Mathematical Excalibur, Volume 8, Number 3

Volume 8, Number 3 June 2003 – July 2003

容斥原則和 Turan 定理

梁 達 榮

Olympiad Corner

The XV Asia Pacific Mathematics Olympiad took place on March 2003. The time allowed was 4 hours. No calculators were to be used. Here are the problems.

Problem 1. Let a, b, c, d, e, f be real numbers such that the polynomial

P(x) = x8 _{– 4x}7 _{+ 7x}6 _{+ ax}5 _{+ bx}4 _{+ cx}3 _{+ dx}2 _{+ ex + f}

factorises into eight linear factors x – xi, with xi> 0 for i = 1, 2, …, 8. Determine all possible values of f.

Problem 2. Suppose ABCD is a square piece of cardboard with side length a. On a plane are two parallel lines l and 1

2

l , which are also a units apart. The square ABCD is placed on the plane so that sides AB and AD intersect _{l at E 1} and F respectively. Also, sides CB and

CD intersect l at G and H respectively. ₂ Let the perimeters of AEF∆ and

CGH

∆ be m1 and m2 respectively.

Prove that no matter how the square was placed, m1 + m2 remains constant.

(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 August

10, 2003.

For individual subscription for the next five issues for the 03-04 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin LI Department of Mathematics

The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong

Fax: (852) 2358 1643 Email: [email protected] 設 A 為有限集，以| A|表示它含元素 的個數。如果有兩個有限集 A 和 B， 以A∪B表示 A 和B 的并集（它包含 屬於 A 或 B 的元素），以A∩B表示 A 和 B 的交集 (它包含同時屬於 A 和 B 的元素)。眾所周知，如果 A 和 B 之間 沒 有 共 同 元 素 ， 則|A∪B|=|A|+ | | B ，但是如果 A 和 B 之間有共同元 素 x，當數算 A 元素的數目時，x 被算 了一次，但數算 B 元素的數目時，x 又再被算了一次。為了抵消這樣的重 覆，在計算|A∪B|時，我們要減去重 覆 數 算 的 次 數 ， 即|A∩B|。 因 此 | | | | | | | |A∪B= A + B − A∩B 。 對於三個集的并集A∪B∪C，我們可 以先數算 A，B 和 C 的個數，相加起 來，發覺是太大了，必須減去一些交 集的個數，現在 A，B 和 C 中任兩個 集 的 交 集 可 以 是 A∩B ， A∩C 和 C B∩ ，當我們減去這些交集的元素 個數時，發覺又變得太少了，最後我 們還要加上三個集的交集的元素個 數 ， 最 後 得 |A∪B∪C|=|A|+ | | | | | | | | | |B + C − A∩B − A∩C − B∩C | |A∩B∩C +

。

一般來說，如果有 n 個有限集 ,A1 n A A ,...,₂ ， 則|A₁∪A₂∪L∪An| = n k j i n j i i j n i∑=1|Ai|−1≤p∑≤ |A ∩A |+1≤p∑p ≤ 2 1 1_| ) 1 ( | |Ai∩Aj∩Ak −L+ − n− A ∩A | n A ∩ ∩L 等式中右邊第一個和式代 表A1至An各集元素個數的總和，第二 個和式代表任何兩個集的交集元素個 數 的 總 和 ， 餘 此 類 推 ， 直 到 考 慮 L , , 2 1 A A 至A_n的交集為止。 上 面 的 等 式 ， 一 般 稱 為 容 斥 原 則 (Inclusion-Exclusion Principle)，其命 意義相當明顯。證明可以採歸納法， 但也可以利用二項式定理加以證明。 過程大概如下。設 x 屬於A1∪A2∪ n A ∪ L ， 則 x 屬 於 其 中 k 個 ,Ai ) 1 (k≥ ，為方便計，設 x 屬於A1,A2,..., k A，但不屬於A_k+1,...,A_n。這樣的話， x 在A1∪A2∪L∪A_n的“ 貢獻” 為 1。在右邊第一個和式中，x 的“ 貢獻” 為_{k C}k 1 = 。在第二個和式中，由於 x 在A1,A2,...,A_k中出現，則 x 在它們任 兩個集的交集中出現，但不在其他兩 個集的交集中出現，因此，x 在第二個 和式中的“ 貢獻” 為_Ck 2。這樣分析下 去，我們發覺 x 在右邊的“ 貢獻” 總 和 是 k k k k k k _{C C C} C₁ − ₂+ ₃ −L+(−1) +1 1 ) 1 1 ( 1− − = = k _。 留意我們用到了二項式定理，由於 x 在兩邊的貢獻相等，我們獲得了容斥 原則成立的證明。 再者二項式系數有以下的性質。 k m C 在 2 k m≤ 時遞增，在_m≥2k _{時遞減。（例} 如 k = 5，有 5 4 5 3 5 2 5 1 5 0 C C C C C < < = > 5 5 C > ， 5 m C 在 m = 2, 3 時取最大值，k = 6，C₀6<C₁6<C₂6<C₃6>C₄6>C₅6>C₆6 ， 6 m C 在 m = 3 時最大值。）利用這個 關係，讀者可以証明，如果在容斥原 則的右邊，略去一個正項及它以後各 項，則式的左邊大於右邊，這是因為 x 對於右邊的貢獻非正，或者被略去的 貢獻非負。同理，如果在容斥原則的 右邊略去一個負項及它以後各項，則 式的左邊變為小於右邊。這是一個有 用的估計。 容斥原則作為數算集的大小的用途上 時常出現，應用廣泛。

Mathematical Excalibur, Vol. 8, No. 3, Jun 03- Jul 03 Page 2 例一： 這是一個經典的題目，將 1, 2, …, n重新安排次序，得到一個排 列，如果沒有一個數字在原先的位置 上，則稱之為亂序，(例如，4321 是 一個亂序，4213 不是)，現在問，有 多少個亂序？ 解答: 顯而易見，所有的排列數目是 1 ) 1 ( !=n× n− ×L× n 。但如果直接找 尋亂序的數目，卻不是很容易。因此 我們定義Ai為 i 在正確位置的排 列，1≤i≤n。易見|A_i|=(n−1)!，同 理 A_i∩A_j =(n−2)!，此處i≠ j，等 等。因此 ! ! 1 ! 3! ! 2! 1 3 2 2 1 1 1 1 1 2 1 ) 1 ( ! 1 ) 1 ( )! 3 ( )! 2 ( )! 1 ( ) 1 ( n n n n n n n n n n n k j i i j k n j i i j n i i n n n C n C n n A A A A A A A A A A A A − − − ≤ ≤ ≤ ≤ = − + − + − = − + − − + − − − = ∩ ∩ ∩ − + − ∑ ∩ ∩ + ∑ ∩ − ∑ = ∪ ∪ ∪ L L L L L p p p 最後，亂序的數目是

(

!

)

1 ! 3 1 ! 2 1 2 1 ) 1 ( ! ! n n n n A A A n − + + − = ∪ ∪ ∪ − L L 。 例二： (IMO 1991) 設 S = {1, 2, …, 280}。求最小的自然數 n，使得 S 的 每一個 n元子集都含有 5 個兩兩互素 的數。 解 答 : 首 先 利 用 容 斥 原 則 求 得 217 ≥ n 。設A1,A2,A3,A4是 S 中分 別為 2, 3, 5, 7 的倍數的集，則 1 , 2 , 4 , 6 , 9 , 8 , 13 , 18 , 20 , 28 , 46 , 40 , 56 , 93 , 140 4 3 2 1 4 3 2 4 3 1 4 2 1 3 2 1 4 3 4 2 3 2 4 1 3 1 2 1 4 3 2 1 = ∩ ∩ ∩ = ∩ ∩ = ∩ ∩ = ∩ ∩ = ∩ ∩ = ∩ = ∩ = ∩ = ∩ = ∩ = ∩ = = = = A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A 因此 216 1 2 4 6 9 8 13 18 20 28 46 40 56 93 140 4 3 2 1 = − + + + + − − − − − − + + + = ∩ ∩ ∩A A A A 對於這個 216 元的集，任取 5 個數， 必有兩個同時屬於A1,A2,A3或A4， 因此不互素 。按題意， 所以必須有 217 ≥ n 。現在要證明 S 中任一 217 元集 必有 5 個互素的數，方法是要構造適當 的“ 抽屜” 。其中一個比較簡潔的構造 是這樣的。設 A 是 S 的一個子集，並且 217 ≥ A 。定義 1 B ={1 和 S 中的素數}，B1 = 60， , 6 }, 13 , 11 , 7 , 5 , 3 , 2 { 2 2 2 2 2 2 ₂ 2= B = B . 5 }, 13 11 , 23 7 , 41 5 , 73 3 , 109 2 { , 5 }, 17 11 , 27 7 , 43 5 , 79 3 , 113 2 { , 6 }, 17 13 , 19 11 , 31 7 , 47 5 , 87 3 , 127 2 { , 6 }, 19 13 , 23 11 , 37 7 , 53 5 , 89 3 , 131 2 { 6 6 5 5 4 4 3 3 = × × × × × = = × × × × × = = × × × × × × = = × × × × × × = B B B B B B B B 易見B1至B6互不相交，並且| B1∪ 88 | 6 5 4 3 2∪B ∪B ∪B ∪B = B 。去掉這 88個數，S 中尚有 280 – 88 = 192 個數。 現在 A 最小有 217 個元素，217 – 192 = 25，即是說 A 中最小有 25 個元屬於B1 至B6。易見，不可能每個Bi只含 A 中 4 個或以下的元素，即是說最少有 5 個或 以上的元素屬於同一個Bi，因此互素。 注意這裏我們用到另一個原則：抽屜原 則。 例三： (1989 IMO) 設 n 是正整數。我 們說集{1, 2, 3, …, 2n}的一個排列(x1, ) ,..., 2 2 x n x 具有性質 P，如果在{1, 2, 3, …, 2n – 1}中至少有一個 i，使得 |xi− n xi+1|= 成立。証明具有性質 P 的排列 比不具有性質 P 的排列多。 解答: 留意如果|xi− xi+1|=n，其中一 個xi或xi+1必小於 n + 1。因此對於 k = 1, 2, …, n, 定義Ak為 k 與 k + n 相鄰的 排列的組合，易見|A_k|=2×(2n−1)!。 （這是因為 k 與 k + n 并合在一起，但 位置可以互 相交換，想 像它是一個 “ 數” ，而另外有 2n - 2 個數，這(2n – 2) + 1 個數位置隨意。）同時|Ak∩Ah| =₂2×_{(2n – 2)!, 1≤k<h≤n, (k 與 k + n} 合在一起成為一個“ 數” h 與 h + n 合 在一起成為一個“ 數” 。)因此具性質 P 的排列的數目 )! 2 2 ( 2 )! 1 2 ( 2 | | | | 2 2 1 1 2 1 − × × − × − × = ∑ ∩ − ∑ ≥ ∪ ∪ ∪ ≤ ≤ = n C n n A A A A A A n n h k k h n k k n p L 2 1 1 2 (2 )! )! 2 ( )! 2 2 ( 2 × − × = × > × = n n n n _nn₋ n 這個數目超過(2n)!的一半，因此具 性質 P 的排列比不具性質 P 的排列 多。 (這一個問題，當年被視為一個難 題，但如果看到它與容斥原則的關 係，就變得很容易了。) 例四：設 n 和 k 為正整數， n>3, ₂n < k < n。平面上有 n 個點，其中任意 三點不共線，如果其中每個點至少與 其它 k 個點用線連結，則連結的線段 中至少有三條圍成一個三角形。 解答: 因為 n >3, k > ₂n_{，則k≥2，} 所以 n 個點中必中兩個點v1和v2相 連結。考慮餘下的點，設與v1相連結 的點集為 A，與v2相連結的點集為 B，則 |A|≥k−1|,B|≥k−1。另外 | | 2 2 | | | | | | | | 2 B A k B A B A B A n ∩ − − ≥ ∩ − + = ∪ ≥ − 即|A∩B|≥2k−n>0。因此，存在 點v3與v1和v2相連結，構成一個三 角形。 例五： 一次會議有 1990 位數學家參 加，其中每人最少有 1327 位合作 者。証明，可以找到 4 位數學家，他 們中每兩人都合作過。 証明: 將數學家考慮為一個點集，曾 經合作過的連結起來，得到一個圖。 如上例，v1互v2曾合作過，所以連 結起來，餘下的，設 A 為和v1合作 過的點集，B 為和v2合作過的點集， 則|A|≥1326|,B|≥1326，同樣，|A∪ ≤ | B 1990 – 2 = 1998，因此 | | | | | | | |A∩B= A + B − A∪B 0 664 1998 1326 2× − = > ≥ 即是說，可以找到數學家v3，與v1和 2 v 都合作過。設 C 為除v1和v2以 外 ， 與v3合 作 過 的 數 學 家 ， 即 1325 | |C ≥ 。同時 | | | | | ) ( | 1998≥ A∩B ∪C = A∩B + C - |A∩B∩C| 即 1988 | | | | | |A∩B∩C ≥ A∩B + C − 0 1 1988 1325 664+ − = > ≥ 。 因 此 A∩B∩C 非 空 ， 取 v4∈ C B A∩ ∩ ，則v1,v2,v3,v4都曾經合 作過。 (continued on page 4)

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.

The deadline for submitting solutions is August 10, 2003.

Problem 181. (Proposed by Achilleas

PavlosPorfyriadis, AmericanCollege of Thessaloniki “Anatolia”, Thessaloniki, Greece) Prove that in a

convex polygon, there cannot be two sides with no common vertex, each of which is longer than the longest diagonal.

Problem 182. Let a0, a1, a2, … be a

sequence of real numbers such that

an+1 ≥ an2+ 1/5 for all n ≥ 0. Prove that an+5 ≥an2−5 for all

. 5

≥ n

Problem 183. Do there exist 10 distinct integers, the sum of any 9 of which is a perfect square?

Problem 184. Let ABCD be a rhombus with ∠B=60o. M is a point inside ∆ADC such that ∠AMC=

o

120 . Let lines BA and CM intersect at

P and lines BC and AM intersect at Q.

Prove that D lies on the line PQ. Problem 185. Given a circle of n lights, exactly one of which is initially on, it is permitted to change the state of a bulb provided one also changes the state of every d-th bulb after it (where d is a divisor of n and is less than n), provided that all n/d bulbs were originally in the same state as one another. For what values of n is it possible to turn all the bulbs on by making a sequence of moves of this kind?

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

Solutions

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

Problem 176. (Proposed by Achilleas

PavlosPorfyriadis,AmericanCollege of Thessaloniki “Anatolia”, Thessaloniki, Greece) Prove that the fraction

n n m n m − + + + ) 1 ( 1 ) 1 (

is irreducible for all positive integers m and n.

Solution. CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form5), TAM Choi Nang Julian (Teacher, SKH Lam Kau Mow Secondary School), Anderson TORRES (Colegio Etapa, Brazil, 3rd _{Grade) and Alan T. W.} WONG (Markham, ON, Canada). If the fraction is reducible, then m (n + 1) + 1 and m (n + 1) – n are both divisible by a common factor d >1. So their difference

n + 1 is also divisible by d. This would

lead to

1 = ( m ( n + 1) + 1) – m ( n + 1 ) divisible by d, a contradiction.

Other commended solvers: CHEUNG Tin (STFA Leung Kau Kui College, Form 4), CHUNG Ho Yin (STFA Leung Kau Kui College, Form 6), D. Kipp JOHNSON (Teacher, Valley Catholic High School, Beaverton, Oregon, USA), LEE Man Fui (STFA Leung Kau Kui College, Form 6), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7), Alexandre THIERY (Pothier High School, Orleans, France), Michael A. VEVE (Argon Engineering Associates, Inc., Virginia, USA) and Maria ZABAR (Trieste College, Trieste, Italy).

Problem 177. A locust, a grasshopper and a cricket are sitting in a long, straight ditch, the locust on the left and the cricket on the right side of the grasshopper. From time to time one of them leaps over one of its neighbors in the ditch. Is it possible that they will be sitting in their original order in the ditch after 1999 jumps? Solution. CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form5), D. Kipp JOHNSON (Teacher, Valley Catholic High School, Beaverton, Oregon, USA), Achilleas Pavlos PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7) and Anderson TORRES (Colegio Etapa, Brazil, 3rd

Grade).

Let L, G, C denote the locust, grasshopper, cricket, respectively. There are 6 orders:

LCG, CGL, GLC, CLG, GCL, LGC.

Let LCG, CGL, GLC be put in one group and CLG, GCL, LGC be put in another group. Note after one leap, an order in one group will become an order in the other

group. Since 1999 is odd, the order LGC originally will change after 1999 leaps. Problem 178. Prove that if x < y, then there exist integers m and n such that

x < m + n 2 < y.

Solution. SIU Tsz Hang (STFA Leung Kau Kui College, Form 7).

Note 0 < 2 −1 < 1. For a positive integer , ) 1 2 log( ) log( − − > b a k we get 0 < ( 2 −1)k _{< b – a. By the} binomial expansion, x = ( 2 −1)k _{= p + q} 2 for some integers p and q. Next, there is an integer r such that

r – 1 ≤ x

a a−[ ] _{< r.} Then a is in the interval

[

[a] (r 1)x,[a] rx

)

.

I = + − +

Since the length of I is x < b – a, we get

a < [a] + rx = ([a] + rp) + rq 2 < b.

Other commended solvers: CHEUNG

Yun Kuen (Hong Kong Chinese

Women’s Club College, Form 5), D. Kipp JOHNSON (Teacher, Valley Catholic High School, Beaverton, Oregon, USA), Alexandre THIERY (Pothier High School, Orleans, France) and Anderson TORRES (Colegio Etapa, Brazil, 3rd _Grade).

Problem 179. Prove that in any triangle, a line passing through the incenter cuts the perimeter of the triangle in half if and only if it cuts the area of the triangle in half.

Solution. CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5), LEE Man Fui (STFA Leung Kau Kui College, Form 6), Achilleas Pavlos PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7), TAM Choi Nang Julian (Teacher, SKH Lam Kau Mow Secondary School), and Alexandre THIERY (Pothier High School, Orleans, France). Let ABC be the triangle, s be its semiperimeter and r be its inradius. Without loss of generality, we may assume the line passing through the incenter cuts AB and AC at P and Q respectively. (If the line passes through a vertex of ∆ABC, we may let Q = C.)

Mathematical Excalibur, Vol. 8, No. 3, Jun 03- Jul 03 Page 4 Let [XYZ] denote the area of XYZ∆ .

The line cuts the perimeter of ABC∆

in half if and only if AP + AQ = s, which is equivalent to

[APQ] = [API] + [AQI] = (r·AP) / 2 + (r·AQ) / 2 = rs/2 = [ABC] /2.

i.e. the line cuts the area of ABC∆ in half.

Problem 180. There are n ≥ 4 points in the plane such that the distance between any two of them is an integer. Prove that at least 1/6 of the distances between them are divisible by 3. Solution. CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5), D. Kipp JOHNSON (Teacher, Valley Catholic High School, Beaverton, Oregon, USA) and SIU Tsz Hang (STFA Leung Kau Kui College, Form 7).

We will first show that for any 4 of the points, there is a pair with distance divisible by 3. Assume A, B, C, D are 4 of the points such that no distance between any pair of them is divisible by 3. Since x ≡ 1 or 2 (mod 3) implies x2 _≡

1 (mod 3), AB2_{, AC}2_{, AD}2_{, BC}2_{, BD}2 _and CD2 _{are all congruent to 1 (mod 3).}

Without loss of generality, we may assume that ∠ACD = α + β, where α = ∠ACB and β=∠BCD. By the cosine

law,

AD2 _{= AC}2 _{+ CD}2 _{– 2AC·CDcos∠ACD.}

Now

cos ∠ACD = cos(α + β)

= cos α cos β – sin α sin β. By cosine law, we have

BC AC AB BC AC ⋅ − + = 2 cos α 2 2 2 and . 2 cos 2 2 2 CD BC BD CD BC ⋅ − + = β

Using sinx= 1−cos2x, we can also find sin α and sin β. Then

2BC2_·AD2 _{= 2BC}2 _(AC2 _{+ CD}2₎ – (2AC·BC)(2BC·CD) cos∠ACD = P + Q, where P = 2BC2 _(AC2 _{+ CD}2₎ – (AC2_+BC2_–AB2_)(BC2_+CD2_–BD2₎ and Q2 _{= ( 4 AC}2_·BC2 _{– ( AC}2 _{+ BC}2 _{– AB}2₎2 ₎ × ( 4 BC2_·CD2_{– ( BC}2 _{+ CD}2 _{– BD}2₎2 _). However, 2BC2⋅AD2≡2(mod 3),P≡0 (mod 3) and Q≡0 (mod 3). This lead to a contradiction.

For n ≥ 4, there are _Cn

4 groups of 4 points. By the reasoning above, each of these groups has a pair of points with distance divisible by 3. This pair of points is in a total of 2 2− n C groups. Since 1 2 4n /Cn−

C ₌ 1_6C2n, _{the result follows.}

Olympiad Corner

(continued from page 1)

Problem 3. Let k ≥ 14 be an integer, and let pk be the largest prime number which is strictly less than k. You may assume that

pk ≥ 3k/4. Let n be a composite integer. Prove:

(a) if n = 2pk, then n does not divide (n–k)!;

(b) if n > 2pk, then n divides (n–k)!. Problem 4. Let a, b, c be the sides of a triangle, with a + b + c =1, and let n ≥ 2 be an integer. Show that

n n n n n n n _an _+bn _{+ b +c + c +a .} 2 2 1+ n <

Problem 5. Given two positive integers

m and n, find the smallest positive integer

k such that among any k people, either there are 2m of them who form m pairs of mutually acquainted people or there are 2n of them forming n pairs of mutually unacquainted people.

容斥原則和 Turan 定理

(continued from page 2)

套用圖論的語言，例四和例五的意義 正如，給定一個 n 點的圖，最少有多 少條線，才可以保證有一個三角形 ) (K3 或一個K4（四點的圖，任兩點 都相連），或者換另一種說法，設有 一個 n 點的圖沒有三角形，則該圖最 多有多少條線段，等等。這一範圍的 圖論稱為極端圖論。最先的結果是這 樣的： Mantel定理 (1907): 設 n 點的簡單圖 不含K3，則其邊數最大值為_ _4 2 n _。 (此處 [x] 是小於或等於 x 的最大整 數。在例四中，邊數和多於

( )

n₂ ×n× > 2 1     4 2 n

_，

_{因此結果立即成立}

_。)

比較精緻的命題是這樣的。 定理： 如果 n 點的圖有 q 條邊，則 圖至少有 n q q n 3 ) ( 4 − ₄2 個三角形。 例六： 在圓周上有 21 個點，由其中 二點引伸至圓心所成的圓心角度，最 多有 110 個大於₁₂₀o_。 解答： 如果兩點與圓心形成的圓心 角度大於₁₂₀o_{，則將兩點連結起來，} 得到一個圖，這個圖沒有三角形，因 此 邊 數 最 多 有

[ ]

441₄ 4 212_=      = 110 條，或者最多有 110 個引伸出來的圓 心角度大於₁₂₀o_。 如上所說，定義Kp是一個 p 個點的 完全圖，即 p 個點任兩點都相連，對 於一個 n 點的圖 G，如果沒有包含 p K ，則 G 最多有多少條邊呢? Turan定理(1941): 如果一個 n 點的 圖 G 不含Kp，則該圖最多有 ) 1 ( 2 ) 1 ( 2 ) 1 ( 2 2 −− − − − ₋ p r p r p p n 條邊，其中 r 是 由 n = k(p – 1) + r, 0 r≤ < p - 1所定義 的。如 Mantel 定理的情況，這個定 理是極端圖論的一個起點。 Paul Turan (1910-1976) 猶太裔匈牙 利人，當他在考慮這一類問題時，還 是被關在一個集中營內的呢！

1. Let a, b, c, d, e, f be real numbers such that the polynomial

p(z) = z8 - 42'

+

7z8

+

az5

+

bz'

+

cz3

+

dz'

+

ez

+

f

factorises into eight h e w factors z - zi, with zi > 0 for i = 1 , 2 , . . . ;8. Determine all possible values of f .

Solution.

From

z8 - 42'

+

72'

+

az5

+

bz4

+

c z 3

+

dz2

+

e z

+

f = (z

-

q ) ( z

-

f z ) . . . (z

-

18)

we have

;=I

where the second s u m is over all pairs ( i , j ) of integers where 1 5 i < j 5 8. Since this sum can also be written

we get / 8 \ 2 8 8 so Now 8 while

cz,

= 4. (3 marks] i=l l=I 8 * c ( 2 z , - 1)2 = 4 x 2 ; - 4 c z , + 8 = 4(2) - 4(4) + 8 = 0, 1x1 ,=I *=I

which forces z, = 1/2 for all i. [3 marks] Therefore

Alternote solution: After obtaining (1) (3 marks], use Cauchy's inequality to get

16 = (21 . 1

+

2 2 . 1

+

... + ~ 8 . 1 ) ' 5 ($+ z: + .. . + z~)(lz + '1 + .. . + 1') = 8 . 2 = 16 ;

or the power mean inequality to get

Either way, equality must hold, which can only happen if all the terms z; are equal, that is, if z, = 1/2 for all

i. [1 mark] Thus f = 1/256 as above. [l mark]

2. Suppose ABCD is a square piece of cardboard with side length a. On a plane are two parallel lines

Cl and &, which are also a units apart. The square ABCD is placed on the plane so that side3 AB and

AD intersect (1 at E and F respectively. Also, sides CB and CD intersect 6 at G and H respectively. Let the perimeters of A A E F and A C G X be ml and ma respectively. Prove that no matter how the square was placed, m l

+

m2 remains constant.

LEFD. Similarly, EH bisects LEEF. So 0 is an excentre of A A E F . Similarly, 0 is an excentre of A C G H . [2 marks] Construct these excirdes with centre 0. Let M , N, P, Q be on sides A E , BC, CD, D A respectively, where these excircles touch the square. Then OM I AB, O N I E C , O P I C D , and OQ I DA. Since

A S 11 CD and AD 11 BC, M, 0, P are collinear and N, 0, Q are collinear. Now M P = N Q = a. [2 marks!

Using the fact that the two tangents from a point to a circle have the same length, we get EF = EM

+

FQ

and GH = GN

+

HP. [I mark] Then

ml = A E

+

AF

+

EF = AE

+

AF + ( E M

+

FQ) = A M

+

AQ = OQ

+

OM

i

and Therefore mz = CG + C H

+

GH = C G + C H

+ (GN

M P ) = C N

+

CP = O P + O N . ml +m2 = (09 + O M )

+

( O P + O N ) = M P

+

N Q = 2a. [l mark] [I mark]

i

Solution 2 .

Extend A B to I and DC to J so that AE = BI = CJ. Let

I

intersect I J a t M , and let K lie on I J so that G K I I J . Then, since AE = G K , AAEF and A K G M are congruent. [I mark] Thus, since G K = C J

and GC = K J ,

ml

+

mz = perimeter(KGA4)

+

perimeter(CGH) = perimeter(XMJ). [2 marks]

Let L lie on C D so that E L I CD. Then a circle with centre E and radius a will touch DC at L, I J at I, and the interior of HM at some point N , so

perimeter(HA4.T) = J H

+

( H N

+

N M )

+

JM = (JH

+

HL)

+

( M I

+

J M ) = J L

+

I J = a

+

a = 2a.

[4 marks] Thus ml

+

mz = 2a.

Solution 3

Without loss of generality, assume the square has side a = 1. Let 0 be the acute angle between (1 (or fz) and the sides AB and CD of the square. Then, letting EF = z and G H = y, we have

E A =zcosO, AF = zsmS, CH = ycosS, CG = ysin8. Thus

Draw lines parallel t o fl, L2 through A and C respectively. The distance between these lines is sine

+

cose

[l mark], as can be seen by drawing a mutual perpendicular to these lies through B, say. Also, the altitudes

from A to EF and from C to GH have lengths zsinOcos0 and ysinecos0 respectively (1 mark]. Therefore

the distance between (1 and (2 must be

m l

+

mz = (z

+

y)(sinB+ cosS

+

1). [2 marks] (1)

(sin 0

+

cos 0)

-

z sine CM 0

-

y sin 0 cos 0

But we are given that this distance is a = 1, so

(Z

+

y) sin 0 cos 0

+

1 = sin 0

+

cos 8, or

sin .9

+

cos 0

-

1

.. . + .

. .

in G, say a , b and c, such that a, b are acquainted but a, c are not. Now consider the group A obtained by- removing a, b and c from G. A has t

-

3 = r(m - 1, n

-

1) people, so by the definition of r(m

-

1, n - l ) ,

A either contains 2(m

-

1) people forming m

-

1 mutually acquainted pairs, or eke contains 2(n - 1) people forming n

-

1 mutually unacquainted pain. In the former case, we add the acquainted pair a, b to A to form m

\

-3. Let k 2 14 be an integer, and let pk be the largest prime number which is strictly less than k. You may assume that Pk 2 3k/4. Let n be a composite integer. Prove:

(a) if n = 2pk, then n does not divide (n

-

k)! ;

(b) if n > 2pk, then n divides (n

-

k)! . Solution

(a) Note that n

-

k = 2 p k

-

k c 2pt

-

pk = p t , so p t ,," ( n

-

k)! , so 2pk ,," ( n

-

k)! . [1 mark]

(b) Note that n > 2Pk 2 3 k / 2 implies k < 2n/3, so n - k > n/3. So if we can find integers a, b 2 3 such that n = ab and a # b, then both a and b will appear separately in the product (n

-

k)! = 1 x 2 x . . . x (n

-

k),

which means nl(n - k)!

.

Observe that k 2 14 implies Pk 2 13, so that n > 2pk 2 26.

If n = 29 for some integer a 2 5 , then take a = 22, 6 = 2O-'. 11 mark] Otherwise, since n 2 26 > 16, we can take a to be an odd prime factor of n and b = n/a [I mark], unless b < 3 or b = a.

Case (i): b < 3. Since n is composite, this means b = 2, so that 2a = n > 2Pk. As a is a prime number and p k is the largest prime number which is strictly less than k, it follows that a 2 k. From n - k = 2a

-

k 2

2a

-

a = a > 2 we see that n = 2a divides into (n - k)! . [Z marks]

Caw (ii): b = a. Then n = a' and a > 6 since n 2 26. Thus n

-

k >'n/3 = a2/3 > 2a, so that both a and 2a appear among { 1 , 2 , . . . , n

-

k}. Hence n = az divides into (n

-

k)! . [Z marks]

4. Let a, b, c be the sides of a triangle, with a

+

b

+

c = 1, and let n 2 2 be an integer. Show that

Solution.

Without loss of generality, assume a 5 b 5 c. As a

+

b > c, we have

~ = ~ ( a + b + c ) > ~ ( c + c ) - ~ > 2 2 2

m.

[Zmarks] As a 5 c and n 2 2, we have Thus Likewise m < b + i . [lmark] ~ ' 5 a a V ? i ~ + ~ + ~ < - + c + - + b + - = I + - - . [lmark] 2 2 2 2 . .

5. Given two positive integers m and n , find the smallest positive integer k such that among any k people, either there are 2m of them who form m pairs of mutually acquainted people or there are 2n of them forming

n pairs of mutually unacquainted people. Solution.

Let the smallest positive integer k satisfying the condition of the problem be denoted r(m,n). We shall show that

r ( m , n ) = Z ( m C n ) - m i n { r n , n ) - 1

Observe that, by symmetry, r(m,n) = r(n,m). Therefore it suffices to consider the case where m 5 n, and to prove that

r(m, n ) = 2m

+

n

-

1. 11 mark]

.

(1) First we prove that

r(m,n) 2 2 m + n - 1

by an example. Call a c o u p of k people, every two of whom are mutually acquainted, a k-clique. Consider a set of 2m

+

n

-

2 people consisting of a (2m

-

1)-clique together with an additional n

-

1 people none of whom know anyone else. (Call such people isolated) Then there are not 2m people forming m mutually acquainted pairs, and there also are not 2n people forming n mutually unacquainted pairs. Thus r(m, n) 5

(2m - 1 ) i (n

-

1)

+

1 = 2m

+

n - 1 by the definition of r(m, n). (1 mark]