Mathematical Excalibur, Volume 10, Number 1

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Volume 10, Number 1 March 2005 – April 2005

例析數學競賽中的計數問題(二)

費振鵬（江蘇省鹽城市城區永豐中學 224054 )

Olympiad Corner

Following are the problems of 2004 Estonian IMO team selection contest. Problem 1. Let k > 1 be a fixed natural

number. Find all polynomials P(x) satisfying the condition P(xk) = (P(x))k for all real number x.

Problem 2. Let O be the circumcentre

of the acute triangle ABC and let lines

AO and BC intersect at a point K. On

sides AB and AC, points L and M are chosen such that KL = KB and KM = KC. Prove that segments LM and BC are parallel.

Problem 3. For which natural number n

is it possible to draw n line segments between vertices of a regular 2n-gon so that every vertex is an endpoint for exactly one segment and these segments have pairwise different lengths?

Problem 4. Denote

∑

= + − = m k k m k m f 1 . 1 2 cos ) 1 ( ) ( π

For which positive integers m is f(m) rational?

(continued on page 4)

Editors: ஻ Ի ஶ(CHEUNG Pak-Hong), Munsang College, HK

ଽ υ ࣻ (KO Tsz-Mei)

గ ႀ ᄸ (LEUNG Tat-Wing)

؃ ୊ ፱ (LI Kin-Yin), Dept. of Math., HKUST

֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU Artist: ྆ ؾ ़ (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 May 7, 2005.

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] 3 運用算兩次原理與抽屜原理算兩次原理，就是把一個量從兩個（或更多）方面去考慮它，然後綜合起來得到一個關係式（可以是等式或不等式），或者導出一個矛盾的結論．具體表示為三步：“一方面（利用一部分條件）……，另一方面（利用另一部分條件）……，綜合這兩個方面 ……＂．義大利數學家富比尼（Fubini）首先應用這個思想方法，因此今天我們也稱它為富比尼原理．在解這些問題時，要根據問題的特點選擇一個適當的量，再將這個量用兩（或幾）種不同的方法表達出來．抽屜原理，德國數學家狄利克雷（Dirichlet）提出．對於這個原理的具體解釋，想必很多同學早就知道了，在此不再贅述．例 6 有 26 個不同國家的集郵愛好者，想通過互相通信的方法交換各國最新發行的紀念郵票，為了使這 26 人每人都擁有這 26 個國家的一套最新紀念郵票，他們至少要通多少封信？解答不妨設這 26 個集郵愛好者中的某一個人為組長．一方面，對於組長，要接收到其他25 個國家的最新紀念郵票，必須從這25 個集郵愛好者的手中發出（不管他們是否直接發給組長），至少要通 25 封信；同樣地，其他 25 個集郵愛好者分別要接收到組長的一套紀念郵票，必須由組長發出（不管組長是否直接發給這25 個集郵愛好者），至少要通25 封信．總計至少要通 50 封信．另一方面，其餘25 個集郵愛好者每人將本國的一套最新的紀念郵票 25 份或26 份發給組長，計 25 封信；組長收到這25 封信後，再分別給這 25 個集郵愛好者各發去一封信，每封信中含有25 套郵票（發給某人的信中不含其本國的郵票）或26 套郵票（發給某人的信中包含其本國的郵票），計 25 封信．總計 50 封信．這就是說通 50 封信可以使這 26 人每人都擁有這 26 個國家的一套最新紀念郵票．因此他們至少要通50 封信．例7 從 1，2，3，…，1997 這 1997 個數中至多能選出多少個數，使得選出的數中沒有一個是另一個的 19 倍？解答因為 1997÷19=105…2，所以 106，107，…，1997 這 1892 個數中沒有一個是另一個的19 倍．又因106÷19=5…11，故 1，2，3，4， 5，106，107，…，1997 這 1897 個數中沒有一個是另一個的19 倍．另一方面，從（6，6×19），（7，7× 19），…，（105，105×19）這 100 對互異的數中最多可選出100 個數（每對中至多選1 個），即滿足題意的數至少剔除100 個數．綜上所述，從 1，2，3，…，1997 中至多選出1897 個數，使得選出的數中沒有一個是另一個的19 倍．例8 在正整數 1，2，3，…，1995， 1996，1997 裏，最多能選出多少數，使其中任意兩個數的和不能被這兩個數的差整除．解答在所選的數中，不能出現連續自然數、連續奇數或連續偶數，這是由於連續自然數之和必能被其差 1 整除；連續奇數或連續偶數之和是偶數，必能被其差2 整除．再考慮差值為3 的兩數，不能是 3 的倍數，否則其和仍是3 的倍數，必能被其差 3 整除；而選擇全是 3 除餘 1，或全是 3 除餘2 的數，注意到各自中任意兩數之和非3 的倍數，不能被其差 3 的倍

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Mathematical Excalibur, Vol. 10, No. 1, Mar. 05- Apr. 05 Page 2 數整除，滿足題意．另一方面，從 (1,2,3) ， (4,5, 6 )，…，(1993,1993,1995)，(1996,1997) 中，最多可選出 666 個(每組至多可選一個)，否則會出現連續自然數、連續奇數或連續偶數，而不滿足題意．又間隔4 的所有數的個數較上述滿足題意的所有數的個數少．綜上可知，1，4，7，…，1990， 1993，1996（666 個）或 2，5，8，…， 1991，1994，1997（666 個）均滿足題意．即最多可選出666 個，使其中任意兩數之和不能被這兩數之差整除．例9 設自然數n有以下性質：從 1， 2，…，n中任取 50 個不同的數，這 50 個數中必有兩個數之差等於 7，這樣的n最大的一個是多少？ 解答 n的最大值是 98．說明如下： ⑴一方面當自然數從 1，2，…，98 中任取50 個不同的數，必有兩個數之差等於7．這是因為：首先將自然數1，2，…，98 分成7 組：（1，2，3，4，5，6，7，8， 9，10，11，12，13，14），（15， 16，17，18，19，20，21，22，23， 24，25，26，27，28），…，（85， 86，87，88，89，90，91，92，93， 94，95，96，97，98）．考慮取出的數中不出現某兩個數之差等於7 的情形：由於每組中含有差為7 的兩數，故每組最多可取出 7 個數（即每組中屬於 7 的同一個剩餘類的兩個數只能取其中的任意一個）．並且如果在第1 組中取出了 m （m=1，2，…，14），那麼後面的 每組分別取出m+14n（n=1，2，…， 6），可使所取數中的任意兩個數之差都不是7．這樣從上述 7 組數中最多只能取出7×7=49 個數．根據抽屜原理，知從1，2，…， 98 中任取 50 個不同的數，必有兩個數之差等於7． ⑵另一方面當自然數從 1，2，…， 99 中任取 50 個不同的數，不能保證必有兩個數之差等於7．這是因為：首先將自然數1，2，…，99 分成 8 組：（1，2，3，4，5，6，7，8，9， 10，11，12，13，14），（15，16，17， 18，19，20，21，22，23，24，25，26， 27，28），…，（85，86，87，88，89， 90，91，92，93，94，95，96，97，98），（99）．比如，取出前7 組中每組的前 7 個數，第8 組的 99 這 50 個數，就不含有兩個數之差等於7．綜合⑴、⑵，可得n 的最大值是 98． 例10 某校組織了 20 次天文觀測活動，每次有5 名學生參加，任何 2 名學生都至多同時參加過一次觀測．證明：至少有21 名學生參加過這些觀測活動．證法1 （反證法）假設至多有 20 名學生參加過這些觀測活動．每次觀測活動中的 5 名學生中有 2 5 5 4 10_{2 1} C = ×_× = 個2 人小組，又由題意知 20 次觀測中 2 人小組各不相同，所以 20 次觀測中 2 人小組總共有個． 20 10 200× = 而另一方面，20 名學生中的 2 人小組最多有 2 20 20 19 190_{2 1} C = ×_× = 個．兩者自相矛盾．故至少有 21 名學生參加過這些觀測活動．稍作簡化，即可證明如下：證法2 （反證法）假設至多有 20 名學生參加過這些觀測活動．由題意知：⑴共有 20 次觀測；⑵ 最多有 2 20 2 5 19 C C = 次觀測．兩者自相矛盾．故至少有 21 名學生參加過這些觀測活動．對於低年級學生，還可作出如下證明：證法 3 設參加觀測活動次數最多的學生A參加了a次觀測，共有x名學生參加 過天文觀測活動．由於有 A 參加的每次觀測活動 中，除了A，其他學生各不相同（這 是因為任何 2 名學生都至多同時參加過一次觀測），故x≥ 4a+1．（I）另一方面，學生A 參加觀測的次 數不小於每名學生平均觀測次數．即 20 5 a x× ≥ ．（II） 綜合（I）、（II），得x 400 1 x + ≥ ，．從而x ≥ 21． 0 2 ₄₀₀ x − −x ≥ 即至少有21 名學生參加過這些觀測活動．例11 2n名選手參加象棋循環賽，每 一輪中每個選手與其他2n−1人各賽一場，勝得1 分，平各得1₂分，負得 0 分．證明：如果每個選手第一輪總分與第二輪總分至少相差n分，那麼 每個選手兩輪總分恰好相差n分． 證明令集A={第二輪總分>第一輪總分的人}，集B={第二輪總分<第一輪總分的人}，並且|A|=k，|B|=h， k+h=2n． 不妨設k ≥ n ≥ h．考慮 A 中選手 第二輪總分之和S（若 h ≥ n ≥ k，則 考慮B 中選手第一輪總分之和 T）． 另一方面，對於每輪 A 中選手和 B 中選手的 kh 場比賽中，所得總分之 和為kh，充其量全為 A 中選手取勝， 則S ≤ +kh．如 A 中選手第一輪總 分之和為 S’ ，那麼 S−S’ ≥ kn ， +kh−kn ≥ S−kn ≥ S’ ≥ ．從而得 h ≥ n，所以 n=h=k，並且以上不等式 均為等式． 2 k C 2 k C 2 k C 所以 A 中每個選手第二輪總分 恰比第一輪總分多n 分，B 中每個選 手第一輪總分恰比第二輪總分多 n 分．因此，原命題成立． (to be continued)

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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

submitting solutions is May 7, 2005.

Problem 221. (Due to Alfred Eckstein, Arad, Romania) The Fibonacci

sequence is defined by F0 = 1, F1 = 1 and Fn = Fn−1+Fn−2 for n ≥ 2. Prove that is divisible by F 3 1 3 3 2 7Fn+ − Fn − Fn+ n+3.

Problem 222. All vertices of a convex

quadrilateral ABCD lie on a circle ω. The rays AD, BC intersect in point K and the rays AB, DC intersect in point

L.

Prove that the circumcircle of triangle AKL is tangent to ω if and only if the circumcircle of triangle CKL is tangent to ω.

(Source: 2001-2002 Estonian Math

Olympiad, Final Round)

Problem 223. Let n ≥ 3 be an integer

and x be a real number such that the numbers x, x2_{and x}n_{have the same}

fractional parts. Prove that x is an integer.

Problem 224. (Due to Abderrahim Ouardini) Let a, b, c be the sides of

triangle ABC and I be the incenter of the triangle. Prove that 3 3 abc IC IB IA ⋅ ⋅ ≤

and determine when equality occurs.

Problem 225. A luminous point is in

space. Is it possible to prevent its luminosity with a finite number of disjoint spheres of the same size? (Source: 2003-2004 Iranian Math

Olympiad, Second Round)

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

Solutions

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

Problem 216. (Due to Alfred Eckstein, Arad, Romania) Solve the equation

6 2

4x −6x +2 2= 0.

Solution. Kwok Sze CHAI Charles

(HKU, Math Major, Year 1), CHAN Tsz

Lung, HUDREA Mihail (High School

“Tiberiu Popoviciu” Cluj-Napoca Romania), MA Hoi Sang (Shun Lee Catholic Secondary School, Form 5),

Achilleas P. PORFYRIADIS (American

College of Thessaloniki “Anatolia”, Thessaloniki, Greece), Anna Ying PUN (STFA Leung Kau Kui College, Form 6),

Badr SBAI (Morocco), TAM Yat Fung

(Valtorta College, Form 5), WANG Wei Hua and WONG Kwok Cheung (Carmel Alison Lam Foundation Secondary School, Form 6).

We have ₈_x6 ₋12 _x2 ₊ 4 2 ₌ 0. Let t = 2x2_{. We get} ). 2 2 )( 2 )( 2 ( ) 4 2 )( 2 ( ) 2 6 6 ( ) 2 ( 2 4 6 0 2 3 3 3 + − − = − + − = − − − = + − = t t t t t t t t t t Solving 2x2 ₌ ₂ _{and 2x}2 _{= −2} ₂ _, we get _x₌_±₁4₂_or_±_i4 ₂_.

Other commended solvers: CHAN Pak Woon (Wah Yan College, Kowloon, Form

7), Kin-Chit O (STFA Cheng Yu Tung Secondary School) and WONG Sze Wai (True Light Girls’ College, Form 4).

Problem 217. Prove that there exist

infinitely many positive integers which cannot be represented in the form

3 5 7 9

1 2 3 4 ,

11 5

x +x +x +x +x

where x1, x2, x3, x4, x5 are positive integers.

(Source: 2002 Belarussian Mathematical

Olympiad, Final Round)

Solution. Achilleas P. PORFYRIADIS

(American College of Thessaloniki “Anatolia”, Thessaloniki, Greece) and Tak Wai Alan

WONG (Markham, ON, Canada).

On the interval [1, n], if there is such an integer, then ]. [ , ], [ ], [ 1/11 5 5 / 1 2 3 / 1 1 n x n x n x ≤ ≤ _L ≤

So the number of integers in [1, n] of the required form is at most n1/3_n1/5_n1/7_n1/9_n1/11

= n3043/3465_{. Those not of the form is at}

least n − n3043/3465_{, which goes to infinity}

as n goes to infinity.

Problem 218. Let O and P be distinct

points on a plane. Let ABCD be a

parallelogram on the same plane such that its diagonals intersect at O. Suppose P is not on the reflection of line AB with respect to line CD. Let M and N be the midpoints of segments AP and BP respectively. Let Q be the intersection of lines MC and ND. Prove that P, Q, O are collinear and the point Q does not depend on the choice of parallelogram ABCD. (Source: 2004

National Math Olympiad in Slovenia, First Round)

Solution. HUDREA Mihail (High

School “Tiberiu Popoviciu” Cluj-Napoca Romania) and Achilleas P.

PORFYRIADIS (American College of

Thessaloniki “Anatolia”, Thessaloniki, reece). G A B C D O P M _N Q

Let G1 be the intersection of OP and MC.

Since OP and MC are medians of triangle APC, G1 is the centroid of

triangle APC. Hence OG1=⅓OP.

Similarly, let G2 be the intersection of

OP and ND. Since OP and ND are

medians of triangle BPD, G2 is the

centroid of triangle BPD. Hence

OG2=⅓OP. So G1=G2 and it is on both

MC and ND. Hence it is Q. This implies P, Q, O are collinear and Q is the unique

point such that OQ=⅓OP, which does not depend on the choice of the parallelogram ABCD.

Other commended solvers: CHAN Pak Woon (Wah Yan College,

Kowloon, Form 7) and CHAN Tsz

Lung, Anna Ying PUN (STFA Leung

Kau Kui College, Form 6) and WONG

Tsun Yu (St. Mark’s School, Form 5). Problem 219. (Due to Dorin Mărghidanu, Coleg. Nat. “A.I. Cuza”, Corabia, Romania) The sequences a0,a1,a2,… and b0,b1,b2,… are defined

as follows: a0,b0 > 0 and 1 1 , 2 n n n a a b + = + 1 1 2 n n n b b a + = + for n = 1,2,3,…. Prove that 2004 2004 max{a ,b }> 2005.

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Mathematical Excalibur, Vol. 10, No. 1, Mar. 05- Apr. 05 Page 4

O (STFA Cheng Yu Tung Secondary

School), Achilleas P. PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece) and

Anna Ying PUN (STFA Leung Kau Kui

College, Form 6). We have . 1 4 1 2 4 1 4 1 1 4 1 ) 2 1 )( 2 1 ( 0 0 0 1 1 1 1 1 1 + + + = = + + + = + + = + + =

∑

= − − − − + + n b a b a b a b a b a b a b a a b b a b a n i i i n n n n n n n n n n n n n n n n L Then 2005 2004 4 1 2 2004 4 1 }) , (max{ 0 0 0 0 0 0 0 0 2004 2004 2 2004 2004 = + ≥ + + > ⋅ ≥ b a b a b a b a b a b a

and the result follows.

Other commended solvers: HUDREA Mihail (High School “Tiberiu

Popoviciu” Cluj-Napoca Romania).

Problem 220. (Due to Cheng HAO, The Second High School Attached to Beijing Normal University) For i = 1,

2, …, n, and k ≥ 4, let Ai = (ai1, ai2, … , aik) with aij = 0 or 1 and every Ai has at

least 3 of the k coordinates equal 1. Define the distance between Ai and Aj

to be 1 | | . k im jm m a a = −

∑

If the distance between any Ai and Aj

(i ≠ j) is greater than 2, then prove that

n ≤ 2k–3 _{– 1.}

Solution.

Let |Ai−Aj| denote the distance between Ai and Aj. We add A0 = (0,…,0) to the n

Am’s. Then |Ai−Aj| ≥ 3 still holds for A0,

A1, …, An.

Next we put the coordinates of A0 to

An into a (n + 1) × k table with the

coordinates of Ai in the (i + 1)-st row.

Note if we take any of the k columns and switch all the 0’s to 1’s and 1’s to

0’s, then we get n + 1 new ordered k-tuples that still satisfy the condition |Ai−Aj| ≥ 3.

Thus, we may change A0 to any

combination with 0 or 1 coordinates. Then the problem is equivalent to showing

n + 1 ≤ 2k−3_{for n + 1 sets satisfying |A} i−Aj|

≥ 3, but removing the condition each Ai

has at least 3 coordinates equal 1.

For k = 4, we have n + 1 ≤ 2. Next, suppose k > 4 and the inequality is true for the case k−1.

In column k of the table, there are at least [(n + 2)/2] of the numbers which are the same (all 0’s or all 1’s). Next we keep only [(n+2)/2] rows whose k-th coordinates are the same and we remove column k. The condition |Ai−Aj| ≥ 3 still

holds for these new ordered (k−1)-tuples. By the case k − 1, we get [(n + 2)/2] + 1 ≤ 2k−4_{. Since (n + 1)/2 < [(n + 2)/2] + 1, we}

get n + 1≤ 2k−3_{and case k is true.}

Generalization of Problem 203

Naoki Sato

We prove the following generalization of problem 203:

Let a1, a2, …, an be real numbers, and let si

be the sum of the products of the ai taken i

at a time. If s1 ≠ 0, then the equation

0 2 2 2 1 1xn− + s xn− + +nsn = s _L

has only real roots.

Proof. Let . 2 ) ( 2 2 1 1xn s xn nsn s x f = − + − + + L

We can assume that none of the ai are equal

to 0, for if some of the ai are equal to 0, then

rearrange them so that a1, a2, …, ak are

nonzero and ak+1, ak+2, …, an are 0. Then sk+1 = sk+2 = … = sn = 0, so n n n _s _x _ns x s x f = − + − + + L 2 2 1 1 2 ) ( k n k n n _s _x _ks _x x s − ₊ − ₊ ₊ − = 2 L 2 1 1 2 ). 2 ( 2 2 1 1 k k k k n _s _x _s _x _ks x + + + = − − − L

Thus, the problem reduces to proving the same result on the numbers a1, a2, …, ak.

Let g(x) = (a1x+1)(a2x+1)…(anx+1). The

roots of g(x) = 0 are clearly real, namely −1/a1, −1/a2, …, −1/an. We claim that the

roots of g' (x)=0 are all real.

Suppose the roots of g(x) = 0 are distinct. Let r1 < r2< … <rn be these roots. Then

by Rolle’s theorem, the equation g' (x) = 0 has a root in each of the intervals (r1,r2), (r2,r3), …, (rn−1,rn), so it has n − 1

real roots.

Now, suppose the equation g(x) = 0 has j distinct roots r1 < r2 < … < rj, and root ri

has multiplicity mi so m1+m2+…+mj = n.

Then ri is a root of the equation g' (x) = 0

having multiplicity mi−1. In addition,

again by Rolle’s theorem, the equation has a root in each of the interval (r1,r2),

(r2,r3), …, (rj−1,rj), so the equation g' (x)

= 0 has the requisite

(m1−1)+(m2−1)+…+ (mj−1) + j−1 = n−1

real roots.

Expanding, we have that

g(x) = (a1x+1)(a2x+1)…(anx+1)

= snxn + sn−1xn−1 + … + 1,

So g' (x) = nsnxn−1+ (n−1)sn−1xn−2 +…+s1.

Since s1 ≠ 0, 0 is not a root of g' (x) = 0.

Finally, we get that the polynomial n n n n _s_x _s _x _ns x g x − ′ = − + − + + L 1 2 1 1 1 ₍1₎ ₂

has all real roots.

Olympiad Corner

(continued from page 1)

Problem 5. Find all natural numbers n

for which the number of all positive divisors of the number lcm(1,2,…,n) is equal to 2k_{for some non-negative}

integer k.

Problem 6.

Call a convex polyhedron a footballoid if it has the following properties.

(1) Any face is either a regular pentagon or a regular hexagon.

(2) All neighbours of a pentagonal face are hexagonal (a neighbour of a face is a face that has a common edge with it).

Find all possibilities for the number of a pentagonal and hexagonal faces of a footballoid.

(5)

t k > 1 be a fixed natural number. Find all polynomials P ( z ) satisfymg the on ~ ( x ' ) = ( ~ ( z ) ) ' for all real numbers z.

Answer; P(z) = 0 and P(z) = z", where n is an arbitrary non-negative integer; in the case of odd k also P(x) = -x".

0.

Let i be the largest index smaller than n for which ai

#

0 (suppose that such an index i exists), then

ai-1zi-'+

...

+

a l z +no)'

.

Find next the coefficient of the term z('-')~+~ in both polynomials. As i < n, we get

kn > (k - l)n

+

i > ki, and the coefficient of this term in the polynomial P ( z k ) is

therefore 0. On the other hand, we get the term a,,z" in (P(z))' iff we take the term

a,,z" from k

-

1 factors and aizi from one factor; therefore in the polynomial (P(z))', the coefficient of this term is kak-'ai

#

0. This contradiction shows that there is no such index i and the polynomial P ( z ) has the form P ( z ) = a,,z". Also notice that if the degree n of the polynomial P ( z ) is 0, then also P ( z ) =

From the equality ~ ( z ' ) = (~(z))', we now get a,,xn' = at." for real z, i.e. a,, = a: or an(&' -1) = 0. Hence a,, E {-l,O, 1) for odd k (a,, = Oispossible only if n = 0) and

a,, E {0,1} for even k.

= a,,z".

t 0 be the circumcentre of the acute triangle ABC and let lines A 0 and BC inter-

sect at point K . On sides AB and AC, points Land M are chosen such that IKLI = IKBI

and IKMI = IKCl. Prove that segments LM and BC are parallel.

Solution. Draw heights for triangles K B L and K C M from the vertex K and let theh bases be S and T, respectively.

Also lengthen the segment AK until it intersects the circumcircle of ABC at point P (see Figure 13). As segment

AP is a diameter of the circumcirde of ABPC, the trian-

gles ABP and ACP are right-angled. Triangle ASK is

similar to triangle ABP and triangle ATK is similar to triangle ACP (their corresponding sides are parallel), so

lASl

-

lAKl

-

lATl

~ m e triangle AST is similar to

lABl (API

-

m'

triangle ABC and therefore ST

11

BC. As lLSl = (SBI and

lMTl= [TCl, we obtain LM

11

BC.

.

Figure 13

r which natural number n is it possible to draw n line segments between vertices of a regular 2n-gon so that every vertex is an endpoint for exactly one segment and these segments have pairwise different lengths?

Solution. Colour the vertices of the 2n-gon alternately black and white. Call the smallest number of sides needed to pass when moving from one vertex to another the wdghf of the segment with endpoints at these vertices. We see that segments with odd weigths connect vertices with different colours, but segments with even weights connect vertices with the same colour.

Suppose that the required construction exists for a given n. As there is an equal number of vertices of both colours and all segments with odd weights occupy an equal number of vertices of both colours, also all segments with even weights must take occupy an equal number of vertices of both colours. Therefore the number of segments connecting

two white vertices equals the number of segments connecting two black vertices, and the number of segments with even weights is an even number. Therefore there must be an even number of even numbers among numb&s 1 , 2 , .

...

n, which is possible only if

n

=

0 (mod 4) or n

=

1 (mod 4).

We show next that such sets exist for n = 4k and n = 4k

+

1. In the following tables, the segments are grouped into blocks of parallel segments. In each TOW of a block, (z, v )

shows that a segment is drawn between vertices z and y, next comes the weight of this

segment and, after the end of a block, the number of segments in this block is shown.

Construction for n = 4k: (0,4k) 4k } 1 (2k - 1,2k) (6k - 1,Gk

+

1) ( 2 k - 3 , 2 k + 2 )

:).

( 6 k - - 3 , 6 k + 3 )

(6)

For which positive integers m is j ( m ) rational?

Answer: for all positive integers m.

Solution Fix a positive integer m arbitrarily and take

a = cos- U # 0 . Using the formula

2(2m

+

1) _,

we get the chain of equations coso cos y = L(cos(z 2 - y)

+

cos(z

+

y)) ,

(2k - 1 ) ~ (2k

+

.)1 = - C ( - l ) k ( c o s 1 " +Cos--) = 2(2m

+

1) 2(2m

+

1) k=I 2 = A(-cos r + (2m + 11,) = 2 2 ( 2 m + l ) 2(2m

+

1) 1 = '(-a+ (-1)m.o) = - - a . 2 2 1 2

Hence f(m) =

--

for all m.

d all natural numbers n for which the number of all' positive divisors of the number I a n (1,2,

. .

.

, n) i s equal to 2' for some non-negative integer k.

Answer; 1,2,3 and 8.

Solution. Let P be the set of all prime numbers. Let b(m) denote the number of positive divisors of natural number m and let A(n) = d(lm (1,.

. .

,

n)). Denote by p o m the exponent of prime number p in the canonical representation of m. Notice that

A(n) = d(lm (1,.

.

. , n)) = =bl,...@n)) ,=

PEP

Therefore A(n) is a power of 2 iff all numbers in the form [log,, n]

+

1, where p E P, are powers of 2. Let [log, nJ

+

1 = 2k and [log, nJ

+

1 = 2'. As log, n 2 log, n, we get k 2 1.

Consider two cases. If k = I then

this holds for n = 1 and n = 3. With immediate check we 8ee that, for n = 2 and

4 5 n < 8, the equation (14) does not hold. If n 2 8 then

4 +

1 , (15)

this holds for n = 2 and n = 8. With immediate check we see that for 4 5 n 5 7 and

9 5 n < 27 the inequality (15) does not hold. If n 2 27 then

log,n-log227= log,3(log,n-log327) < 2(log3n-3) = 2log,n-6,

hence 2 log, n - log, n > 6

-

log, 27 > 6

-

5 = 1. Therefore [log2 n] < 12 log, n] 5

2[iog,nJ +1foralln~27.Hence(15)holdsiffn=2orn=8.

Therefore the only possible values for n are 1,2,3 and 8. For them, we get A(l) = 1 = 2',

4 2 ) = 2 = 2*, 4 3 ) = 4 = 2' and A(8) = 32 = 25. Hence all these values suit. a convex polyhedron a footbulloid if it has the follaving properties.

(1) Any face is either a regular pentagon or a regular hexagon.

(2) All neighbours of a pentagonal face are hexagonal (a neighbour of a face is a face that Find all possibilities for the number of pentagonal and hexagonal faces of a footballoid.

Answer: there are 12 pentagonal and 20 hexagonal faces.

Solution. We show first that there exists a footballoid with 12 pentaganal and 20 hexag- onal faces. Start with a regular icosahedron and abstract from every vertex a regular pyramid with lateral edge

-

of the edge of the icosahedron In such a way, we get 12 regular pentagons instead of 12 vertices of icosahedron and w e get 20 regular hexagons instead of 20 faces of icosahedron. All neighbours of any pentagonal face are hexagonal.

Now show that it is the only possibility. Let B be a footballoid. Consider an arbitrary vertex of B; let it belong to z pentagonal and y hexagonal faces. Then z

+

y 2 3 as every

vertex of a polyhedron belongs to at least 3 faces. As the sizes of the interior angles of pentagonal and hexagonal faces are 108" and 120", respectively, we getz~108"+~-120" < 360". Hence x

+

y 5 3 and x > 0. Therefore z

+

y = 3, which means that every vertex of a footballoid belongs to exactly 3 faces, at least one of which is pentagonal. As these 3 faces are painvise neighbours and pentagonal faces cannot be neighbours, every vertex must belong to exactly one pentagonal and two hexagonal faces.

Consider an arbitrary hexagonal face. All its vertices belong to one pentagonal and one hexagonal face. Therefore the neighbours of a hexagonal face are alternately pentagonal and hexagonal, so there are exactly three of both kinds.

Now cover every pentagonal face with a regular pentagonal pyramid, whose lateral edges are continuations of the (hexagonal) neighbours of this face. In this way, hexagonal faces become equilateral triangles and pentagonal faces are replaced with vertices in which five edges meet. As any two neighbours of a pentagonal face which are neigh- bowing themselves meet under the Same angle (two regular hexagons and one regular pentagon can meet in one vertex in principle in only one way), also the triangles meet- ing in a vertex of our new polyhedron meet under the same angle. Therefore the new polyhedron is a regular icosahedron. It has 12 vertices and 20 faces, so the footballoid B had to have 12 pentagonal and 20 hexagonal faces.

has a common edge with it).

1