Mathematical Excalibur, Volume 15, Number 2

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Volume 15, Number 2 July - September, 2010

Lagrange Interpolation Formula

Kin Y. Li

Olympiad Corner

Below are the problems used in the selection of the Indian team for IMO-2010.

Problem 1. Is there a positive integer n, which is a multiple of 103, such that 22n+1_{≡2 (mod n)?}

Problem 2.

Let a, b, c be integers such that b is even. Suppose the equation x3_+ax2_{+bx+c=0 has roots α, β, γ such}

that α2 _{= β+γ. Prove that α is an integer}

and β≠γ.

Problem 3. Let ABC be a triangle in which BC < AC. Let M be the midpoint of AB; AP be the altitude from A on to BC; and BQ be the altitude from B on to AC. Suppose QP produced meet AB (extended) in T. If H is the orthocenter of ABC, prove that TH is perpendicular to CM.

Problem 4. Let ABCD be a cyclic quadrilateral and let E be the point of intersection of its diagonals AC and BD. Suppose AD and BC meet in F. Let the midpoints of AB and CD be G and H respectively. If Γ is the circumcircle of triangle EGH, prove that FE is tangent to Γ.

(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 October 20, 2010.

For individual subscription for the next five issues for the 10-11 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]

Let n be a positive integer. If we are given two collections of n+1 real (or complex) numbers w0, w1, …, wn and

c0, c1, …, cn with the wk’s distinct, then

there exists a unique polynomial P(x) of degree at most n satisfying P(wk) = ck for k = 0,1,…,n. The uniqueness is clear since if Q(x) is also such a polynomial, then P(x)−Q(x) would be a polynomial of degree at most n and have roots at the n+1 numbers w0, w1, …, wn, which leads

to P(x)−Q(x) be the zero polynomial. Now, to exhibit such a polynomial, we define f0(x)=(x−w1)(x−w2)⋯(x−wn) and

similarly for i from 1 to n, define fi(x)=(x−w0)⋯(x−wi−1)(x−wi+1)⋯(x−wn).

Observe that fi(wk) = 0 if and only if i≠k. Using this, we see

∑

= = n i i i i i _f _w x f c x P 0 ( ) ) ( ) (

satisfies P(wk) = ck for k = 0,1,…,n. This is the famous Lagrange interpolation formula.

Below we will present some examples of using this formula to solve math problems.

Example 1. (Romanian Proposal to 1981 IMO) Let P be a polynomial of degree n satisfying for k = 0,1,…,n,

. 1 ) ( 1 − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = k n k P Determine P(n+1).

Solution. For k = 0,1,…,n, let wk=k and . )! 1 ( )! 1 ( ! 1 1 + − + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = − n k n k k n c_k

Define f0, f1, … , fn as above. We get

fk(k) = (−1)n−k_k!(n−k)! and . ) 1 ( )! 1 ( ) 1 ( k n n n f_k − + + = +

By the Lagrange interpolation formula, , ) 1 ( ) ( ) 1 ( ) 1 ( 0 0

∑

= = − − = + = + n k n k k n k k k k f n f c n P

which is 0 if n is odd and 1 if n is even. Example 2. (Vietnamese Proposal to 1977 IMO) Suppose x0, x1, …, xn are

integers and x0 > x1> ⋯ > xn. Prove that

one of the numbers |P(x0)|, |P(x1)|, … ,

|P(xn)| is at least n!/2n_{, where P(x) = x}n ₊ a1xn–1 + ⋯ + an is a polynomial with real

coefficients.

Solution. Define f0, f1, … , fn using x0,

x1, …, xn. By the Lagrange interpolation

formula, we have , ) ( ) ( ) ( ) ( 0

∑

= = n i i i i i x f x f x P x P

since both sides are polynomials of degrees at most n and are equal at x0,

x1, …, xn. Comparing coefficients of xn, we get

∑

= = n i i i i x f x P 0 . ) ( ) ( 1

Since x0, x1, …, xn are strictly decreasing

integers, we have

∏

Let the maximum of |P(x0)|, |P(x1)|, … ,

|P(xn)| be |P(xk)|. By the triangle inequality, we have . ! ) ( 2 ! ) ( ) ( | ) ( 1 0 0 n x P i n n x P x f x P n n k i k n i i i i ₌ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ≤ ≤

∑

= = Then |P(xk)| ≥ n!/2n_.

Example 3. Let P be a point on the plane of ∆ABC. Prove that

. 3 ≥ + + AB PC CA PB BC PA

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Mathematical Excalibur, Vol. 15, No. 2, Jul. - Sep. 2010 Page 2 Solution. We may take the plane of

∆ABC to be the complex plane and let P, A, B, C be corresponded to the complex numbers w, w1, w2, w3 respectively.

Then PA=|w–w1|, BC=|w2–w3|, etc.

Now the only polynomial P(x) of degree at most 2 that equals 1 at

w

1, w2,

w3 is the constant polynomial P(x) ≡ 1.

So, expressing P(x) by the Lagrange interpolation formula, we have

) )( ( ) )( ( ) )( ( ) )( ( 3 1 2 1 3 2 2 3 1 3 2 1 w w w w w x w x w w w w w x w x − − − − + − − − − ₁_. ) )( ( ) )( ( 1 2 3 2 1 3 _≡ − − − − + w w w w w x w x

Next, setting x = w and applying the triangle inequality, we get

. 1 ≥ + + BC PA AB PC AB PC CA PB CA PB BC PA _(*) The inequality (r+s+t)2 _{≥ 3(rs+st+tw),}

after subtracting the two sides, reduces to [(r–s)2_+(s–t)2_+(t–r)2_{]/2 ≥ 0, which is}

true. Setting r= PA/BC, s=PB/CA and t=PC/AB, we get . 3 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ ₊ ≥ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ ₊ BC PA AB PC AB PC CA PB CA PB BC PA AB PC CA PB BC PA

Taking square roots of both sides and applying (*), we get the desired inequality.

Example 4. (2002 USAMO) Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree n with real coefficients is the average of two monic polynomials of degree n with n real roots.

Solution. Suppose F(x) is a monic real polynomial. Choose real y1, y2, … ,yn

such that for odd i, yi < min{0,2F(i)} and for even i, yi > max{0,2F(i)}. By the Lagrange interpolation formula, there is a polynomial of degree less than n such that P(i) = yi for i=1,2,…,n. Let

G(x) = P(x)+(x−1)(x−2)_⋯(x−n) and

H(x) = 2F(x)−G(x).

Then G(x) and H(x) are monic real polynomials of degree n and their average is F(x).

As y1, y3, y5, … < 0 and y2, y4, y6, ⋯ > 0,

G(i)=yi and G(i+1)=yi+1 have opposite signs (hence G(x) has a root in [i,i+1]) for i=1,2,…,n−1. So G(x) has at least n−1 real roots. The other root must

also be real since non-real roots come in conjugate pair. Therefore, all roots of G(x) are real.

Similarly, for odd i, G(i) = yi < 2F(i) implies H(i)=2F(i)−G(i) > 0 and for even i, G(i) = yi > 2F(i) implies H(i) = 2F(i)−G(i) < 0. These imply H(x) has n real roots by reasoning similar to G(x).

Example 5. Let a1, a2, a3, a4, b1, b2, b3, b4

be real numbers such that bi–aj≠0 for i,j=1,2,3,4. Suppose there is a unique set

of numbers X1, X2, X3, X4 such that

, 1 4 1 4 3 1 3 2 1 2 1 1 1 ₌ − + − + − + − b a X a b X a b X a b X , 1 4 2 4 3 2 3 2 2 2 1 2 1 ₌ − + − + − + − b a X a b X a b X a b X , 1 4 3 4 3 3 3 2 3 2 1 3 1 ₌ − + − + − + − b a X a b X a b X a b X . 1 4 4 4 3 4 3 2 4 2 1 4 1 ₌ − + − + − + − b a X a b X a b X a b X

Determine X1+X2+X3+X4 in terms of the

ai’s and bi’s. Solution. Let .) ( ) ( ) ( 4 1 4 1

∏

= = − − − = i i i i x b a x x P

Then the coefficient of x3_{in P(x) is}

4 _. 1 4 1

∑ ∑

= = − i i i i a b Define f1, f2,f3, f4 using a1, a2, a3, a4 as

above to get the Lagrange interpolation formula . ) ( ) ( ) ( ) ( 4

∑

= = i i i i i i a f x f a P x P

Since the coefficient of x3_{in fi(x) is 1, the}

coefficient of x3_{in P(x) is also} . ) ( ) ( 4 1

∑

= i i i i a f a P

Next, observe that P(bj)/fi(bj) = bj – ai, which are the denominators of the four given equations! For j = 1,2,3,4, setting x = bj in the interpolation formula and dividing both sides by P(bj), we get

. ) ( / ) ( ) ( ) ( ) ( ) ( 1 4 4 1

∑

= = − = = i i i j i i i i i i j i j i a b a f a P a f b f b P a P

Comparing with the given equations, by uniqueness, we get Xi=P(ai)/fi(ai) for i =

1,2,3,4. So . ) ( ) ( 4 1 4 4 1 4 1

∑

∑ ∑

∑

= = = = − = = i i i i i i i i i i b a a f a P X

Comment: This example is inspired by problem 15 of the 1984 American Invitational Mathematics Examination. Example 6. (Italian Proposal to 1997 IMO) Let p be a prime number and let P(x) be a polynomial of degree d with integer coefficients such that:

(i) P(0) = 0, P(1) = 1;

(ii) for every positive integer n, the remainder of the division of P(n) by p is either 0 or 1.

Prove that d ≥ p − 1.

Solution. By (i) and (ii), we see P(0)+P(1)+⋯+P(p − 1)≡k (mod p) (#) for some k∈{1,2,…, p − 1}.

Assume d ≤ p − 2. Then P(x) will be uniquely determined by the values P(0), P(1), …, P(p − 2). Define f0, f1, …, fp−2

using 0, 1, …, p − 2 as above to get the Lagrange interpolation formula

. ) ( ) ( ) ( ) ( 2 0

∑

− = =p k k k k f x f k P x P As in example (1), we have fk(k) = (−1)p−2−k_{k!(p−2−k)!,} k p p p fk ₋ ₋ − = − 1 )! 1 ( ) 1 ( and so . 1 ) 1 )( ( ) 1 ( 2 0

∑

− = − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = − p k k p k p k P p P

Next, we claim that

. 2 0 ) (mod ) 1 ( 1 − ≤ ≤ − ≡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − p k for p k p _k

This is true for k = 0. Now for 0 < i < p, ) (mod 0 )! ( ! ! _p i p i p i p ≡ − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛

because p divides p!, but not i!(p−i)!. If the claim is true for k, then

) (mod ) 1 ( 1 1 1 1 ₁ p k p k p k p _k₊ − ≡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + −

and the induction step follows. Finally the claim yields

∑

− = − ≡ − 2 0 ). (mod ) ( ) 1 ( ) 1 ( p k p _P _k _p p P So P(0)+P(1)+_{⋯+P(p − 1)≡ 0 (mod p),} a contradiction to (#) above.

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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 20, 2010.

Problem 351. Let S be a unit sphere with center O. Can there be three arcs on S such that each is a 300_{° arc on} some circle with O as center and no two of the arcs intersect?

Problem 352. (Proposed by Pedro

Henrique O. PANTOJA, University of Lisbon, Portugal) Let a, b, c be real numbers that are at least 1. Prove that

. 2 3 1 1 1 2 2 2 ≥ + + + + + ab ab c ca ca b bc bc a

Problem 353. Determine all pairs (x, y) of integers such that x5_−y2_=4.

Problem 354. For 20 boxers, find the least number n such that there exists a schedule of n matches between pairs of them so that for every three boxers, two of them will face each other in one of the matches.

Problem 355. In a plane, there are two similar convex quadrilaterals ABCD and AB1C1D1 such that C, D are inside

AB1C1D1 and B is outside AB1C1D1

Prove that if lines BB1, CC1 and DD1

concur, then ABCD is cyclic. Is the converse also true?

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

Solutions

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

Problem 346. Let k be a positive integer. Divide 3k pebbles into five piles (with possibly unequal number of pebbles). Operate on the five piles by selecting three of them and removing one pebble from each of the three piles. If it is possible to remove all pebbles after k operations, then we say it is a harmonious ending.

Determine a necessary and sufficient condition for a harmonious ending to exist in terms of the number k and the distribution of pebbles in the five piles.

(Source: 2008 Zhejiang Province High School Math Competition)

Solution. CHOW Tseung Man (True Light Girl’

s

College), CHUNG Ping Ngai (MIT Year 1), HUNG Ka Kin Kenneth (CalTech Year 1).

The necessary and sufficient condition is every pile has at most k pebbles in the beginning.

The necessity is clear. If there is a pile with more than k pebbles in the beginning, then in each of the k operations, we can only remove at most 1 pebble from that pile, hence we cannot empty the pile after k operations.

For the sufficiency, we will prove by induction. In the case k=1, three pebbles are distributed with each pebble to a different pile. So we can finish in one operation. Suppose the cases less than k are true. For case k, since 3k pebbles are distributed. So at most 3 piles have k pebbles. In the first operation, we remove one pebble from each of the three piles with the maximum numbers of pebbles. This will take us to a case less than k. We are done by the inductive assumption. Problem 347. P(x) is a polynomial of degree n such that for all w∈{1, 2, 22_{, …,}

2n_{}, we have P(w) = 1/w.} Determine P(0) with proof.

Solution 1. Carlo PAGANO (Università di Roma “Tor Vergata”, Roma, Italy).

William CHAN Wai-lam (Carmel Alison

Lam Foundation Secondary School) and

Thien Nguyen (Nguyen Van Thien

Luong High School, Dong Nai Province, Vietnam). Let Q(x) = xP(x)−1 = a(x−1)(x−2)_⋯(x−2n_). For x_{≠1, 2, 2}2_{, …, 2}n_, . 2 1 2 1 1 1 ) ( ) ( ' n x x x x Q x Q − + + − + − = L Since Q(0)= −1 and Q’(x)=P(x)+xP’(x),

∑

= − = = − = = n k k n Q Q Q P 0 . 2 1 2 2 1 ) 0 ( ) 0 ( ' ) 0 ( ' ) 0 (

Solution 2. CHUNG Ping Ngai (MIT Year 1), HUNG Ka Kin Kenneth (CalTech Year 1), Abby LEE (SKH Lam Woo Memorial Secondary School, Form 5) and WONG Kam Wing (HKUST, Physics, Year 2).

Let Q(x) = xP(x)−1 = a(x−1)(x−2)_⋯(x−2n_). Now Q(0) = −1 = a(−1)n+1₂s_{, where s =} 1+2+_{⋯+n. So a = (−1)}n ₂−s_{. Then P(0) is}

the coefficient of x in Q(x), which is

∑

= − − ₊ ₊ ₌ ₌ ₋ + − n k k n n s s s n a 0 1 _. 2 1 2 2 1 ) 2 2 2 ( ) 1 ( _L

Other commended solvers: Samuel

Lilό ABDALLA

(ITA-UNESP, S

ão

Paulo, Brazil),

Problem 348. In ∆ABC, we have

∠BAC = 90° and AB < AC. Let D be the foot of the perpendicular from A to side BC. Let I1 and I2 be the incenters

of ∆ABD and ∆ACD respectively. The circumcircle of ∆AI1I2 (with

center O) intersects sides AB and AC at E and F respectively. Let M be the intersection of lines EF and BC. Prove that I1 or I2 is the incenter of the

∆ODM, while the other one is an excenter of ∆ODM.

(Source: 2008 Jiangxi Province Math Competition)

Solution. CHOW Tseung Man (True Light Girl’

s

College).

A B D C I₁ I2 O F E M

We claim EF intersects AD at O. Since ∠EAF=90°, EF is a diameter through O. Next we will show O is on AD.

Since AI1, AI2 bisect ∠BAD, ∠CAD

respectively, we get _∠I1AI2=45°. Then

∠I1OI2=90°. Since OI1=OI2, ∠OI1I2=45°.

Also, DI1, DI2 bisect ∠BDA, ∠CDA

respectively implies _∠I1DI2=90°. Then

D, I1, O, I2 are concyclic. So . 45 2 2 1 2 OII ADI ODI =∠ = =∠ ∠ o

Then O is on AD and the claim is true. Since ∠EOI1 = 2∠EAI1 = 2∠DAI1 =

∠DOI1 and I1 is on the angle bisector of

∠ODM, we see I1 is the incenter of

∆ODM. Similarly, replacing E by F and I1 by I2 in the last sentence, we see

I2 is an excenter of ∆ODM.

Other commended solvers: CHUNG

Ping Ngai (MIT Year 1), HUNG Ka

Kin Kenneth (CalTech Year 1) and

Abby LEE (SKH Lam Woo Memorial

Secondary School, Form 5).

Problem 349. Let a1, a2, …, an be

rational numbers such that for every positive integer m, m n m m _a _a a1 + 2 +L+

is an integer. Prove that a1, a2, …, an

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Mathematical Excalibur, Vol. 15, No. 2, Jul. - Sep. 2010 Page 4 Solution. CHUNG Ping Ngai (MIT

Year 1) and HUNG Ka Kin Kenneth (CalTech Year 1).

We may first remove all the integers among a1, a2, …, an since their m-th

powers are integers, so the rest of a1,

a2, …, an will still have the same

property. Hence, without loss of generality, we may assume all a1, a2, …,

an are rational numbers and not integers. First write every ai in simplest term. Let Q be their least common denominator and for all 1≤i≤n, let ai=ki/Q. Take a prime factor p of Q. Then p is not a prime factor of one of the ki’s. So one of the remainders ri when ki is divided by p is nonzero! Since ki_≡ri(mod p), so for every positive integer m, ). (mod 0 1 1 1 m m n i m i n i m i n i m i k a Q p r ⎟ ≡ ⎠ ⎞ ⎜ ⎝ ⎛ = ≡

∑

= = = This implies . 1

∑

= ≤ n i m i m _r p Since ri < p, , 0 lim 1 lim 1 1 1 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ≤

∑

= ∞ → = ∞ → n i m i m n i m i m m _p r r p which is a contradiction.

Comments: In the above solution, it does not need all positive integers m, just an infinite sequence of positive integers m with the given property will be sufficient.

Problem 350. Prove that there exists a positive constant c such that for all positive integer n and all real numbers a1, a2, …, an, if P(x) = (x − a1)(x − a2) ⋯ (x − an), then . ) ( max ) ( max ] 1 , 0 [ ] 2 , 0 [ P x c x P x n x∈ ≤ ∈

(Ed.-Both solutions below show the conclusion holds for any polynomial!)

Solution 1. LEE Kai Seng.

Let S be the maximum of |P(x)| for all x∈[0,1]. For i=0,1,2,…,n, let bi=i/n and ). ( ) )( ( ) ( ) ( 0 i1 i1 n i x x b x b x b x b f = − L − − − + L −

By the Lagrange interpolation formula, for all real x,

. ) ( ) ( ) ( ) ( 0

∑

= = n i i i i i b f x f b P x P

For every w_{∈[0,2], |w−b}k| ≤ |2−bk| for all k = 0,1,2,…,n. So

_∏

= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ≤ n i i i _n i f w f 0 2 ) 2 ( ) ( n n n n n n(2 1)(2 2) ( 1) 2 − − + = L _. ! )! 2 ( n n n n =

Also, |P(bi)| ≤ S and

. )! ( ! ) ( i _n i n i n i b f = −

By the triangle inequality,

. 2 2 ) ( ) ( ) ( ) ( 0 0

∑

= = ⎟⎟⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ≤ ≤ n i i i i n i i n i n i n S b f w f b P w P Finally, . 2 2 2 2 2 2 2 0 4 2 0

∑

= = ≤ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ≤ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ n i n n n i n n n n i n n i n i n Then . ) ( max 16 2 ) ( max ] 1 , 0 [ 4 ] 2 , 0 [ P w S x P x n n w∈ ≤ = ∈

Solution 2. G.R.A.20 Problem Solving Group (Roma, Italy).

For a bounded closed interval I and polynomial f(x), let ||f||I denote the maximum of |f(x)| for all x in I. The Chebyschev polynomial of order n is defined by T0(x) = 1, T1(x) = x and

Tn(x) = 2xTn−1(x)−Tn−2(x) for n ≥ 2. (Ed.-By induction, we can obtain

Tn(x) = 2n_xn_+cn−1_xn−1₊_{⋯ + c}

0

and Tn(cos θ)=cos nθ. So Tn(cos(πk/n))= (−1)k_{, which implies all n roots of Tn(x)} are in (−1,1) as it changes sign n times.) It is known that for any polynomial Q(x) with degree at most n>0 and all t_∉[−1,1],

|Q(t)| ≤ ||Q||[−1,1] |Tn(t)|. (!)

To see this, we may assume ||Q||[−1,1] = 1 by

dividing Q(x) by such maximum. Assume x0∉[−1,1] and |Q(x0)| > |Tn(x0)|. Let

a = T(x0)/Q(x0) and R(x) = aQ(x)−Tn(x).

For k = 0, 1, 2, _{⋯, n, since T}n(cos(πk/n)) = (−1)k_{and |a|<1, we see R(cos(πk/n)) is} positive or negative depending on whether k is odd or even. (In particular, R(x)_≢0.) By continuity, R(x) has n+1 distinct roots on [−1,1]_∪{x0}, which contradicts the

degree of R(x) is at most n.

Next, for the problem, we claim that for every t∈[1,2], we have |P(t)| ≤ 6n_||P||

[0,1].

(Ed.-Observe that the change of variable

t = (s+1)/2 is a bijection between s∈[−1,1] and t∈[0,1]. It is also a bijection between s_{∈[1,3] and t∈[1,2].)} By letting Q(s) = P((s+1)/2), the claim is equivalent to proving that for every s∈[1,3], we have |Q(s)|≤ 6n_||Q||

[−1,1].

By (!) above, it suffices to show that |Tn(s)| ≤ 6n_{for every s}_∈[1,3].

Clearly, |T0(s)|=1=60. For n=1 and

s_{∈[1,3], |T}1(s)|=s≤3<6. Next, since the

largest root of Tnis less than 1, we see all Tn(s) > 0 for all s_{∈[1,3]. Suppose} cases n−2 and n−1 are true. Then for all s_{∈[1,3], we have 2sT}n−1(s), Tn−2(s) > 0 and so

|Tn(s)| = |2sTn−1(s)−Tn−2(s)| ≤ max(2sTn−1(s), Tn−2(s)) ≤ max(6·6n−1_{, 6}n−2_{) = 6}n_. This finishes everything.

Olympiad Corner

(continued from page 1)

Problem 5. Let A=(ajk) be a 10×10 array of positive real numbers such that the sum of the numbers in each row as well as in each column is 1. Show that there exist j<k and l<m such that

. 50 1 ≥ + jm kl km jla a a a

Problem 6. Let ABC be a triangle. Let AD, BE, CF be cevians such that ∠BAD=∠CBE=∠ACF. Suppose these cevians concur at a point Ω. (Such a point exists for each triangle and it is called a Brocard point.) Prove that

. 1 2 2 2 2 2 2 ≥ Ω + Ω + Ω AB C CA B BC A

(Ed.-A cevian is a line segment which joins a vertex of a triangle to a point on the opposite side or its extension.) Problem 7. Find all functions f:_ℝ→ℝ such that

f(x+y) + xy = f(x)f(y) for all reals x,y.

Problem 8. Prove that there are

infinitely many positive integers m for which there exist consecutive odd positive integers pm, qm (=pm+2)such that the pairs (pm, qm) are all distinct and 2 2 2 2 _, m m m m m m m m p q q p mp q q p + + + +

(5)

CO

_{We show that there}_is_{no such positive integer}_n._{Suppose the contrary; assume that}_a._{positive integer}

.

_n

exists such tha.t 2271.H ₌₌_{2 (mod}_n)_{and 103Jn. Then 2}2_71.+1_:=_{2 (mod 103) as well; and as such}_2271.₌₌₁

(mod 103). Since 103 is prime, Fermat's little t,heorem gives 2102 ₌₌_{1 (mod 103).}_,If_d

1

=

gcd(102,2n), it

follows that 2d1 ₌₌_{1 (mod 103).}_But₁₀₂₌₌_{2 x 3 x 17.}_It_{is easy}_to_{rule out d}

1 "" 2,.3,6. Hence 17Jd1 • In

tum 171n.

Again, using that 17 is a factor of n, we get 22n₊₁_:=_{2 (mod 17) or 2}2_n.₌₌_{1 (mod 17). Now 17 being}

a prime, Fermat's little theorem implies that 216 =. 1 (mod 17). If d2 =: gcd(16,2n), we see that d2

ta

!l.

power of2 and 2d~ a 1 (mod 17). We see that d2

=

2,4 do not fit in. Hence d2 =: 8 or 16. But then d212n ~s tha.t 4Jn. Hence 2~n+l := 2 (mod 4), which may be seen to be impossible. Renee no such n exists.

~ ,

Let Ct, (1, / be the roots ofthe cubic. Then we have

Q

+

(1

+,

=: -a, a(1

+

(1'Y

+

'YC:X

=

b, a(1,

=

-c.

Thus 0<2 =: (1

+

'Y

=

.-a - C/. We also observe that Cl: :j:. OJ otherwise c=:O contradicting that c is odd., ' Hence (1"(

=

-cia and

b:::: a((J

+

'Y)

+

(1'Y == ci - :.,

0<

or C/4-bCl-C= O. This gives (a.+a)2 -bee-c= 0, vihich,simplifies to a(2a-b-l);=: c+0,(a-1). Note ~hat the right side is odd and hence is not equal to O. We thus get

c+a(a.~l)

a= Za-b-l .

This shows .that a is a rational number. Now a is a rational root of a monic polynomial with integer coefficients. Hence it must be an integer. Since a=< -a(a

+

1), a is even.

Suppose(J = ,. Then 2(1

=

(1 +/ =; -a - a, so that {3 is a rational number. Hence (1(= /) is an integer. Again 2(1 = -a - 0: shows that a is even. But then c

=

-0:(1"( shows that c is even. Hence (1 =: 'Y is not

possible. ' '

(j)

Solution: Complete the parallelogram ADBC. Join CD, CH and liD. Let Sand L be the midpoints of CR and HD respectively. Observe that OP HQ is a cyclic quadrilateral and OH is a. diameter of the \ circumscribing circle. Thus S is the centre of a circle

r

1 passing through H,P"Q,O. Similarly, L is. the

centre of a circle

r2

passing through A, D, B, H. Hence the radical axis of these circles, which passes

through H is perpendicular to SL. However SL is parallel to CD (which, passes through M) and hence the, radical axis of I't and

r

2 is perpendicular to CM. But BPQA is also cyclic so that TP. TQ = TB· TA.

This shows t;hat T has same power with respect to

r

₁and

r

2 • Thus T is on the radica,l axis of these.

circles. It follows that TR is perpendicular to CM.

@

Solution: Let AB and DC meet in Mi let FE l;l1eet HO in T, AB in P and DC in Q respectively. We ,

use the following filet: in a triangle ABO with circumcircle ,,(, given a. point D outside " AD is tangent' :

to / if and only if :

F

M

DB AB2

. ,DC

=

AQ2' Thus it suffices to prove that

TO OEd

TH = EHz'

.

\

Let us put AB = a, DC

=

b, OM =: ,x, ' HM

=

J1.. Observe that DB, OA and FQ are concurrent cemns in the triangle FDC. Ceva's theorem gives . .

Looking at the transversal A-B-M in the tri-angle FDC, we can apply Menelaus' theorem to get . FA DM CB

\

I 1 -A-D "-M-C . -B-F

=

1. .\ I It follows that

.sg

=

fJt;. ,

'\

Similarly,

we

get' :~

=

f!1J.

Observe "

, . DH+HQ DQ DM DH+HM

. CH-HQ

=

QC =: MO =: HM-HC' , '

Slmplifica.tion giVes (using DH

=

CR) HeP.

=

HQ· HM. In similar way, we get OP· OM =: OB2. Thus •

, OP

PM

Th us GP PM =~. 4),'-a' S' '1 I Iml at ,/, 9M HQ

=

4!!='ijb' b • U' smg the transversal T - P - Q in the triangle MON, we . have

% .

¥fl.

tJB

=: 1. Thus

OT PO QM a2 _{4p.2 _}_b2

TH =: MP' HQ

=

4).2 - a2 • ~.

Suppose we show that 4>.2_a2 ₌_{4J1.2_b'i.}_{We get}_GTITH

=

_a2

1b

_2•_{We observe that triangles}_EBA

a.nd

ECD are similar; and they havearespe;tive medians EO a.nd EH. Hence EG2IEHz

=

ABa ICD2 "" a,"lb2•

It follows that GTITH ;:; OE IER , giving what we required. We use coordinate geometry to prove

4'A2_-a2_=4Jl.2-b2

" ' ,

Let us fix M = (0,0). The circumcircle of ABCD has equation a;2

+

y2

+

29a;

+

2fy

+

c = O. Let

the equation of MA be y ",,'mx. Let A"" (:l:l,Vl). B =: (X2,V2)' Then Xl,X2 are the solutions of (1

+

m~):t~

+

(29

+

21m):!:

+

c

=

0, so that

rei'

+:1;2

= _

20

+

21m

1+m2 '

(6)

, We also have

I

0.2

=

AB2 ;:: (1

+

m2)(xl _ :&2)2 "" (1

+

m2 ) (Xl

+

:1:2)2 _ _4X1X2)== '4 (9 1:

~)2

_ 4a. \ Thus 4),2 _ 0.2

=

4c Similarly, we get 4p.2 - b2 '" 4c. It follows that 4),2 - ,0.2

₌

4J.t2 ~ b2.

®

' .

,

Solution: Consider the fust coltunn. Let M be the maximum value of the entries in this columIl;. Then M;::: 1/10, by pigeonhole principle .. Writing the first column 0.1,1,0.2,1, ••• ,0.10,1; let M

=-

aj.1, for some j. Consider the Bum

10 .

2:)aj,la/<,/

+

aj,la",l), 1=2'

wherek

1:

j. This is equal to

10 , 1 0

aJ,1

(.I:>",/)

+

ak,l

(2:

a

j,l)

=

aj,1 (1-0.1<,1)

+

4k,1 (1 - aj,l)'.

1=2 1=2

Varying k from 1 to 10, Iv =/> j, we get

10

L:

I)aj,lak,/

+

aj,La",1)

=

l<-,6jl=2

:L

aj,1(1-ak,1)

+

ak,l(1-aj,1)

k-,6j

= aj,l

L

(1 - ak,l)

+

(1 - aj,l)

E

(11:,1

"~j b~j

aj,l (8

+

ai,l)

+

(1 - aj,l)2

M2

+

8M

+

(1 - M)2

=

2M2

+

6~

+

1,~

2

(1~0

+

1 3

0)

+

1

=

:~.'

, There are 9 X 9

=

81 summands in the above sum. Hence we

can

find le, I such that

1

0.;,10.10,1

+

aj,lak.l ;:::

50'

Solution: Note th<Lt LBfW "'"

L&n +

BAn

=

B -

~

+

~

,,;.

B, where

~

is the Brocard angle.

I

Thus triangles BnD and ABD are similar, It follows that BD2

=

AD, f/.D. Let BD : DO

=-

x : 11, OE: EA

=

Z! x, AF: FB

=-

y: z. Then BD:= axl(x +V), We a.1so have '

nD

z

An

x+v

, AD

=

x+y+z' AD = z+y+z'

Using this we get x2a2/(x + y)2 = zAD2/(x + 11 + z,), This implies that

( ' )2 2 Ana

=

a;

+

y AD2

=

1 . ::.-. . ~2 (x

+

y

+

z)2 , a;

+

11

+

z z • Thus An2 ₁ x2 B02 = x+y+z'

z·

We obtain Ana 1 x2 ~-=

""'-»1

LJ B02 x

+

y

+

z L.J z - , oyolio cyolic by Cauchy-Schwarz inequality.

CD

Solution: Taking x

=

11

=

'0, we get 1(0)

=

f(0)2. Thus 1(0)

=

0 or 1. If 1(0) = 0, then 11 = 0 gives

' f(x) = 0 for all x. But f(x) sO is not a solution as ma.y be verified. We conclude that f(O)

=

1. Taking

x

=

1,11

=

-1, we see that f(l)f(-l)

=

O. Thus f(l) =: 0 or f( -1) = o. We consider these cases .. ,

If f(l)

=

0, then y

=

1 gives

f(w

+

1)

+

w =' f(m)f(1)

=

0,

for all x. It follows that f(x

+

1)

=

-x or f(a;) <=: 1-w for all x.

Suppose 1(-1) =< O. Ta.ke y

=

-1 to get

f(x - 1) - x

=-

0:-.' This gives I(x)

=

1

+

a: for all x,

It is easy to verify that f(x) "" 1 - fC and I(x)

=-

1

+

x are both solutions.

@

Solutiollt Consider therelationsp2+pq+q2

=

u2 _andp2+mpq+q2₌_v_2•_Thus_{(m-l)pq";; (v-u)(v+u.).}

Suppose we choose m - 1

=

1"2 where l' is a positive integer at:ld v - u = rp and v

+

u = rq. Then

u= (rq-rp)/2 and 4(p2 +pq+ q2)

=

(rq _rp)2. This ~eads to

(1'2 - 4)p2 - (21'2 -I:-4)pq

+

(1"2 - 4)q2

=

O.

Solving for

plq,

we get

E

=

21'2

+

4 d: /4(1"2

+

2)2 - 4(1'2 - 4)2

q 2(1"2

-4)

We want the discriminant to be a perfect square. This forces 3(1.2 - 1) to be perfect square, which leads to the Pell's equation 1'2 - 3t2

₌

1. fhe equation has infinitely many solutions (1'", tn) given by rn.

+

tn

va ""

(2

+

va)".

We also have recurrence relations:

rn+l = 2rn

+

St;.. tn+l = Tn

+

2tnl

where 1'1 = 2 arid t 1

=

1. Induction shows that tal is even for all 1 2 1. We can express pi q in terms of t:

P (td:1)2 t+l t-1

-==~=--Or--.

q t2 _- ₁ _{t -} ₁ _t

₊

₁

Take m =: r~1

+

1, Pm

=

tal -1, qm

==

t21

+

Ii yte see that PI, ql are co~'ecutive o'dd integers. Moreover, p~

+

Pmqm

+

q! r~lt

p~+mPmqm+q!

=

(t2Ir21)2.