Mathematical Excalibur, Volume 2, Number 1

Olympiad Corner

First Day

Taipei, April 13, 1995

Problem 1. Let P(x) = a0 + a1x + +

an-1xn-1 + anxn be a polynomial with complex coefficients. Suppose the roots of P(x) are 1, 2, , n with | 1| > 1, | 2| > 1, , | j| > 1, and | j+1| 1, , | n| 1. Prove: i i j _n n a a a a = + + + 1 0 2 1 2 2 .

Problem 2. Given a sequence of integers x1, x2, x3, x4, x5, x6, x7, x8. One constructs a second sequence |x2 x1|, |x3 x2|, |x4

x3|, |x5 x4|, |x6 x5|, |x7 x6|, |x8 x7|, |x1 x8|. Such a process is called a single operation. Find all the 8-terms integral sequences having the following property: after finitely many single operations it becomes an integral sequence with all terms equal.

(continued on page 4) In mathematics, often we are interested in finding a solution to equations. Consider the following two problems:

Problem 1. Given real numbers m1,

m2, …, mn (all distinct) and a1, a2, …, an, find a polynomial v(x) such that

v(m1) = a1, v(m2) = a2, …, v(mn) = an.

Problem 2. Given positive integers m1,

m2, …, mn (pairwise relatively prime) and integers a1, a2, …, an, find an integer v such that v a1 (mod m1), v a2 (mod

m2), …, v an (mod mn).

Problem 1 comes up first in algebra and analysis (later in engineering and statistics). It is an interpolation problem, where we try to fit the values ai at mi (i.e., to find a polynomial whose graph passes through the points (m1,a1), (m2,a2), …, (mn,an)). Problem 2 comes up in number theory. It is a congruence problem, where we try to count objects by inspecting the remainders (i.e., to find a number which has the same remainder as ai upon division by mi).

There is a technique that can be applied to both problems. The idea is to solve first the special cases, where exactly one of the ai's is 1 and all others 0. For problem 1, this is easily solved by defining (for i = 1, 2, …, n) the polynomial Pi(x) to be (x• m1)(x• m2) (x• mn) with the factor (x• mi) omitted, i.e., P xi x mj j j i n ( )= ( ) =1 , and vi(x) = Pi(x)/Pi(mi). Then vi(mi) = 1 and vi(mk) = 0 for k i because Pi(mk) = 0 (for k i ).

For problem 2, this is solved similarly by first defining (for i = 1, 2, …, n) the integer Pi to be m1m2 mn with the factor

mi omitted. Consider Pi, 2Pi, …, miPi. Upon division by mi, no two of these will have the same remainder because the difference of any two of them is not divisible by mi. So one of these, say ciPi, has remainder 1. Let vi = ciPi, then vi 1 (mod mi) and vi 0 (mod mk) for k i because Pi 0 (mod mk).

Finally to solve problem 1 or 2 in general, we use the special case solutions

v1, v2, …, vn to form v = a1v1 + a2v2 + +

anvn. It is now easy to check that the expression v solves both problems 1 and 2. For problem 1, v x a P x P m a P x P m n n n n ( ) ( ) ( ) ( ) ( ) = 1 1 + + 1 1

is called Lagrange's interpolation

formula. For problem 2, although the ci's may be tedious to find, we know a solution v = a1c1P1 + + ancnPn exists. This is the assertion of the Chinese

remainder theorem. Note also that if we

add to v any multiple of (x• m1)(x• m2) (x• mn) in problem 1 or any multiple of

m1m2 mn in problem 2, we get other solutions.

The expression of v, involving a sum of multiples of v1, v2, …, vn, is so common in similar problems that it is now come to be called a linear combination of v1, v2, …,

vn. In passing, note that the ai's are

numbers. However, the vi's are

Mathematical Excalibur

Volume 2, Number 1 January-February, 1996

Solution by Linear Combination

Kin-Yin Li

Editors: Cheung, Pak-Hong, Curr. Studies, HKU Ko, Tsz-Mei, EEE Dept, HKUST Leung, Tat-Wing, Appl. Math Dept, HKPU Li, Kin-Yin, Math Dept, HKUST Ng, Keng Po Roger, ITC, HKPU Artist: Yeung, Sau-Ying Camille, MFA, CU

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 February 28, 1996.

For individual subscription for the remaining two issues for the 95-96 academic year, send us two stamped self-addressed envelopes. Send all correspondence to:

Dr. Tsz-Mei Ko

Department of Electrical and Electronic Engineering Hong Kong University of Science and Technology

Clear Water Bay, Kowloon, Hong Kong Fax: 2358-1485

polynomials in problem 1 and numbers in problem 2. Like vectors expressed in coordinates, the vi's are objects that may take on different values at different positions. So functions corresponding to solutions of equations are often viewed as vectors (with infinitely many coordinates). Concepts like these are the foundation of Linear Algebra, which studies the properties of solutions of these kind of problems in an abstract manner.

Example 1. If f(x) is a polynomial of

degree at most n and f(k) = (n+1• k)/(k+1) for k = 0, 1, …, n, find

f(n+1).

Solution 1. Applying Lagrange's

interpolation formula, we define Pk(x) =

x(x• 1) (x• n) with the factor (x• k)

omitted. Then Pk(n+1) = (n+1)!/(n+1• k), Pk(k) = (• 1) n-kk!(n• k)! and f n n k n k n k k n n ( ) ( ) ( ) ( )! ( )!( )! ( ) + = + + = + + = 1 1 1 1 1 1 1 1 0

where we used the binomial expansion of (1• 1)n+1 in the last step.

Solution 2. The polynomial g(x) =

(x+1)f(x) • (n+1• x) has degree at most

n+1. We are given that g(0) = g(1) = =

g(n) = 0. So g(x) = Cx(x• 1) (x• n). To find C, we set x = • 1 and get g(• 1) = • (n+2) = C(• 1)n+1(n+1)! . Therefore, C = (• 1)n(n+2)/(n+1)! and g(n+1) = (n+2)f(n+1) = (• 1)n(n+2), which implies

f(n+1) = (• 1)n.

Example 2. Prove that for each positive

integer n there exist n consecutive positive integers, none of which is an integral power of a prime number. (Source: 1989 IMO.)

Solution. Let p1, p2, …, p2n be 2n distinct prime numbers and consider the congruence problem v • 1 (mod p1p2),

v • 2 (mod p3p4), …, v • n (mod

p2n-1p2n). Since p1p2, p3p4, …, p2n-1p2n are pairwise relatively prime, by the Chinese remainder theorem, there is a positive integer solution v. Then each of the n consecutive numbers v+1, v+2, …, v+n is divisible by more than one prime number. So each is not a power of a prime number.

N 2 mod 3

N 3 mod 5

N 2 mod 7

Mathematical Excalibur, Vol. 2, No. 1, Jan-Feb, 96 Page 2

Solution by Linear Combination:

(continued from page 1)

N N 70 2 21 3 15 2 105 2 23 7 5 3 1 70 70 2 3 2 21 7 3 5 1 21 3 5 3 15 2 3 5 7 2 3 5 7 N

Problem Corner

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions should be preceded by the solver’s name, address, school affiliation and grade level. Please send submissions to Dr.

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

The deadline for submitting solutions is February 28, 1996.

Problem 26. Show that the solutions of

the equation cos x= 1

3 are all irrational numbers. (Source: 1974 Putnam Exam.)

Problem 27. Let ABCD be a cyclic

quadrilateral and let IA, IB, IC, ID be the incenters of BCD, ACD, ABD,

ABC, respectively. Show that IAIBICID is a rectangle.

Problem 28. The positive integers are

separated into two subsets with no common elements. Show that one of these two subsets must contain a three term arithmetic progression.

Problem 29. Suppose P(x) is a

nonconstant polynomial with integer coefficients and all coefficients are

greater than or equal to 1. If P(2) = 0, show that P(1) 0.

Problem 30. For positive integer n > 1,

define f(n) to be 1 plus the sum of all prime numbers dividing n multiplied by their exponents, e.g., f(40) = f(23 51) = 1 + (2 3 + 5 1) = 12. Show that if n > 6, the sequence n, f(n), f(f(n)), f(f(f(n))), … must eventually be repeating 8, 7, 8, 7, 8, 7, ….

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

Solutions

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

Solution 1: Independent solution by LIU Wai Kwong (Pui Tak Canossian

College) and YUNG Fai (CUHK). Construct a sequence u1 = 1, u2 = 1 and

un =

un 1+1

2 for n 3. We have

un < un+1 < 1 for n 2 and P(un) = (P(un+1)2/2) 1 for n 1. Note that P(un) 0 for n 1 (otherwise P(un) = 0 would imply P(un-1), P(un-2), , P(u1) are

rational, but P( )1 =1 3.)

Differentiating the functional equation for P, we get 4xP (2x2 1) = P(x)P (x). Since P(1) 4, we get P (u1) = P (1) = 0. This implies 0 = P (u2) = P (u3) = . Therefore, P (x) is the zero polynomial and so P(x) is constant.

Comments: This problem was from the

1991 USSR Math Winter Camp. Below we will provide a solution without calculus.

Solution 2: Suppose P(x) = a0xn + a1xn-1 + + an is such a polynomial with degree n 1. Then a x a x a a x a x a n n n n n n 0 2 1 2 1 0 1 1 2 2 1 2 1 2 1 ( ) ( ) ( ) . + + + = + + +

Comparing the coefficients of x2n, we find

a02n = a02/2, so a0 = 2n+1. Suppose a0, a1,

…, ak are known to be rational. Comparing the coefficients of x2n-k-1, the left side yields a rational number involving a0, …, ak, but the right side yields a number of the form a0ak+1 plus a rational number involving a0, …, ak. So

ak+1 is also rational. Hence a0, a1, …, an are all rational. Then P(1) = a0 + a1 + + an is rational. However, P(1) = (P(1)2/2)

1 forces P( )1 =1 3, a contradiction. Therefore P(x) must be constant.

Other commended solver: William CHEUNG Pok Man (S.T.F.A. Leung

Kau Kui College).

Problem 22. An acute-angled triangle ABC is given in the plane. The circle

with diameter AB intersects altitude CE and its extension at points M and N, and the circle with diameter AC intersects altitude BD and its extension at P and Q. Prove that the points M, N, P, Q lie on a common circle. (Source: 1990 USA Mathematical Olympiad).

Solution: William CHEUNG Pok Man

(S.T.F.A. Leung Kau Kui College). If M, N, P, Q are concyclic, then A must be the center because it is the intersection

of the perpendicular bisectors of PQ and

MN. So it suffices to show AP = AM.

Considering the similar triangles ADP and APC, we get AD/AP = AP/AC, i.e.,

AP2 = AD AC. Similarly, AM2 = AE AB. Since BEC = BDC, points B, C, D, E are concyclic. Therefore, AD AC =

AE AB and so AP = AM.

Other commended solvers: HO Wing Yip (Clementi Secondary School), LIU Wai Kwong (Pui Tak Canossian

College), Edmond MOK Tze Tao (Queen's College), WONG Him Ting (HKU) and YU Chun Ling (Ying Wa College).

Problem 23. Determine all sequences

{a1, a2, …} such that a1 = 1 and |an am| 2mn/(m2 + n2) for all positive integers m and n. (Source: Past IMO problem proposed by Finland).

Solution: Independent solution by CHAN Wing Sum (HKUST), LIU Wai Kwong (Pui Tak Canossian College) and YUNG Fai (CUHK).

For fixed m, lim lim n an am n mn m +n = 2 0 2 2 . So for all m, am a n n = lim .

It follows that all terms are equal (to a1 = 1.)

Problem 24. In a party, n boys and n

girls are paired. It is observed that in each pair, the difference in height is less than 10 cm. Show that the difference in height of the k-th tallest boy and the k-th tallest girl is also less than 10 cm for k = 1, 2, …, n.

Solution: Independent solution by HO Wing Yip (Clementi Secondary School) and YU Chun Ling (Ying Wa College).

Let b1 b2 bn be the heights of the boys and g1 g2 gn be those of the girls. Suppose for some k, |bk gk| 10. In the case bk gk 10, we have bi gj 10 for

1 i k and k j n. Consider the boys of height bi (1 i k) and the girls of height gj (k j n). By the pigeonhole principle, two of these n+1 must be paired originally. However, bi gj 10 contradicts the hypothesis. (The case

gk bk 10 is handled similarly.) So |bk gk| < 10 for all k.

Comments: This was a problem from the

1984 Tournament of the Towns, a competition started in 1980 at Moscow and Kiev and is now participated by students in dozens of cities in different continents.

Other commended solvers: CHAN Wing Sum (HKUST), William CHEUNG Pok Man (S.T.F.A. Leung Kau Kui College, KU Yuk Lun (HKUST), LIU Wai Kwong (Pui Tak Canossian College) and WONG Him Ting (HKU).

Problem 25. Are there any positive

integers n such that the first four digits from the left side of n! (in base 10 representation) is 1995?

Solution 1: LIU Wai Kwong (Pui Tak

Canossian College).

Let [x] be the greatest integer not exceeding x and {x} = x [x]. Also, let aj = 1 + j 10-8 , b0 = log 108! and bj = log 108! + (log a1 + + log aj) for j > 0. (For this solution, log means log10.) Observe that

(i) 0 < log ak log a30000 < log 1996 1995 for k = 1, 2, …, 30000;

(ii) logaj (loga loga ) .

j= > + > 1 30000 1 30000 15000 1

Note the distance between {log1995} and {log1996} is log(1996/1995). Now b0,

b1, ..., b30000 is increasing and b30000 b0 > 1 (by (ii)), but 0 < bj+1 bj < log 1996 1995 (by (i)).

So there is a k 30000 such that {log1995}<{ }bk <{log1996}. Now

log108!+ k_j=1loga_j=log(108+k)! 8k

implies

{log1995} < {log(108+k)!} < {log1996}. Adding [log1995] = [log1996] = 3, we have

log1995 < log(108+k)! m < log1996 for m = [log (108+k)!] 3. Therefore,

1995 10m < (108+k)! < 1996 10m. Consequently, the number (108+k)! begins with 1995.

Comments: With 1995 replaced by 1993,

this problem appeared in the 1993 German Mathematical Olympiad. Below we will provide the (modified) official solution.

Solution 2: Let m = 1000100000. If k <

99999 and (m+k)! = abcd (in base 10 representation), then (m+k+1)! = abcd

10001 = efgh , where efgh equals

abcd or the first four digits of abcd+1. So,

the first four digits of each of (m+1)!, (m+2)!, …, (m+99999)! must be the same as or increase by 1 compared with the previous factorial. Also, because the fifth digit of m+k (k < 99999) is 1, the fifth digit of (m+k)! will be added to the first digit of (m+k)! in computing (m+k+1)!. So, in any ten consecutive factorials among (m+1)!, (m+2)!, …, (m+99999)!, there must be an increase by 1 in the first four digits. So the first four digits of (m+1)!, (m+2)!, …, (m+99999)! must take on all 9000 possible choices. In particular, one of these is 1995.

Olympiad Corner

Problem 3. Suppose n persons meet in a

meeting, every one among them is familiar with exactly 8 other participants of that meeting. Furthermore suppose that each pair of two participants who are familiar with each other have 4 acquaintances in common in that meeting, and each pair of two

(continued on page 4)

Mathematical Excalibur, Vol. 2, No. 1, Jan-Feb, 96 Page 4

Problem Corner

Erratum: In the article in

the last issue, (Chebychev

Theorem) should be corrected as

n n n n n 8 12

each other have only 2 acquaintances in common. What are the possible values of

n?

***************** Second Day

Taipei, April 15, 1995

Problem 4. Given n (where n 2) distinct

integers m1, m2, , mn. Prove that there exist a polynomial f(x) of degree n and with integral coefficients which satisfies the following conditions:

(i) f(mi) = 1, for all 1 i n.

(ii) f(x) cannot be factorized into a product of two nonconstant polynomials with integral coefficients.

Problem 5. Let P be a point on the

circumscribed circle of A1A2A3. Let H be the orthocenter of A1A2A3. Let B1 (B2, B3 respectively) be the point of intersection of the perpendicular from P to A2A3 (A3A1, A1A2 respectively). It is known that the three points B1, B2, B3 are colinear. Prove that the line B1B2B3 passes through the midpoint of the line segmentPH.

Problem 6. Let a, b, c, d be integers such

that ad bc = k > 0, (a,b) = 1, and (c,d) = 1. Prove that there are exactly k ordered pairs of real numbers (x1,x2) satisfying 0

x1, x2 < 1 and both ax1 + bx2 and cx1 + dx2 are integers.