Form 2), LEE
Tak Wing (Carmel Alison Lam
Foundation Secondary School), Gordon
MAN Siu Hang (CCC Ming Yin College) and Emanuele NATALE (Università di Roma “Tor Vergata”, Roma, Italy). We may assume the light that was turned on at t = 0 was at the origin.
Let z = 2005 = 5×401 = (22+12)(202+12) =
|(2+i)(20+i)|2 = |41+22i|2 = 412+222. Let x
= 412−222=1037 and y =2×41×22 = 1716.
Then x2+y2 = z2.
By the Euclidean algorithm, we get gcd(1037, 1716) = 1. By eliminating the remainders in the calculations, we get 84×1716−139×1037=1.
Let V1, V2, V3, V4, V5 be the vectors from
the origin to (2005,0), (1037, 1716), (1037, −1716), (1716, 1037), (1716, −1037) respectively. We have V2 + V3 = (2×1037,0)
and V4 + V5 = (2×1716,0). Then we can get
(1,0) =1003[84(V4+V5)−139(V2+V3)]−V1.
So, from the origin, following these vector movements, we can get to the point (1,0). Similarly, we can get to the point (0,1). As (a,b) = a(1,0) + b(0,1), we can get to any lattice point.
Olympiad Corner
(continued from page 1)
Problem 4. Let n be a fixed positive odd integer. Take m+2 distinct points P0,
P1, …, Pm+1 (where m is a non-negative
integer) on the coordinate plane in such a way that the following 3 conditions are satisfied:
(1) P0=(0,1), Pm+1=(n+1,n), and for each
integer i, 1 ≤ i ≤ m, both x- and y- coordinates of Pi are integers lying in
between 1 and n (1 and n inclusive). (2) For each integer i, 0 ≤ i ≤ m, PiPi+1 is
parallel to the x-axis if i is even, and is parallel to the y-axis if i is odd.
(3) For each pair i,j with 0 ≤ i < j ≤ m, line segments PiPi+1 and PjPj+1 share at
most 1 point.
Determine the maximum possible value that m can take.
Problem 5. Find all functions f:ℝ→ℝ, where ℝ is the set of all real numbers, satisfying the following 2 conditions: (1) There exists a real number M such that for every real number x, f(x) < M is satisfied.
(2) For every pair of real numbers x and y, f (x f (y)) + y f (x) = x f (y) + f (xy) is satisfied.
Harmonic Series (II)
(continued from page 2)
Solution By Wolstenholme’s theorem, . 1 1 2 1 1 )! 1 ( 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + + + − p p p L So, 2 1 1 1 ... , 2 1 x p p y + + + = −
where x, y are integers with y not divisible by p. So we have , 1 2 y x p p b a = −
which implies ap−b = p3bx/y. Finally, ! ) 1 ( 2 1 4 3 1 3 2 p p p p b a= ⋅ L + ⋅ ⋅ L +L+ ⋅ L −
and the numerator of the right side is of the form mp+(p−1)!. Hence, it is not divisible by p. So p | b and p4 | p3bx/y =
ap−b.
Example 13: Let p be an odd prime, then prove that
2 2 2 2 1 1 1 1 ... ( 1) 0(mod ). 2 3 ( 1) p p p − − + − + − ≡ −
Solution The proof is not hard. Indeed,
∑
− = − − 1 1 2 1 1 ) 1 ( p k k k∑
− = − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − + − − = ( 1)/2 1 2 ( )2 1 ) 1 ( 1 ) 1 ( p k k p k k p k∑
− = − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − + − − ≡ ( 1)/2 1 2 1 2 ( ) 0(mod ). 1 ) 1 ( 1 ) 1 ( p k k k p k k5
olIN-h;oY\.s
~ obt\f~
C00't1bV )
V~.
t {;
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SOLUTIONS FOR 2011 APMO PROBLEMSProblem 1.
Solution: Suppose all of the 3 numbers a2
+
b+
c, b2+
c+
a and c'J.
+
a+
bare perfect squares. Then from the fact that a2+
b+
c is a perfect square bigger thana2 it follows that a2
+
b+
c 2: (a+
1)2, and therefore, b+
c 2: 2a+
1. Similarly we obtain c+
a 2: 2b+
1 and a+
b 2: 2c+
1.Adding the corresponding sides of the preceding 3 inequalities, we obtain
2(a
+
b+
c) 2: 2(a+
b+
c)+
3, a contradiction. This proves that it is impos-sible to have all the 3 given numbers to be perfect squares.Alternate Solution:, Since the given conditions of the problem are symmetric in a, b, c, we may assume that a 2: b 2: c holds. From the assumption that a2
+
b+ c is a perfect square, we can deduce as in the solution above the inequality b+
c 2: 2a+
1. But then we have2a 2: b
+
c 2: 2a+
1,a contradiction, which proves the assertion of the problem. Problem 2.
Solution: We will show that 36° is the desired answer for the problem.
First, we obs~erve that if the given 5 points form a regular pentagon, then the minimum of the angles formed by~ any triple among the five vertices is 36°, and therefore, the answer we seek must be bigger than or equal to 36°.
Next, we show that for any configuration of 5 points satisfying the condition of the problem, there must exist an angle smaller than or equal to 36° formed by a triple chosen from the given 5 points. For this purpose, let us start with any 5 points, say AI, A 2, A3 , A4, A5 , on the plane satisfying the condition of the problem, and c~nsider the smallest convex subset, call it
r,
in the plane containing all of the 5 points. Since this convex subsetr
must be either a triangle or a quadrilateral or a pentagon, it must have an interior angle with 108° or less. We may assume without loss of generality that this angle is LAIA2A3' By the definition ofr
it is clear that the remaining 2 points A4 and A5 lie in the interior of the angular region determined by LA IA2A3, and therefore, there must be an angle smaller than or1 .
equal to
3' .
10So = 36°, which is formed by a triple chosen from the given 5 points, and this proves that 36° is. the desired maximum.Problem 3.
Solution: Since LBIBB2 = 90°, the circle having B}B2 as its diameter goes through the points B, B I, B2. From BIA : BIC = B2A : B 2C = BA : BC, it follows that this circle is the Apolonius circle with the ratio of the dh3tances from the points A and 0 being BA : BO. Since the point P lies on this circle, we have
PA: PO
=
BA: BO=
sinO: sinA,from which it follows that P A sin A = PC sin O. Similarly, we have P A sin A =
P B sin B, and therefore, P A sin A = P B sin B ::=; PO sin O.
Let us denote by D, E, F the foot of the perpendicular line drawn from P to the line segment BO, OA and AB, respectively. Since the points E, F lie on a circle .having PA as its diameter, we have by the law oi'sines EF
=
PAsinA. Similarly, we have FD = PBsinB and DE=
POsinO. Consequently, we conclude thatDEF is an equilateral triangle. Furthermore, we have LOPE
=
LODE since the quadrilateral 0 D P E is cyclic. Similarly, we have LF P B=
LF DB. Pu;ting these together, we getLBPO = 360° - (LOPE
+
LFPB+
LEPF)=
360° - {(LODE+
LFDB)+
(180° - LFAE)}=
360° - (120°+
150°)=
90°,which proves the asserti~n o~_~he_~robl~em.~,
Problem 4.
Soh~tion: We will show that the desired maximum value for m is n(n ~ 1).
. FIrst, let us show that . . . . m :::: n(n~ 1) always holds for any sequence R P ... 0, 1, , p m+l
satlsfymg the condItIOns of the problem.
Call a point a turning p~int if it coincides with
Pi
fOr'some
~ 'i WIth I ::.; ~::::m.
Let us say also that 2 points {P,Q} are adjacent if {P,Q}=
{Pi-I, Pi} for somei with 1 :::: i :::: m, and vertically adjacent if, in addition, PQ is parallel to the
~ax~. .
Any turning point is vertically adjacent to exactly one other turning point. Therefore, the set of all turning points is partitioned into a set of pairs of points using the relation of "vertical adjacency". Thus we can conclude that if we fix
k E {1, 2, ... , n}, the number of turning points having the x-coordinate k must be even, and hence it is less than or equal to n - 1. Therefore, altogether there are less than or equal to n(n - 1) turning points, and this shows that m :::: n(n - 1) must be s a t i s f i e d . · .
It remains now to show that for any positive odd number n one can choose a sequence for which m = n(n - 1). We will show this by using the mathematical induction on n. For n = 1, this is clear. For n
=
3, choosePo
=
(0,1), PI = (1,1), P4 = (2,1), P5=
(3,1), P2 = (1,2), P6 = (3,3), P3 = (2,2), P7=
(4,3).It
is easy to see that these points satisfy the requirements (See fig. 1 below).Let n be an odd integer 2: 5, and suppose there exists a sequence satisfying the desired conditions for n - 4. Then, it is possible to construct a sequence which gives a configuration indicated in the following diagram (fig. 2), where the configuration inside of the dotted square is given by the induc,tion hypothesis:
"
-.~
· .-·
.
·
..
' ..
,.
.
_1_: :·
-... --... ..J1StJ
figure 2By the induction
hypoth~sis,
there are exactly (n - 4)(n - 5) turning points for the configuration inside of the dotted square in the figure 2 above, and all of the lattice points in the figure 2 lying outside of the dotted square except for the 4points (n, 2), (n -1, n - 2), (2,3), (1, n -1) are turning points. Therefore, the total' number of turning points in this configuration is
, (n - 4)(n - 5) + (n
2
- (n - 4)2 - 4) =n(n - I),
showing that for this n there exists a sequence satisfying the desired properties, and thus completing the induction process.
Problem 5.
Solution: By substituting x
=
1 and y=
1 into the given identity we obtainf(1(1» = f(I). Next, by substituting x
=
1 and y=
f(l) into the given identity~ . and using f(1(I» = f(I), we get f(I)2 = f(I), from which we conclude that eitherf(l) = 0 or f(1)
=
1. But if f(l) = 1, then substituting y=
1 into the given identity, we get f(x) = x for all x, which contradicts the condition (1). Therefore, we must have f(l)=
O.By substituting x = 1 into the given identity and using the fact f(l) = 0, we then obtain f(1(y»
=
2f(y) for all y. This means that if a number t belongs to the range of the functionf,
then so does 2t, and ~y induction we can conclude that for any non-negative integer n, 2nt belongs to the range off
if t does. Now suppose that there exists a real number a for which f(a)>
0, then for any non-negative integer n 2n f(a) must belong to the range of f, which leads to a contradiction to the condition (1). Thus we conclude that f(x) :::; 0 for any real numberx.By substituting ~ for x and f(y) for y in the given identity and using the fact that f(1(y» = 2f(y), we obtain
f(xf(y»
+
f(y)f(~)
=
xf(y)+
f(~f(Y»)
,from which it follows that xf(y) - f(xf(y»
=
f(y)f (~) - f (~f(Y») 2: 0, since the values off
are non-positive. Combining this with the given identity, we conclude that y f (x)2':
f (xy). When x>
0, by letting y to be~
and using the fact thatx .
f(l)
=
0, we get f(x) 2: O. Since f(x) ::::; 0 for any real number x, we conclude thatf(x)
=
0 for any positive real number x. We also have f(O)=
f(1(I» = 2f(l)=
O.If f is identically 0, i.e., f(x)
=
0 for all x, then clearly, this f satisfies the given identity. Iff
satisfies the given identity but not identically 0, then there exists ab
<
0 for which f(b)<
O. If we set c=
f(b), then we have f(e)=
f(1(b» :;= 2f(b) =. 2e. For any negative real number x, we have ex
>
0 so that f(ex) = f(2ex) = 0, and by substituting y = e into the given identity, we getf(29X )
+
ef(x) = 2ex+
f(ex),from wlJ.ich it follows that f(x) = 2x for any negative real x .
. We therefore conclude that if
f
satisfies the given identity and is not identically{
o
if x
>
0 .0, then f is of the form f(x) = . - Finally, let us show that the
. 2x If x
<
o.
function
f
of the form shown above does satisfy the conditions of the problem. Clearly, it satisfies the condition (1). We can check thatf
satisfies the condition(2) as well by separating into the following 4 cases depending on whether x, yare non-negative or negative.
• when both x and yare non-negative, both sides of the given identity are O.
• when x is non-negative and y is negative, we have xy :::; 0 and both sid!is of the given identity are 4xy.
• when x is negative and y is non-negative, we have xy :::; 0 and both sides of the given identity are 2xy.
• when both x and yare negative, we have xy
>
0 and both sides of the given identity are 2xy.Summarizing the arguments above, we conclude. that the functions
f
satisfying the conditions of the problem aref(x) = 0 and f(x) =
{o
2x
if x 2: 0 if x
