Mathematical Excalibur, Volume 5, Number 5

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Olympiad Corner

British Mathematical Olympiad, January 2000:

Time allowed: 3 hours 30 minutes

Problem 1. Two intersecting circles C1 and C2 have a common tangent which touches C1 at P and C2 at Q. The two circles intersect at M and N, where N is nearer to PQ than M is. The line PN meets the circle C2 again at R. Prove that MQ bisects angle PMR.

Problem 2. Show that for every positive

integer n, n n n n ₂₅ ₁₉₀₀ ₍ ₄₎ 121 − + − − is divisible by 2000.

Problem 3. Triangle ABC has a right

angle at A. Among all points P on the perimeter of the triangle, find the position of P such that

AP + BP + CP is minimized.

Problem 4. For each positive integer k,

define the sequence

{ }

a_n by a0 = 1 and an = kn + (−1)n an-1 for each n ≥ 1. (continued on page 4) 我們知道，圓錐曲線是㆒些所謂㆓ 次形的曲線，即㆒條圓錐曲線會滿足 以㆘的㆒般㆓次方程：Ax2_{+ Bxy + Cy}2 + Dx + Ey + F = 0，其㆗ A、B 及 C 不 會同時等於 0。假設 A ≠ 0，那麼我們 可以將㆖式除以 A，並化簡成以㆘模 式： x2 + bxy + cy2 + dx + ey + f = 0。以㆖的方程給了我們㆒個啟示：就是五點能夠定出㆒個圓錐曲線。因為如果我們知道了五個不同點的坐標，我們可以將它們分別代入 ㆖ 面的方程 ㆗，從而得到㆒個有 5 個未知數（即 b、 c、d、e 和 f ）和 5 條方程的方程組。 祇要解出各未知數的答案，就可以知道該圓錐曲線的方程了。不過，㆖述方法雖然明顯，但真正操作時又困難重重！這是由於有 5 個未知數的聯立方程卻不易解！而且我 們在計算之初假設 x2_{的係數非零，但} 萬㆒這假設不成立，我們就要改設 B 或 C 非零，並需要重新計算㆒次了。 幸好，我們可以通過「圓錐曲線族」的想法來解此問題。方法見㆘例：例例例例:::: 求穿過 A(1, 0), B(3, 1), C(0, 3), D(−4, −1), E(−2, −3) 五點的圓錐曲線方程。解解解解:::: 利用兩點式，先求出以㆘各直線的方程： AB : 1 3 0 1 1 0 − − = − − x y ，即 x − 2y − 1 = 0 CD : 0 4 3 1 0 3 − − − − = − − x y ，即 x − y + 3 = 0 AC : 1 0 0 3 1 0 − − = − − x y ，即 3x + y − 3 = 0 BD : 3 4 1 1 3 1 − − − − = − − x y ，即 2x − 7y + 1 = 0 然後將 AB 和 CD 的方程「相乘」， 得㆒條圓錐曲線的方程： (x − 2y − 1)(x − y + 3) = 0，即 x2− 3xy + 2y2 + 2x − 5y − 3 = 0。注意：注意：注意：注意：雖然㆖述的方程是㆒條㆓次形「曲線」，但實際㆖它是由兩條直線所組成的。同時，亦請大家留意，該曲線 同時穿過 A、B、C 和 D ㆕點。 類似㆞，我們又將 AC 和 BD「相 乘」，得： (3x + y − 3)(2x − 7y + 1) = 0，即 6x2− 19xy − 7y2− 3x + 22y − 3 = 0。考慮圓錐曲線族： x2− 3xy + 2y2 + 2x − 5y − 3 + k(6x2− 19xy − 7y2− 3x + 22y − 3) = 0。很明顯，無論 k 取任何數值，這圓錐曲線族都會同樣 穿過 A、B、C 和 D ㆕點。 最後，將 E 點的坐標代入曲線族 ㆗，得：12 + k (−216) = 0，即 k = 1/18， 由此得所求的圓錐曲線方程為 18(x2− 3xy + 2y2 + 2x − 5y − 3) + (6x2− 19xy − 7y2− 3x + 22y − 3) = 0，即 24x2− 73xy + 29y2 + 33x − 68y − 57 = 0。

Volume 5, Number 5 November 2000 – December 2000

五點求圓錐曲線

梁子傑

香港道教聯合會青松㆗學

Editors: 張百康 (CHEUNG Pak-Hong), Munsang College, HK 高子眉 (KO Tsz-Mei)

梁達榮 (LEUNG Tat-Wing), Appl. Math Dept, HKPU 李健賢 (LI Kin-Yin), Math Dept, 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 February 4, 2001.

For individual subscription for the next five issues for the 01-02 academic year, send us five 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

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Majorization Inequality

Kin Y. Li

The majorization inequality is a generalization of Jensen's inequality. While Jensen's inequality provides one extremum (either maximum or minimum) to a convex (or concave) expression, the majorization inequality can provide both in some cases as the examples below will show. In order to state this inequality, we first introduce the concept of majorization for ordered set of numbers. If

n x x x₁≥ ₂≥≥ , n y y y₁≥ ₂≥≥ , ..., , , ₁ ₂ ₁ ₂ 1 1 y x x y y x ≥ + ≥ + 1 1 1 1+ +xn− ≥ y + +yn− x and n n y y x x₁++ = ₁++ , then we say (x₁,x₂,...,x_n) majorizes

) ..., , , (y₁ y₂ y_n and write ) ..., , , (x₁ x₂ x_n (y₁,y₂,...,y_n). Now we are ready to state the inequality.

Majorization Inequality. If the function f

is convex on the interval I = [a, b] and ) ..., , , (x₁ x₂ x_n (y₁,y₂,...,y_n) for x_i,y_i∈I, then ) ( ) ( ) (x₁ f x₂ f x_n f + ++ ) ( ) ( ) (y₁ f y₂ f y_n f + + + ≥ .

For strictly convex functions, equality holds if and only if x_i = y_i for i = 1, 2, …, n. The statements for concave functions can be obtained by reversing inequality signs.

Next we will show that the majorization inequality implies Jensen's inequality. This follows from the observation that if

n x x

x₁≥ ₂≥≥ , then (x₁,x₂,...,x_n) (x, x, …, x), where x is the arithmetic mean of x , ₁ x , …, ₂ x . (Thus, applying the _n majorization inequality, we get Jensen's inequality.) For k = 1, 2, …, n - 1, we have to show x₁++x_k ≥kx. Since ) )( (n−k x₁++x_k 1 ) ( ) ( − ≥ − ₊ ≥ n k kxk k n k xk ). (x_k ₁ x_n k + ≥ ₊

Adding )k(x₁++x_k to the two extremes, we get . ) ( ) (x₁ x k x₁ x knx n ++ _k ≥ ++ _n = Therefore, .x₁++x_k ≥kx

Example 1. For acute triangle ABC,

show that 2 3 cos cos cos 1≤ A+ B+ C≤ and determine when equality holds.

Solution. Without loss of generality,

assume .A≥B≥C Then A≥π/3 and 3 / π ≤ C . Since π/2≥A≥π/3 and 3 / 2 ) ( π π π ≥A+B = −C ≥ , we have (π/2 ,π/2 ,0) (A, B, C) ) 3 / , 3 / , 3 / (π π π . Since f(x) = cos x is strictly concave on I = [0, π/2], by the majorization inequality, ) 0 ( 2 2 1 f f + f      +       = π π ≤ f(A) + f(B) + f(C) = cos A + cos B + cos C ≤ 2 3 3 3 3 =     +       +      π π π f f f .

For the first inequality, equality cannot hold (as two of the angles cannot both be right angles). For the second inequality, equality holds if and only if the triangle is equilateral.

Remarks. This example illustrates the equilateral triangles and the degenerate case of two right angles are extreme cases for convex (or concave) sums.

Example 2. Prove that if a, b ≥ 0, then

.

3 3

3_a₊3_a ₊ _b₊ _b _≤ _a₊ _b₊ _b₊ _a (Source: Math Horizons, Nov. 1995, Problem 36 of Problem Section, proposed by E.M. Kaye)

Solution. Without loss of generality, we

may assume b≥a≥ 0. Among the numbers 3 1 b b x = + , x₂=b+3a, 3 3 a b x = + , x₄=a+3a, 1

x is the maximum and x₄ is the minimum. Since x +₁ x = ₄ x +₂ x , we ₃ get (x , ₁ x )4 (x , 2 x ) or (3 x , 3 x ) 2 (depends on which of x or ₂ x is larger). ₃ Since f(x) = 3x is concave on the interval [0, ∞), so by the majorization inequality,

) ( ) ( ) ( ) (x₄ f x₁ f x₃ f x₂ f + ≤ + ,

which is the desired inequality.

Example 3. Find the maximum of a12+ 12

12 _c

b + if −1≤ a, b, c ≤1 and a + b + c = .−1/2

Solution. Note the continuous function

f(x) = x12 is convex on [-1, 1] since ) ( '' _x f = 132x10≥ 0 on (-1, 1). If 1 ≥ a ≥ b ≥ c ≥ −1 and 2 1 − = + +b c a ,

then we get (1, -1/2, -1) _{(a, b, c). This} is because 1 ≥ a and . 2 1 2 1 1 2 1 b a c− = + − ≥ − =

So by the majorization inequality, a12+b12+c12 = f(a) + f(a) + f(c) ) 1 ( 2 1 ) 1 ( + −     − + ≤ f f f = 2 + 12 2 1 .

The maximum value 2 + (1/ 212 ) is attained when a = 1, b = -1/2 and c = -1. Remarks. The example above is a simplification of a problem in the 1997 Chinese Mathematical Olympiad.

Example 4. (1999 IMO) Let n be a fixed

integer, with n ≥ 2.

(a) Determine the least constant C such that the inequality

(

)

4 1 1 2 2

∑

≤ < ≤ ≤≤     ≤ + n j i i j i j i ni x C x x x x

holds for all real numbers x , ₁ x , …, 2 0

≥ n x .

(b) For this constant C, determine when equality holds.

Solution. Consider the case n = 2 first.

Let x = m + h and ₁ x = m - h, then m = 2 (x + ₁ x )/2, h = (₂ x - ₁ x )/2 and ₂

(

2

) (

4 4

)

2 2 1 2 1x x x 2m h x + = −

(

)

4 2 1 4 8 1 2m = x +x ≤

with equality if and only if h = 0, i.e. x ₁ = x . ₂

(continued on page 4)

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Problem Corner

We welcome readers to submit their solutions to the problems posed below for publication consideration. Solutions should be preceeded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, Hong Kong University of Science & Technology, Clear Water Bay, Kowloon. The deadline for submitting solutions is February 4, 2001.

Problem 116. Show that the interior of a

convex quadrilateral with area A and perimeter P contains a circle of radius A/P.

Problem 117. The lengths of the sides of

a quadrilateral are positive integers. The length of each side divides the sum of the other three lengths. Prove that two of the sides have the same length.

Problem 118. Let R be the real numbers.

Find all functions f : R → R such that for all real numbers x and y,

f (xf (y) + x) = xy + f (x).

Problem 119. A circle with center O is

internally tangent to two circles inside it at points S and T. Suppose the two circles inside intersect at M and N with N closer to ST. Show that OM ⊥MN if and only if S, N, T are collinear. (Source: 1997 Chinese Senior High Math Competitiion)

Problem 120. Twenty-eight integers are

chosen from the interval [104, 208]. Show that there exist two of them having a common prime divisor.

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

Solutions

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

Problem 111. Is it possible to place 100

solid balls in space so that no two of them have a common interior point, and each of them touches at least one-third of the others? (Source: 1997 Czech-Slovak Match)

Solution 1. LEE Kai Seng (HKUST).

Take a smallest ball B with center at O and radius r. Any other ball touching B at x

contains a smaller ball of radius r also touching B at x. Since these smaller balls are contained in the ball with center O and radius 3r, which has a volume 27 times the volume of B, there are at most 26 of these other balls touching B.

Solution 2. LEUNG Wai Ying (Queen

Elizabeth School, Form 6).

Consider a smallest ball S with center O and radius r. Let S and _i S_j (with centers O and _i O_j and radii r and _i r_j, respectively) be two other balls touching S at P and _i P_j, respectively. Since r , _i r_j

≥ r, we have O_i O_j ≥ r + _i rj ≥ r + r = i OO and similarly _i O_i O_j ≥ OO_j. So

i

O O_j is the longest side of ∆OO_i O_j. Hence ∠PiOPj =∠OiOOj ≥60$.

For ball S , consider the solid cone with _i vertex at O obtained by rotating a 30 $ angle about OP as axis. Let _i A be the _i part of this cone on the surface of S. Since

$

60

≥

∠PiOPj , the interiors of A and i j

A do not intersect. Since the surface area of each A_i is greater than

2 ) 30 sin (r $ π = 4πr2/ , which is 1/16 of the surface area of S, S can touch at most 15 other balls. So the answer to the question is no.

Other recommended solvers: CHENG

Kei Tsi (La Salle College, Form 6). Problem 112. Find all positive integers (x,

n) such that x + n 2 + 1 is a divisor of n 1

+

n

x + 2n+1 + 1. (Source: 1998 Romanian Math Olympiad)

Solution. CHENG Kei Tsi (La Salle

College, Form 6), LEE Kevin (La Salle College, Form 5) and LEUNG Wai Ying (Queen Elizabeth School, Form 6). For x = 1, 2(1 + n 2 + 1) > n 1n+1 + 1 2n+ + 1 > 1 + n 2 + 1. For x = 2, n 2(2 + n 2 + 1) > n 2n+1 + 2n+1 + 1 > n 2 + 2 + 1. For x = 3, 3(n 3 + n 2 + 1) n > 3n+1 + 2n+1 + 1 > 2(3 + n 2 + 1). n So there are no solutions with x = 1, 2, 3. For 4x≥ , if n≥2 , then we get x(x +n 2 + 1) > n xn+1+2n+1 + 1. Now 1 + n x +2n+1+ 1 = (x - 1)(x +n 2 + 1) n + x - (n 2 + 1) x +n 3⋅2n+ 2 > (x - 1)(x +n 2 + 1) n because for n = 2, x - (n 2 + 1)x + n 2n+1 = x - 5x + 8 > 0 and for n 2 ≥ 3, x - n (2 + 1)x n ≥ x( 4n−1 - 2 - 1) > 0. n Hence only n = 1 and x ≥ 4 are possible. In that case, x + n 2 + 1 = x + 3 is a n divisor of xn+1+2n+1+ 1 = x + 5 = (x - 2 3)(x + 3) + 14 if and only if x + 3 is a divisor of 14. Since x + 3 ≥ 7, x = 4 or 11. So the solutions are (x, y) = (4, 1) and (11, 1).

Problem 113. Let a, b, c > 0 and abc ≤ 1. Prove that c b a b c a b c a + + ≥ + + .

Solution. LEUNG Wai Ying (Queen

Since abc ≤ 1, we get 1/(bc) ≥ a, 1/(ac) ≥ b and 1/(ab) ≥ c. By the AM-GM inequality, a b c c a c a b c c a 3 bc a 3 2 ₊ ₌ ₊ ₊ _≥ ₃ 2 _≥ . Similarly, 2b/a + a/c ≥ 3b and 2c/b + b/a

≥ 3c. Adding these and dividing by 3, we get the desired inequality.

Alternatively, let x = 9a4b/ c2 , y = 9 4 2 / b a c and z = 9b4c/ a2 . We have a = x y, b = 2 z x, c = 2 y z and xyz = 2 ≤

3_abc _{1. Using this and the} re-arrangement inequality, we get

zx y xy z yz x b c a b c a 2 2 2 + + = + + 3 3 3 2 2 2 z y x zx y xy z yz x xyz _= + +       + + ≥ c b a x z z y y x + + = + + ≥ 2 2 2 _.

Problem 114. (Proposed by Mohammed

Aassila, Universite Louis Pasteur, Strasbourg, France) An infinite chessboard is given, with n black squares and the remainder white. Let the collection of black squares be denoted by

0

G . At each moment t = 1, 2, 3, …, a simultaneous change of colour takes place throughout the board according to the following rule: every square gets the colour that dominates in the three square configuration consisting of the square

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itself, the square above and the square to the right. New collections of black squares G , ₁ G , 2 G , … are so formed. 3 Prove that G is empty. _n

Solution. LEE Kai Seng (HKUST).

Call a rectangle (made up of squares on the chess board) desirable if with respect to its left-lower vertex as origin, every square in the first quadrant outside the rectangle is white. The most crucial fact is that knowing only the colouring of the squares in a desirable rectangle, we can determine their colourings at all later moments. Note that the smallest rectangle enclosing all black squares is a desirable rectangle. We will prove by induction that all squares of a desirable rectangle with at most n black squares will become white by t = n. The case n = 1 is clear. Suppose the cases n < N are true. Let R be a desirable rectangle with N black squares. Let R be the smallest rectangle in R ₀ containing all N black squares, then R is ₀ also desirable. Being smallest, the leftmost column and the bottom row of

0

R must contain some black squares. Now the rectangle obtained by deleting the left column of R and the rectangle ₀ obtained by deleting the bottom row of

0

R are desirable and contain at most n - 1 black squares. So by t = n - 1, all their squares will become white. Finally the left bottom corner square of R will be ₀ white by t = n.

Comments: This solution is essentially the same as the proposer's solution. Other commended solvers: LEUNG Wai

Ying (Queen Elizabeth School, Form 6). Problem 115. (Proposed by Mohammed

Aassila, Universite Louis Pasteur, Stras-bourg, France) Find the locus of the points P in the plane of an equilateral triangle ABC for which the triangle formed with PA, PB and PC has constant area.

Solution. LEUNG Wai Ying (Queen

Without loss of generality, assume PA ≥ PB, PC. Consider P outside the circumcircle of ∆ABC first. If PA is between PB and PC, then rotate ∆PAC about A by 60 so that C goes to B and P $ goes to P'. Then ∆APP' is equilateral

and the sides of ∆PBP' have length PA, PB, PC.

Let O be the circumcenter of ∆ABC, R be the circumradius and x = AB = AC =

AO

3 = 3R. The area of ∆PBP' is the sum of the areas of ∆PAP' , ∆PAB,

∆P'AB (or ∆PAC), which is PAB x PA PA + ⋅ sin∠ 2 1 4 3 2 PAC x PA sin 2 1 _⋅ + . Now

sin∠PAB + sin∠PAC = 2 sin150 cos($ ∠PAB−150$) = )−cos(∠PAB+30$ R PA R PA PO PAO ⋅ − − = ∠ − = 2 cos 2 2 2 . Using these and simplifying, we get the area of ∆PBP' is 3(PO2−R2)/4. If PC is between PA and PB, then rotate

PAC

∆ about C by 60 so that A goes to $ B and P goes to P'. Similarly, the sides of

'

PBP

∆ have length PA, PB, PC and the area is the same. The case PB is between PA and PC is also similar.

For the case P is inside the circumcircle of ABC∆ , the area of the triangle with sidelengths PA, PB, PC can similarly computed to be 3(R2−PO2)/4 . Therefore, the locus of P is the circle(s) with center O and radius R2±4c/ 3, where c is the constant area.

Comments: The proposer's solution only differed from the above solution in the details of computing areas.

Olympiad Corner

(continued from page 1)

Problem 4. (cont’d)

Determine all values of k for which 2000 is a term of the sequence.

Problem 5. The seven dwarfs decide to

form four teams to compete in the Millennium Quiz. Of course, the sizes of the teams will not all be equal. For instance, one team might consist of Doc alone, one of Dopey alone, one of Sleepy, Happy and Grumpy as a trio, and one of Bashful and Sneezy as a pair. In how many ways can the four teams be made up? (The order of the teams or of the

dwarfs within the teams does not matter, but each dwarf must be in exactly one of the teams.)

Suppose Snow White agreed to take part as well. In how many ways could the four teams then be formed?

Majorization Inequality

(continued from page 2) For the case n > 2, let a = _i xi/(x1++

) n

x for i = 1, …, n, then a₁++a_n = 1. In terms of a 's, the inequality to be _i proved becomes

(

a a

)

C a a _i _j n j i i j ≤ +

∑

≤ < ≤ 2 2 1 . The left side can be expanded and regrouped to give

(

)

∑

= + + − + + + n i i i i n a a a a a 1 1 1 1 3 ). 1 ( ) 1 ( ₁ 3 3 1 a an an a − + + − = Now f(x) = x3(1−x)=x3−x4 = is strictly convex on     2 1 , 0 because the

second derivative is positive on 

     2 1 , 0 . Since the inequality is symmetric in the

i a 's, we may assume a₁≥a₂≥≥a_n. If 2 1 1≤ a , then since

(

a ,a ,...,an

)

0 ..., , 0 , 2 1 , 2 1 2 1       _,

by the majorization inequality, ) ( ) ( ) (a₁ f a₂ f a_n f + ++ 8 1 ) 0 ( ) 0 ( 2 1 2 1 = + + +       +       ≤ f f f f . If 2 1 1> a , then 1−a₁,a₂,...,a_n are in [0, 2 1 ]. Since ) ..., , ( ) 0 ..., , 0 , 1 ( −a₁ a₂ a_n , by the majorization inequality and case n = 2, we have f(a₁) + f(a₂) + + f(a_n) ) 0 ( ) 0 ( ) 1 ( ) (a₁ f a₁ f f f + − + + + ≤ = 8 1 ) 1 ( ) (a₁ + f −a₁ ≤ f .

Equality holds if and only if two of the variables are equal and the other n−2 variables all equal 0.