Mathematical Excalibur, Volume 4, Number 3

Olympiad Corner

39th International Mathematical Olympiad, July 1998:

Each problem is worth 7 points.

Problem 1. In the convex

quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel. Suppose that the point P, where the perpendicular bisectors of AB and DC meet, is inside ABCD. Prove that ABCD is a cyclic quadrilateral if and only if the triangles ABP and CDP have equal areas.

Problem 2. In a competition, there

are a contestants and b judges, where b

≥

3 is an odd integer. Each judge rates each contestant as either "pass" or "fail". Suppose k is a number such that, for any two judges, their ratings coincide for at most k contestants. Prove that b b a k 2 1 − ≥ .

Problem 3. For any positive integer

n, let d(n) denote the number of positive divisions of n (including 1 and n itself).

(continued on page 4)

The rearrangement inequality (or the permutation inequality) is an elementary inequality and at the same time a powerful inequality. Its statement is as follow. Suppose a1≤a2≤ ≤an and

n b b

b1≤ 2 ≤ ≤ . Let us call

A = a1b1+a2b2+ +anbn the ordered sum of the numbers and

B = a1bn+a2bn−1+ +anb1

the reverse sum of the numbers. If n

x x

x1, 2,..., is a rearrangement (or permutation) of the numbers

n b b

b₁, ₂,..., and if we form the mixed sum , 2 2 1 1x a x anxn a X = + + +

then the rearrangement inequality asserts that A ≥ X ≥ B. In the case the

i

a 's are strictly increasing, then equality holds if and only if the b 's are all equal._i We will look at A ≥ X first. The proof is by mathematical induction. The case n = 1 is clear. Suppose the case n = k is true. Then for the case n = k + 1, let

i

k x

b ₊1= and xk+1=bj. Observe that 0 ) )( (a_k+1−a_i b_k+1−b_j ≥ . We get j k k i k k j ib a b ab a b a + +1 +1≥ +1+ +1 .

So in X, we may switch x and i xk+1 to

get a possibly larger sum. After switching, we can apply the case n = k to the first k terms to conclude that A ≥ X. The inequality X ≥ B follows from A ≥ X using −bn ≤−bn−₁≤ ≤−b₁ in place of b1≤b2 ≤ ≤bn.

Now we will give some examples.

Example 1. (Chebysev's Inequality) Let

A and B be as in the rearrangement inequality, then

(

)(

)

. 1 1 _B n b b a a A≥ + + n + + n ≥

Proof. Cyclically rotating the b 's, wei get n mixed sums

+ + + 2 2 1 1b a b a a_nb_n, , 1 3 2 2 1b a b a b a + + + n ..., n b a1 +a2b1+ +anbn−1.

By the re-arrangement inequality, each of these is between A and B, so their average is also between A and B. This average is just the expression given in the middle of Chebysev's inequality.

Example 2. (RMS-AM-GM-HM

In-equality) Let c₁,c₂,...,c_n ≥0. The root mean square (RMS) of these numbers is

1/2 2 2

1 )/ ]

[(c + +c_n n , the arithmetic mean (AM) is (c1+c2+ +cn)/n and the geometric mean (GM) is

(

)

n

n c c

c_{1 2} 1/ . We have RMS ≥ AM

≥ GM. If the numbers are positive, then the harmonic mean (HM) is

(

)

(

)

[

c cn

]

n 1/ 1 + + 1/ . We have GM

≥ HM.

Proof. Setting ai =bi =ci in the left half of Chebysev's inequality, we easily get RMS ≥ AM. Next we will show AM

≥ GM. The case GM = 0 is clear. So suppose GM > 0. Let a₁=c₁/GM, 2 2 1 2 cc / GM a = , ..., a_n =c1c2 c_n 1 /GMn = and b_i =1/a_n−i+1 for i = 1, 2, ... n. (Note the a 's may not be_i increasing, but the b 's will be in thei reverse order as the a 's). So the mixed_i sum = + + + 2 2 1 1b a b a b a n n GM c GM c GM c1/ + 2/ + + n/

is greater than or equal to the reverse sum a₁bn + +anb₁=n. The AM-GM inequality follows easily. Finally GM ≥ HM follows by applying AM ≥ GM to the numbers 1/c1 ,...,1/cn.

Volume 4, Number 3 January, 1999 - March, 1999

Rearrangement Inequality

Kin-Yin Li
          ! " " # $!%&   ' ( "  $ )   $* + ((,& - ' (  " )." / & - ' ( "  $   "   0  .$  "  1 234536 /    /  7 ,, & 8+ 

Acknowledgment: Thanks to Elina Chiu, Math Dept,

Catherine NG, EEE Dept, HKUST and Tam Siu Lung 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 April 30, 1999.

For individual subscription for the two remaining issues for the 98-99 academic year, send us two stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin Li Department of Mathematics Hong Kong University of Science and Technology

Clear Water Bay, Kowloon, Hong Kong Fax: 2358-1643 Email: [email protected]

Example 3. (1974 USA Math

Olympiad) If a, b, c > 0, then prove that

3 / ) ( ) ( a b c c b a_{b c abc} a ≥ + + .

Solution. By symmetry, we may assume c b a≤ ≤ , then lna≤lnb≤lnc. By Chebysev's inequality, c c b b a aln + ln + ln . 3 ) ln ln )(ln (a+b+c a+ b+ c ≥

The desired inequality follows from exponentiation.

Example 4. (1978 IMO) Let c , ₁ c , ...,₂ n

c be distinct positive integers. Prove that n n c c c n 1 2 1 1 4 2 2 1+ +9 + ≥ + +9 + .

Solution. Let a , ₁ a , ..., ₂ a be the n c 'si arranged in increasing order. Since a 'si are distinct positive integers, ai ≥i. Since 1>1/4>...>1/n2, by the re-arrangement inequality, 2 2 1 4 n c c c + +9 + n 2 2 1 4 _n a a a + + + n ≥ 9 n 1 2 1 1+ + + ≥ 9 .

Example 5. (1995 IMO) Let a, b, c > 0

and abc = 1. Prove that

+ + + + ( ) 1 ) ( 1 3 3 a c b c b a 2 3 ) ( 1 3 ₊ ≥ b a c . Solution. (HO Wing Yip, Hong Kong Team Member) Let x = bc = 1/a, y = ca = 1/b, z = ab = 1/c. The required inequality is equivalent to 2 3 2 2 2 ≥ + + + + + y x z z x y y z x . By symmetry, we may assume x≤y≤z, then x2≤ y2≤z2 and 1/(z + y) ≤ 1/(x + z) ≤ 1/(y + x). The left side of the required inequality is just the ordered sum A of the numbers. By the rearrangement inequality, z x z y z y x y x A + + + + + ≥ 2 2 2 , y z z x y y z x x A + + + + + ≥ 2 2 2 . (continued on page 4)

Intersecting Chords Theorem. Let two

lines through a point P not on a circle intersect the inside of the circle at chords AA' and BB', then PA × PA' = PB × PB'. (When P is outside the circle, the limiting case A = A' refers to PA tangent to the circle.)

This theorem follows from the observation that triangles ABP and A'B'P are similar and the corresponding sides are in the same ratio. In the case P is inside the circle, the product PA × PA' can be determined by taking the case the chord AA' passes through P and the center O. This gives PA × PA' =

2 2

d

r − , where r is the radius of the circle and d = OP. In the case P is outside the circle, the product PA × PA' can be determined by taking the limiting case PA is tangent to the circle. Then PA × PA' = d2−r2.

The power of a point P with respect to a circle is the number d2−r2 as mentioned above. (In case P is on the circle, we may define the power to be 0 for convenience.) For two circles C₁ and C with different centers 2 O and1

2

O , the points whose power with respect to C and ₁ C are equal form a₂ line perpendicular to line O₁O . (This₂ can be shown by setting coordinates with line O₁O as the x-axis.) This line is₂ called the radical axis of the two circles. In the case of the three circles C , 1 C ,2

3

C with noncollinear centers O , ₁ O ,₂

3

O , the three radical axes of the three pairs of circles intersect at a point called the radical center of the three circles. (This is because the intersection point of any two of these radical axes has equal power with respect to all three circles, hence it is on the third radical axis too.) If two circles C and 1 C intersect, their2

radical axis is the line through the intersection point(s) perpendicular to the line of the centers. (This is because the intersection point(s) have 0 power with respect to both circles, hence they are on the radical axis.) If the two circles do not intersect, their radical axis can be found by taking a third circle C₃ intersecting both C and 1 C . Let the2

radical axis of

1

C , C intersect the radical axis of 3 C ,2 3

C at P. Then the radical axis of C ,₁

2

C is the line through P perpendicular to the line of centers of C , ₁ C .₂

We will illustrate the usefulness of the intersecting chords theorem, the concepts of power of a point, radical axis and radical center in the following examples.

Example 1. (1996 St. Petersburg City

Math Olympiad) Let BD be the angle bisector of angle B in triangle ABC with D on side AC. The circumcircle of triangle BDC meets AB at E, while the circumcircle of triangle ABD meets BC at F. Prove that AE = CF.

Solution. By the intersecting chords theorem, AE × AB = AD × AC and CF

× CB = CD × CA, so AE/CF = (AD/CD)(BC/AB). However, AB/CB = AD/CD by the angle bisector theorem. So AE = CF.

Example 2. (1997 USA Math

Olympiad) Let ABC be a triangle, and draw isosceles triangles BCD, CAE, ABF externally to ABC, with BC, CA, AB as their respective bases. Prove the lines through A, B, C, perpendicular to the lines EF, FD, DE, respectively, are concurrent.

Solution. Let C be the circle with1

center D and radius BD, C be the₂ circle with center E and radius CE, and

3

C be the circle with center F and radius AF. The line through A perpendicular to EF is the radical axis of

2

C , C , the line through B3

perpendicular to FD is the radical axis of

3

C , C and the line through C₁ perpendicular to DE is the radical axis of

1

C , C . These three lines concur at the₂ radical center of the three circles.

Example 3. (1985 IMO) A circle with

center O passes through vertices A and C of triangle ABC and intersects side AB at K and side BC at N. Let the circumcircles of triangles ABC and KBN intersect at B and M. Prove that OM is perpendicular to BM.

(continued on page 4)

Mathematical Excalibur, Vol. 4, No. 3, Jan, 99 - Mar, 99 Page 2

Power of Points Respect to Circles

Kin-Yin Li

Problem Corner

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions should be preceeded by the solver’s name, home address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, Hong Kong University of Science and Technology, Clear Water Bay, Kowloon. The deadline for submitting solutions is April 30, 1999.

Problem 81. Show, with proof, how

to dissect a square into at most five pieces in such a way that the pieces can be reassembled to form three squares no two of which have the same area. (Source: 1996 Irish Mathematical Olympiad)

Problem 82. Show that if n is an

integer greater than 1, then n4+4n cannot be a prime number. (Source: 1977 Jozsef Kurschak Competition in Hungary)

Problem 83. Given an alphabet with

three letters a, b, c, find the number of words of n letters which contain an even number of a's. (Source: 1996 Italian Mathematical Olympiad)

Problem 84. Let M and N be the

midpoints of sides AB and AC of

∆ABC, respectively. Draw an arbitrary line through A. Let Q and R be the feet of the perpendiculars from B and C to this line, respectively. Find the locus of the intersection P of the lines QM and RN as the line rotates about A.

Problem 85. Starting at (1, 1), a

stone is moved in the coordinate plane according to the following rules: (a) From any point (a, b), the stone

can be moved to (2a, b) or (a, 2b). (b) From any point (a, b), the stone can be moved to (a - b, b) if a > b, or to (a, b - a) if a < b.

For which positive integers x, y, can the stone be moved to (x, y)? (Source: 1996 German Mathematical Olympiad)

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

Solutions

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

Problem 76. Find all positive

integers N such that in base 10, the digits of 9N is the reverse of the digits

of N and N has at most one digit equal 0. (Source: 1977 unused IMO problem proposed by Romania)

Solution. LAW Ka Ho (Queen

Elizabeth School, Form 6) and Gary

NG Ka Wing (STFA Leung Kau Kui

College, Form 6).

Let [a₁a₂...a_n] denote N in base 10 with a1≠0. Since 9N has the same number of digits as N, we get a = 1 and₁

n

a = 9. Since 9 ×19≠91, n > 2. Now

2

[

9 a ... an₋₁] + 8 = [an−1 ... a ].2

Again from the number of digits of both sides, we get a2≤ 1. The case a = 12

implies 9an₋₁ + 8 ends in a and so2 1

−

n

a = 7, which is not possible because 9[1 ... 7] + 8 > [7 ... 1]. So a = 0 and₂

1 −

n

a = 8. Indeed, 1089 is a solution by direct checking. For n > 4, we now get

] [

9 a3: an−2 + 8 = [8an−2...a ].3

Then a₃≥ 8. Since 9a_n−2 + 8 ends in

3

a , a = 8 will imply ₃ an₋₂ = 0, causing another 0 digit. So a = 9 and₃

2 −

n

a = 9. Indeed, 10989 and 109989 are solutions by direct checking. For n > 6, we again get 9[a₄: a_n−3] + 8 = [8a_n−3...a ]. So ₄ a = ... = 4 an−3 = 9.

Finally direct checking shows these numbers are solutions.

Other recommended solvers: CHAN

Siu Man (Ming Kei College, Form 6), CHING Wai Hung (STFA Leung Kau

Kui College, Form 7), FANG Wai Tong

Louis (St. Mark's School, Form 6), KEE Wing Tao Wilton (PLK Centenary Li

Shiu Chung Memorial College, Form 7),

KWOK Chi Hang (Valtorta College,

Form 7), TAM Siu Lung (Queen Elizabeth School, Form 6), WONG Chi

Man (Valtorta College, Form 4), WONG Hau Lun (STFA Leung Kau

Kui College, Form 7) and WONG Shu

Fai (Valtorta College, Form 7).

Problem 77. Show that if ∆ABC

satisfies , 2 cos cos cos sin sin sin 2 2 2 2 2 2 = + + + + C B A C B A

then it must be a right triangle. (Source: 1967 unused IMO problem proposed by Poland)

Solution. (All solutions received are

essentially the same.)

Using sin2x=(1−cos2x)/2 and x

2

cos = (1 + cos 2x)/2, the equation is equivalent to . 0 1 2 cos 2 cos 2 cos A+ B+ C+ = This yields cos(A + B) cos(A - B) + cos2 C = 0. Since cos(A + B) = -cosC, we get cosC (cos(A - B) + cos(A + B)) = 0. This simplifies to cosC cosA cosB = 0. So one of the angles A, B, C is 900.

Solvers: CHAN Lai Yin, CHAN

Man Wai, CHAN Siu Man, CHAN Suen On, CHEUNG Kin Ho, CHING Wai Hung, CHOI Ching Yu, CHOI Fun Ieng, CHOI Yuet Kei, FANG Wai Tong Louis, FUNG Siu Piu, HUNG Kit, KEE Wing Tao Wilton, KO Tsz Wan, KWOK Chi Hang, LAM Tung Man, LAM Wai Hung, LAM Yee, LAW Ka Ho, LI Ka Ho, LING Hoi Sheung, LOK Chan Fai, LUNG Chun Yan, MAK Wing Hang, MARK Kai Pan, Gary NG Ka Wing, OR Kin, TAM Kwok Cheong, TAM Siu Lung, TSANG Kam Wing, TSANG Pui Man, TSANG Wing Kei, WONG Chi Man, WONG Hau Lun, YIM Ka Wing and YU Tin Wai.

Problem 78. If c₁,c₂,...,cn(n≥2) are real numbers such that

= + + + −1)( ) (n c₁2 c₂2 ; c_n2 , ) (c₁+c₂+; +c_n 2

show that either all of them are negative or all of them are non-positive. (Source: 1977 unused IMO problem proposed by Czechoslovakia)

Solution. CHOY Ting Pong (Ming

Kei College, Form 6).

Assume the conclusion is false. Then there are at lease one negative and one positive numbers, say

k c c

c1≤ 2 ≤; ≤ ≤0<ck+1≤; ≤cn

with 1 ≤ k < n, satisfying the condition. Let w = c + ... + ₁ c , x =k n k c c ₊1+; + , y = +; + 2 1 c c and z2_k = c_k2₊₁+; +c2_n. Expanding 2 w and 2

x and applying the inequality , 2 2 2 _{b ab} a + ≥ we get ky ≥w2 and (n - k) z ≥ x2. So + ≥ + − = +x n y z ky w ) ( 1)( ) ( 2 2 2 ) (n−k z≥w +x . Simiplifying, we get wx ≥ 0, contradicting w < 0 < x.

Other commended solvers: CHAN

Siu Man (Ming Kei College, Form 6), FANG Wai Tong Louis (St. Mark's

School, Form 6), KEE Wing Tao

Wilton (PLK Centenary Li Shiu

Chung Memorial College, Form 7),

Gary NG Ka Wing (STFA Leung

Kau Kui College, Form 6), TAM Siu

Lung (Queen Elizabeth School, Form

6), WONG Hau Lun (STFA Leung Kau Kui College, Form 7) and

YEUNG Kam Wah (Valtorta College,

Form 7).

Problem 79. Which regular polygons

can be obtained (and how) by cutting a cube with a plane? (Source: 1967 unused IMO problem proposed by Italy)

Solution. FANG Wai Tong Louis

(St. Mark's school, Form 6), KEE

Wing Tao (PLK Centenary Li Shiu

Chung Memorial School, Form 7),

TAM Siu Lung (Queen Elizabeth

School, Form 6) and YEUNG Kam

Wah (Valtorta College, Form 7).

Observe that if two sides of a polygon is on a face of the cube, then the whole polygon lies on the face. Since a cube has 6 faces, only regular polygon with 3, 4, 5 or 6 sides are possible. Let the vertices of the bottom face of the cube be A, B, C, D and the vertices on the top face be A', B', C', D' with A' on top of A, B' on top of B and so on. Then the plane through A, B', D' cuts an equilateral triangle. The perpendicular bisecting plane to edge AA' cuts a square. The plane through the mid-points of edges AB, BC, CC', C'D', D'A', A'A cuts a regular hexagon. Finally, a regular pentagon is impossible, otherwise the five sides will be on five faces of the cube implying two of the sides are on parallel planes, but no two sides of a regular pentagon are parallel.

Problem 80. Is it possible to cover a

plane with (infinitely many) circles in such a way that exactly 1998 circles pass through each point? (Source: Spring 1988 Tournament of the Towns Problem)

Solution. Since no solution is

received, we will present the modified solution of Professor Andy Liu (University of Alberta, Canada) to the problem.

First we solve the simpler problem where 1998 is replaced by 2. Consider the lines y = k, where k is an integer, on the coordinate plane. Consider

every circle of diameter 1 tangent to a pair of these lines. Every point (x, y) lies on exactly two of these circles. (If y is an integer, then (x, y) lies on one circle on top of it and one below it. If y is not an integer, then (x, y) lies on the right half of one circle and on the left half of another.) Now for the case 1998, repeat the argument above 998 times (using lines of the form y = k + (j /999) in the j-th time, j = 1, 2, ..., 998.)

Olympiad Corner (continued from page 1)

Determine all positive integers k such that k n d n d ₌ ) ( ) ( 2 for some n.

Problem 4. Determine all pairs (a, b)

of positive integers such that ab + b2 + 7 divides a2b+a+b.

Problem 5. Let I be the incentre of

triangle ABC. Let the incircle of ABC touch the sides BC, CA and AB at K, L and M, respectively. The line through B parallel to MK meets the lines LM and LK at R and S, respectively. Prove that ∠RISis acute.

Problem 6. Consider all functions f

from the set < of all positive integers

into itself satisfying

2 2 _{( )) ( ())}

(t f s s f t

f = ,

for all s and t in < . Determine the least

possible value of f (1998).

Rearrangement Inequality (continued from page 2)

So         + + + + + + + + ≥ z x z x y z y z x y x y A 2 2 2 2 2 2 2 1 . Applying the RMS-AM inequality

, 2 / ) ( 2 2 2 _{s r s}

r + ≥ + the right side is at least (x+y+z)/2, which is at least

2 / 3 2 / ) (

3xyz 1/3 = by the AM-GM inequality.

Power of Points Respect to Circles

(continued from page 2)

Solution. For the three circles mentioned, the radical axes of the three pairs are lines AC, KN and BM. (The centers are noncollinear because two of them are on the perpendicular bisector of AC, but not the third.) So the axes will concur at the radical center P. Since ∠PMN = ∠BKN =

∠NCA, it follows that P, M, N, C are concyclic. By power of a point, BM × BP = BN × BC = BO2−r2 and PM

× PB = PN × PK = PO2−r2, where r is the radius of the circle through A, C, N, K. Then − = − = − 2 2 2 ) (PM BM PM BP BO PO . 2 BM This implies OM is

perpendicular to BM. (See remarks below.)

Remarks. By coordinate geometry, it can be shown that the locus of points X such that PO2−BO2 = PX2−BX2 is the line through O perpendicular to line BP. This is a useful fact.

Example 4. (1997 Chinese Math

Olympiad) Let quadrilateral ABCD be inscribed in a circle. Suppose lines AB and DC intersect at P and lines AD and BC intersect at Q. From Q, construct the tangents QE and QF to the circle, where E and F are the points of tangency. Prove that P, E, F are collinear.

Solution. Let M be a point on PQ such that ∠CMQ=∠ADC. Then

Q M C

D, , , are concyclic and also, P

M C

B, , , are concyclic. Let r be the₁ radius of the circumcircle C of1

ABCD and O be the center of ₁ C . By₁ power of a point, PO₁2 PQ PM PD PC r = × = × − 2 1 and 2 1 QO - r₁2 =QC×QB=QM×PQ. Then PO₁2 , ) ( 2 2 2 1 PM QM PQ PM QM QO = − = − −

which implies O₁M⊥PQ. The circle

2

C with QO as diameter passes1

through M,E,F and intersects C at₁ .

, F

E If r is the radius of ₂ C and₂

2

O is the center of C , then2

PM r

PO₁2− ₁2 = ×PQ=PO₂2−r₂2. So P lies on the radical axis of C ,₁ C ,₂ which is the line EF.