Mathematical Excalibur, Volume 1, Number 4

Volume 1, Number4

Olympiad Corner

The 36th International Mathematical Olympiad wad held in Toronto, Canada on July, 1995. The following six problems were given to the contestants. (I'he country inside the parantheses are the problem proposers.) -Editors

First

Day

Question 1. (Bulgaria)

Let A, B, C and D be four distinct points on a line, in that order. The circles with diameters AC and BD intersect at the pointsXand Y. ThelineXYmeetsBCat the point Z. Let P be a point on the line

XY different from Z. The line CP

intersects the circle with diameter AC at the points C and M, and the line BP intersects the circle with diameter BD at the points B and N. Prove that the lines

AM.

DN and XY are concurrent. Question 2. (Russia)

Let a, b and c be positve real numbers such that abc= 1. Prove that

1

+

l

+

1 ~

!.

a3_{(b+c) b}3_{(c+a) c}3_{(a+b) 2} (continued on page 4)

Editors: Cheung, Pak-Hong, Curr. Studies, HKU Ko, Tsz-Mei, EEE Dept, HKUST Leung. Tat-Wing. Appl MathDept, HKPU Li, Kin- Yin, Math Dept, HKUST Ng, Keng Po Roger, ITC, HKPU Artist: Yeung, Sau-Ying Camille, MF A, CU Acknowledgment: Thanks to Debbie Leung for her help in typesetting.

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 TeX, MS Word and WordPerfect, are encouraged. The deadline for receiving material for 1he next issue is October 15, 1995. Send all correspondence to:

Dr. Tsz-MeiKo

Deparlment ofElectrical and Electronic Engineering Hong Kong University of Science and Technology

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

Email: [email protected]

September-October, 1995

Descartes' Rule of Signs

Andy Liu

University of Alberta, Canada

Let P(x) be a polynomial of degree n with complex coefficients. The Fundamental Theorem of Algebra tells us that it has exactly n complex roots. We are interested in the number of real roots in the case where the coefficients are real. We may assume that the leading coefficient is 1 and the constant term is non-zero.

As an example, consider

p(x) = x6_{-6.x5+ I0x}4_{-2.x3-3x2+4x-l2.} As it turns out, it has four real roots -1, 2 (with multiplicity 2) and 3, and two non-real roots i and-i.

In

general, we may not be able to find the roots of P(x). However, we can obtain some information about the number of positive roots from the number of sign-switches of P(x). If we consider the sequence of the signs of the non-zero coefficients of P(x) in order; a sign-switch is said to occur if a

+

is followed immediately by a - or vice

versa.

For p(x) above, the sequence is+ - +

- - + -.

Hence the number of sign switches is 5.

The first part of Descartes' Rule of Signs states that the number of positive roots of P(x) has the same parity as the number of sign-switches of P(x). Clearly, the latter is even if and only if the constant term of P(x) is positive (because the sign sequence begins and ends with+). What we have to prove is that the same goes for the number of positive roots of P(x).

From the Fundamental Theorem of Algebra, P(x) is a product of linear factors and irreducible quadratic factors. Now the constant term of a quadratic factor with a negative

discriminant must be . positive. The constant term of a linear factor is positive if and only if it corresponds to a negative root. It follows that the sign of the constant term of P(x) is positive if

and only if the number of positive roots of P(x) is even.

Since the number of sign-switches of p(x) is 5, we can tell that it has an odd number of positive roots without trying to find them.

The second part of Descartes' Rule of Signs states that the number of positive roots of P(x) is less than or equal to the number of sign-switches of P(x). We shall build up P(x) as follows. Start with the product of all irreducible quadratic factors and all linear factors corresponding to negative roots. What we have to prove is that the number of sign-switches increases every time we introduce

a

linear factor corresponding to a positive root.

For any polynomial Q(x) with real coefficients, leading coefficient 1 and a non-zero constant term, we group consecutive terms of the same signs together to express Q(x) as an alternating sum of polynomials of positive coefficients. Then the sign-switches occur precisely between summands. We claim that when we multiply Q(x) by x -

t

for some positive number t, the original sign-switches are preseived, while at least one additional

sign-switch occurs.

Consider each summand in tum. The leading coefficient is positive. This does not change after multiplication by x. However, we may have to combine it with -t times the last term of the preceding summand. Since there is a sign-switch (continued on page 2)

Mathematical Excalibur，的／. 1, No. 4, Sept-Oct’的 P~g已 2

Descartes' Rule of Signs (continued from page 1)

betwe叩 thetwo summan血，位1etennwith

which it is to be combined is also positive. 2血 jus位fiesthe first cla油l. 官1e second claim follows since 誼1econstant terms of Q(x) and (x-t)Q(x） 恤rveopposite signs.

τbis completes the proof of Descartes

’

Rule of Signs.

Let us illu甜atethe proof of the second partwi也

p(x)

=

（正＋1）（：計l)(x-2)2伊3).

We first let

q(x)= （.正＋l)(r+ 1)

=

x3+x2+.x-+ 1. Since the number of si伊－switch，臼 is 0, 也ereis only one summand. We have q1(x)

=

(x-2) q(x)

=

(x-2) (r+.x2+r+ 1)

=

x4-:x!-x2-x-2 =x仁（r+x2+.x-+2). qi(x)

=

(x-2) q1(x)

=

(x-2)x4 - (x-2)(r+x2+.x-+2) ＝（~－2x4）－昕一 :x!-x2-4)

=

~

-

3x4 + :x! + x2 + 4.

Note 也atwe have combined the terms -x4 and -2x4 which have the 回me sign. F趴巫ly,

p(x)

=

(x-3) qi(x)

=(x-3昕一（x-3)(3的＋(x-3)(r+x2+4)

=

(x6-3~）－（3封－9x4)+(x4-2r-3.x2+4x﹒12)

=x6-

6r +10x4-

2x3-

3x2 + 4x-12. We point out 由at u血ig the same argument, we 個1prove that 也e nm曲er

of negative roots of P(x) is not g臼t位 也扭曲enumb位 of sign-switches 姐 P(-x), and differs from it by an even number. For example，位1enm曲erof sign-switches inp(-x) =x6+6r+I0x4+1r-3x2-4x-一12is 1, and we 臼n conclude that p(x) has exa叫yone neg甜:veroot

希臘幾何學的發展

林達威，鄧智傑，王俊威

Form 5, St. Paul

’

s Co-ed College

「幾何 J 一詞，垃丁文是 geometr徊，其 中”g的﹒”代表「地」（與 geography, geology 中”geo－”的意思一致），而 ”metria”則與今天英文的＂metric＂相 闕，代表「量度」﹒兩部的合起來，就 是「量地的學問」－一﹒原來古埃及的尼 羅河每年都設濫一次，摧毀河胖的農 地，拱水過後，政府為重新創定的農地 量度面積，以決定每戶所須擻付的賦 稅﹔幾何學就在這情況下應運而生﹒後 來，這些知識輾轉傳入希臘，逐漸發展 成一套完整的學說﹒ 希臘數學史可分為三個時期：第一段 從愛奧尼亞學派到柏拉圖學派為止﹔第 二段是亞歷山大前期，從歐幾里得到羅 馬攻陷希臘為止﹔最後則是亞歷山大後 期﹒ 第一段時期的希臘共有六個主要學 派，其中以舉行素食的畢達哥拉斯 (Pythagoras）學派最負盛名，他們的「畢 氏定理 J (Pythagorean Theorem）是理科 必備的工具﹒這時期，很多數學家都有 重要的發現，雖然沒有建立一套完整的 學說，但為日後阿基米得等人的數學理 論建立了→圓良好的基礎﹒ 第二段時期中最偉大的數學家可算是 歐幾里得(Euclid）•歐幾里得深譜柏拉圖 幾何的精髓，經過最謹的演輝和推論， 寫成了〈〈原本＞＞(Elements）一書﹒〈〈原 本〉〉為聽何建立了一套完整的理論，在 1637年笛卡爾(Descartes）引入「坐標幾 何」前，它佔幾何學的領導地位﹒它也 是用公理法建立起演繹數學體系的最早 典範一一所謂「公理法J ’就是從一些 大家都公認可以接受，毋須加以証明的 「公理」出麓，通過合乎邏輯的推論而 得出極驗証的結果﹒這可說是人頭從直 觀事物邁向抽象思維的重要一步！ 亞歷山大前期的男一位偉大的數學家 是阿基米德﹒在他所著的〈〈圓的量度〉〉 中，阿基米德利用外切與內接九十六邊 形求得圓周率π的兩個近似值： 31_%1 ＜π ＜3后﹒ 假設圓的半徑為 1 •則圓周剛好為站， 此數值盛頓大於內接六邊形的周界，而 六邊形的周界為 6 0 故此求得 π 的下限 為 3 0 同理，利用圓形的外切六邊形， 可求得 π 的上限為 3.4641 °假若我們把 六邊形換作十二邊形、廿四邊形，．．．， 則（內接或外切）多邊形的周界會越來 越接近圓，而相應的上限及下限也會趨 近 π 的真實數值﹒阿基米得利用圓形的 外切與內接九十六遍形，求得 31m ＜π ＜3

y.;

.阿基米得的另一項建樹 是體積的計算，例如圓球體積是它的外 接圓柱體積的三份二﹒阿基米德還發現 圓球的表面積恰巧也是外接圓柱表面積 的三份二，他非常欣賞這定理，吩咐親 人把這個圓形刻在他的墓碑上（見團）． 最後一段是亞歷山大後期﹒這個時期 的數學家，以歐幾里得的〈〈原本均為根 撮，作出了不少增潤修補的工作，而我 們現在所學的平面幾何，前在這時帳逐 漸形成﹒讀到這里，同學們該明白到這 門學間是最多數學家集體智慧的結晶及 長期辛勞的成果﹒

11

I

i !

I

Mathematical Excalibur, Vol. 1, No. 4, Sept-Oct, 95

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 and school affiliation. Please send submissions to Dr. Tsz-Mei Ko, Dept of EEE, Hong Kong University of Science and Technology, Clear Water Bay, Kowloon. The deadline for submitting solutions is October 15, 1995.

Problem 16. Let a, b, c, p be real numbers, with a, b, c not all equal, such

1 1 1

that a+-=b+-=c+-=p.

b C a

Determine all possible values of p and prove that abc + p = 0. (Source: 1983 Dutch Mathematical Olympiad.)

Problem 17. Find all sets of positive

integers x, y and z such that x :5:y :5: z and

xY

+

y"

=

z'.

Problem 18. For real numbers

a,

b,

c,

define

f(a,b,c) = a+b-la-bl-la+b + la-bl

-2cl.

Show that fi:a,b,c) > 0 if and only if fi:b,c,a) > 0 if and only if.f{c,a,b) > 0.

Problem 19. Suppose A is a point inside a given circle and is different from the center. Consider all chords ( excluding the diameter) passing through A. What is the locus of the intersection of the tangent lines at the endpoints of these chords?

Problem 20. For n > 1, let 2n chess pieces be placed on any 2n squares of an n x n chessboard. Show that there are 4 pieces among them that formed the vertices of a parallelogram. (Note that if 2n - 1 pieces are placed on the squares of the first column and the first row, then there is no parallelogram. So 2n is the best possible.)

*****************-Solutions

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

Problem 11. Simplify 1995

:E

tan n tan( n + 1). n=I

(There is an answer with two terms involving tanl, tan1996 and integers.)

School for Girls, Kwun Tong), CHAN

Chi Kin (Pak Kau English School),

CHAN Sze Tai, Angie (Ming Kei

College), CHAN Wing Sum

(HKUST), CHOW Chak On

(HKUST), CHUI Yuk Man (Queen

Elizabeth School), LEUNG

Ka

Fai (Ju

Ching Chu Sceondary School (Yuen Long)), LIU Wai Kwong (Pui Tak Canossian College), Alex MOK Chi

Chiu (Homantin Government

Secondary School),

TAM Tak

Wing

(Delia Memorial School (Yuet Wah)) and WOO Chin Yeung (St. Peter's

Secondary School).

From tanl = tan[(n+l) - n]

=

(tan(n+ 1)-tann )/(1 +tanntan(n+ 1)), we get

~:;tan(n)tan(n+l)=

L

tan(n+l)- tan(n) I

1995 1995( )

n=I n=I tan 1

= _{tanl996- tanl _ 1995} = tanl996 - l9%.

tanl tan}

Comments: This problem illustrates the telescoping method of summing a series, i.e., by some means, write a,, as b n+ 1 - b,, , then summing a,, will result in many cancellations yielding a simple answer.

Problem 12. Show that for any integer

n

> 12, there is a right triangle whose sides are integers and whose area is between n and 2n. (Source: 1993 Korean Mathematical Olympiad.)

Solution: WONG Chun Keung, St.

Paul's Co-ed College.

Consider triangle A with sides 3d, 4d, 5d, which has area 6d2. So for n in the interval (3d2 + 1, 6d2 - 1 ), triangle A has an area between n and 2n. Ford'?. 3, 6tf -1-(3(d+l)2_{+ 1) =3(d-1)}2_{-8> 0. So}

the intervals (3d2 + 1, 6d2 - 1) with d

=

3, 4, 5, ... cover all positive integers n greater than or equal to 28. For d

=

2, triangle A has area 24, which takes care of the cases n

=

13, 14, ... , 23. Finally, the cases n

=

24, 25, 26, 27 are taken care ofby the triangle with sides 5, 12, 13, which has area 30.

Other commended solvers: CHAN Wing Sum (HKUST) and LIU Wai Kwong (Pui Tak Canossian College). Problem 13. Suppose x"' Yk (k

= I,

Solution: Independent solutions by Iris 2, ... , 1995) are positive and x1 + x2

+ ...

CHAN Chau Ping (St. Catherine's + x1995 =Yi+ y2 + ... + y,995

=

1.

Page3 Prove that

1995 X y I

I

k k :5:-. k=t xk + Yk 2

Solution: Independent solution by

CHAN Chi Kin (Pak Kau English

School), CHAN Wing Sum (HK.UST), KWOK Wing Yin (St. Clare's Girls'

School) and LEUNG Ka Fai (Ju Ching

Chu Secondary School (Yuen Long)). Since x,Jl,/(xk+yJ .::;; (xk +yJ/4 (is equivalent to (xk -

yJ

2 _{'?. 0 by simple} algebra), we get

1995 1995

L

XkYk :5:

L

xk

+

Yk =

.!..

k=t xk

+

Yk k=t 4 2 Other commended solvers: Iris CHAN Chau Ping (St. Catherine's School for

Girls, Kwun Tong), CHEUNG Lap Kin (Hon Wah Middle School), CHOW Chak On (HKUST), LIU Wai Kwong (Pui Tak Canossian. College), Alex MOK Chi Chiu (Homan.tin

Government Secondary School),

TAM

Tak

Wing (Delia Memorial School

(Yuet Wah)), WONG Chun Keung

(St. Paul's Co-ed College) and WOO Chin Yeung (St. Peter's Secondary

School).

Problem 14. If llABC, !:ut'B'C' are (directly) similar to each other and MA'A", MB'B", ACC'C" are also (directly) similar to each other, then show that llA "B "C", llABC are (directly) similar to each other.

Solution: Independent solution by CHAN Wing Sum (HK.UST) and LIU Wai Kwong (Pui Tak Canossian College).

We will use capital letters for points and small letters for the corresponding complex numbers. Since AAA 'A", !illB'B", ACC'C" are (directly) similar to each other,

a"-a b"-b c"-c

- - = - - = - - = r .

a'-a b'-b c'-c

Then a"= ra'+(l-r)a, b" = rb'+(l-r)b, c"

=

rc'+(l-r)c. Since llABC, !:ut'B'C' are (directly) similar to each other,

Then

b-a b'-a'

c-a c'-a'

b"-a" r(b'-a')+(l-rXb-a) b-a

c"-a" == r( c'-a') + (1-r)( c - a)

= -;-::;; ,

(continued on page 4)

Mathematical Excalibur, Vol. J, No. 4, Sept-Oct, 95

Problem Corner

(continued from page 3)

which is equilvalent to M"B"C" (directly) simlar to MBC.

Problem 15. Is there an infinite

sequence a0, a1, az, · · · of non-zero real

numbers such that for n == I, 2, '.3, • • ·, the polynomiaJ

P.(x)

=

a0 + a1x +a~+···+ a;,:" has exactly n distinct real roots? (Source: 1990 Putnam Exam.)

Solution: Yes. Take a0

=

1, a1 == -1 and

proceed by induction. Suppose a0, • • ·, a. have been chosen so that P.(x) has n

distinct real roots and P .(x) -,. oo or

-oo as x -,. oo depending upon whether

a. is positive or negative. Suppose the roots of Ph) is in the interval (-T,n.

Let a,,.1 == (-ly,+1_{/M, where Mis chosen}

to be very large so that T,,.1_{/M is very} small. ThenP ,...1(x)

=

P.(x) + (-x),...1/M

is very close to P.(x) on [-T,7] because

IP .,..

1(x)- PnCx)I .:S Tnt1/Mfor every x on

[-T,1]. So, P .,..1(x) has a sign change very close to every root of P .(x) and has the same sign as Pn(x) at T. Since P.(x)

and P ,..1 (x) take on different sign whenx

- oo, there must be another sign change

beyond T. So P nt 1 (x) must have n+ 1 real roots.

Comments: Liu Wai Kwong sent in a more detail solution showing that the numbers can even be chosen to have the

same sign.

Olympiad Corner

(continued from page I) Question 3. (Czech Republic) Determine all integers n > 3 for which

there exist n points A1, Ai, ···,A. in the

plane, and real numbers r1, r1, ·· ·, r.

satisfying the following two conditions: (i) no three of the points A 1, Ai, · · ·, A.

lie on a line;

(ii) for each triple i, j, k (I .:S i < j < k

:,; n ) the triangle A,A1A* has area

equal tor,+ r1 +

'*

.

Second Day

Question 4. (Poland)

Find the maximum value of x0 for which there exists a sequence of positive real numbers xo, X1, ... , x1995 satisfying the

two conditions: (i) Xo = X1995 ; ( ._II . ) _{Xi+I + - =} 2 2 _{X ; + -}1 X;-1 X1 for each i = 1, 2, ···, 1995.

Question 5. (New Zealand)

LetABCDEFbe a convex hexagon with

and

AB=BC=CD, DE=EF=FA,

L BCD = L EFA = 60". Let G and Hbe two points in the interior of the hexagon such that:

L AGB

=

L DHE

=

120°. Prove that

Page4

Question 6. (Poland)

Let p be an odd prime number. Find the numberofsubsetsA oftheset {l, 2, ···,

2p} such that

(i) A has exactly p elements, and (ii) the sum of all the elements in A is

divisible by p.

IM0~95, Toronto, Canada

Kin Y Li

On July 16, the Hong Kong team started their journey to Toronto, Canada for the thirty-sixth International

Mathematical Olympiad. The flight

took about 18 hours with one stop at Anchorage, Alaska. Shortly after arrival, the team

was

interviewed by local Chinese media. The Canadian host certainly publicized the event very

well. During the entire period, the team stayed at the beautiful York University campus. The quarters provided were very comfortable; each person had his own room!

Opening ceremony came two days later and the examination followed. Team leaders and deputy leaders began maoongs and coordination soon afterward, while the students were given tours to Toronto's top attractions, such as Skydome, Ontario Science Center, Downtown Toronto, CN (Canadian National) Tower, Canada's Wonderland and of course, Niagara FaJls. Meanwhile the scores were

quickly decided. This year the team brought home two silvers, three bronzes and one honorable mention. (One silver was actually one mark short of a gold!) In the closing ceremony, the winners

received their medals. Also, for entertainment, there were impressive

perfonnances, which included an awesome laser show. Throughout the

events, there were many opportunities for students from different countries to

get to know each other. Enjoying every moment of the whole trip, the team

finally came home reluctantly on the evening of July 25. Everybody had

fond memories and developed new

friendships.

Photo a1 left: The 199S Hong Kong Math Olympiad Toam takat at the Kai Tak Airport before departure. From left to right are: U Kin-Yin (leader). MOK Tue Tao, HO Wmg Yip. POON Wai Ho~ CHEUNG

Kwok. Koon, YU Oum Ling. WONO Him Tmg. and