Olympiad Corner
Tenth Asian Pacific Mathematics Olympiad, March, 1998:
Each question is worth 7 points.
Problem 1. Let F be the set of all n-tuples (A1, A2,..., An) where each Ai, i =
1, 2, ..., n is a subset of {1,2,...,1998}. Let |A| denote the number of elements of the set A. Find the number
(
∑
) ∪ ∪ n A A A n A A A , , , 2 1 2 1 K K .Problem 2. Show that for any positive
integers a and b, (36a+b) (a+36b) cannot
be a power of 2.
Problem 3. Let a, b, c be positive real
numbers. Prove that
⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + + ≥ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 3 1 2 1 1 1 abc c b a a c c b b a
Problem 4. Let ABC be a triangle and D
the foot of the altitude from A. Let E and F be on a line passing through D such
that AE is perpendicular to BE, AF is
perpendicular to CF, and E and F are
different from D. Let M and N be the
midpoints of the line segments BC and EF, respectively. Prove that AN is
perpendicular to NM.
(continued on page 4)
In topology, there are many abstractions of geometrical ideas, such as continuity and closeness. ‘Topology’ is derived from the Greek words τοποσ , a place and λογοσ, a discourse. It was introduced in 1847 by Johann Benedict Listing (1808-1882), who was a student of Carl Friedrich Gauss (1777-1855). In the early days, people called it analysis situs, that is, analysis of
position. Rubber-sheet geometry is a rather descritpive term to say what it is. (Just think of properties of objects drawn on a sheet of rubber which are not changed when the sheet is being distorted.) Hence, topologists could not distinguish a triangle from a rectangle and they may even consider a basketball as a ping-pong ball. Topologists consider two objects to be the same (homeomorphic) if one can be continuously deformed to look like the other. Continuous deformations include bending, stretching and squashing without gluing or tearing points.
Example 1. The following are
homeomorphic: (See Figure 1.)
Example 2. The following are
non-homeomorphic: (See Figure 2.)
In practise, continuous deformations may not be easy to carry out. In fact, there is a simple method to see two objects are non-homeomorphic, by seeking their Poincaré-Euler characteristics, (in short, Euler numbers). In order to see what the Euler number is, we need to introduce the concept of subdivision on an n-manifold
(here n≤2throughout). (An n-manifold
is roughly an n dimensional object in
which each point has a neighborhood homeomorphic to an open interval (if n = 1)
or an open disk (if n = 2). For example, a
circle is a 1-manifold and a sphere is a 2-manifold.)
Basically, we start with an n manifold and
subdividing it into a finite number of vertices, edges and faces. A vertex is a point. An edge is a curve with endpoints that are vertices. A face is a region with boundary that are edges.
Here are typical pictures of vertex, edge and face, (see Figure 3.)
The Euler number (χ) of a compact
(loosely speaking, bounded) 1-manifold is defined to be the number of vertices(v)
minus the number of edges(e), and for a
compact 2-manifold (surface), it is defined to be the number of vertices(v) minus the
number of edges (e) plus the number of
Volume 4, Number 2 March, 1998 - December, 1998
A Taste of Topology
Wing-Sum Chan
Beauty is the first test: there is no permanent place in the world for ugly mathematics.
(G. H. Hardy )
Editors: Ի ஶ (CHEUNG Pak-Hong), Curr. Studies, HKU
ଽ υ ࣻ (KO Tsz-Mei), EEE Dept, HKUST
గ ႀ ᄸ (LEUNG Tat-Wing), Appl. Math Dept, HKPU
፱ (LI Kin-Yin), Math Dept, HKUST
֔ ᜢ ݰ (NG Keng Po Roger), ITC, HKPU Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU 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 December 31, 1998.
For individual subscription for the three remaining issues for the 98-99 academic year, send us three 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
faces (f) (see Figure 4.) The following
theorem is a test to distinguish non-homeomorphic objects.
Theorem 1. If two n-manifolds are homeomorphic, then they have the same Euler number.
So figure 4 and theorem 1 imply the sphere and the torus are not homeomorphic, i.e. the sphere cannot be continuously deformed to look like the torus and vice versa.
Here are two terms we need before we can state the next theorem. A connected manifold is one where any two points on the manifold can be connected by a curve on the manifold. The manifold is orientable if it has 2 sides, an inside and an outside.
Theorem 2. Two connected orientable n-manifolds (n≤2) with the same number of boundary components are homeomorphic if and only if they have the same Euler number.
Here are some important results that tell us the general pictures of one and two manifolds.
Classification I. Any connected compact one-manifold is either homeomorphic to an open interval or a circle.
Classification II. Any connected, orientable and compact two-manifolds is homeomorphic to one of the followings:
(see Figure 5.)
Finally, we mention a famous open problem (the Poincaré conjecture), which is to show that every compact, simply connected three-manifold is homeomorphic to a three-sphere, where simply connected means any circle on the manifold can be
shrunk to a point on the manifold.
有一類關於求陰影部分面積的問題，我 們可根據題意適當設元，通過一次方程 組求得結果。這種 數形結合，將幾何 面 積 問 題 轉 化 為 解 一 次 方 程 組 代 數問題的方法，由 於方法新穎、思路 清晰，因而頗受師 生重視。現舉三例 分析說明如下： 一﹒列二元一次方程組求陰影面積 例一：如上圖，O 為正三角形 ABC 的中 心，AB=8 3cm，則AOB、BOC、 COA所圍成的陰影部分的面積是ˍˍ 2 cm 。 (1996 年陝西省中考題) 分析：上圖中含有形狀不同的兩類圖 形，分別為 x 和 y，由圓形特徵得知，2 個 x 和 1 個 y 組成一個圓心角為120°的 弓形，而 3 個 x 和 3 個 y 組成一個正三 角形 ABC。由於正三角形 ABC 的高 12 3 8 2 3× = = ，又 O 為正三角形 ABC 的中心，故BO=32×12=9=MB。 MBC MBC S S y x+ = − ∆ ∴2 扇形 3 16 3 64 4 3 8 2 1 360 8 120 2 − π = × × − ⋅ π = 3 48 12 3 8 2 1× × = = 解下列方程組 ⎪⎩ ⎪ ⎨ ⎧ = + − π = + ) 2 ( 3 48 3 3 ) 1 ( 3 16 3 64 2 L L L L L L y x y x ，得3x=64π−93 3。這就是所求陰影 部分的面積。 二﹒列三元一次方程組求陰影面積 例 2：如下圖，在正方形 ABCD 中，有 一個以正方形的中心 為圓心，以邊長一半 為半徑的圓。另分別 以 A、B、C、D 為圓 心，以邊長一半為半 徑畫四條弧。若正方形的邊長為 2a，求 所圍成的陰影部分的面積。（1997 年泰 州市中考模擬題） 分析：圖中含有形狀不同三類圖形，分 別為 x、y、z。由圖形特徵得知：4 個 x 和 1 個 y 組成一個圓；1 個 x 和 1 個 z 組成一個以 a 為半徑、圓心角為直角的 的扇形；4 個 x、4 個 z 和 1 個 y 組成一 個正方形。 故此，可列出方程組 ⎪ ⎩ ⎪ ⎨ ⎧ = + + π = + π = + ) 3 ( 4 4 4 ) 2 ( ) 1 ( 4 2 2 4 1 2 L L L L L L L L L a z y x a z x a y x ) 1 ( ) 3 ( − 得 2 4 1(4 )a z= −π 再代入(2)得 2 2 1) ( a x= π− 。 2 ) 4 2 ( 4x a S = = π− ∴ 陰影 。 三﹒列四元一次方程組求陰影面積 例 3：如左圖，菱形 ABCD 的兩條對角 線長分別為 a、b，分別以每邊為直徑向 形 內 作 半圖。求 4 條 半 圓 弧 圍 成 的 花 瓣 形 面 積（陰影 部分的面積）。（人教版九年義務教材 初中《幾何》第三冊 P. 212） 分析：圖中含有形狀不同的四類圖形， 分別為 x、y、z、u，則由圖形特徵得知： 2x、2y、z、u 組成一個以邊長為直徑的 半圓；x、z、u 組成直角三角形 BOC。 解:設 x、y、z、u 如圖所示，則依題意 得 ⎪ ⎪ ⎩ ⎪⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = + + + ⋅ ⋅ = + + + (2) 2 1 2 1 2 2 ) 1 ( 2 2 2 1 2 4 2 2 L L b a u y z x b a u z x π (2) – (1) 再乘 4 得 （續於第四頁）
巧列一次方程組 妙解陰影面積題
于志洪 江蘇泰州橡膠總廠中學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 Dec 31, 1998.
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)
Problem 77. Show that if
∆
ABCsatisfies , 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)
Problem 78. If c1, c2,…, cn (n ≥2) are real numbers such that
, ) ... ( ) ... )( 1 ( 2 2 1 2 2 2 2 1 n n c c c c c c n + + + = + + + −
show that either all of them are nonnegative or all of them are nonpositive. (Source: 1977 unused IMO
problem proposed by Czechoslovakia)
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)
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)
*****************
Solutions
*****************Problem 71. Find all real solutions of
the system . ) 1 log( , ) 1 log( , ) 1 log( 2 2 2 x z z z z y y y y x x x = + + + = + + + = + + +
(Source: 1995 Israel Math Olympiad) Solution: CHOI Fun Ieng (Pooi To
Middle School (Macau), Form 5).
If x < 0, then 0<x+ x2+1<1. So
0 ) 1
log(x+ x2+ < , which implies y < x
< 0. Similarly, we get z < y < 0 and x < z <
0, yielding the contradiction x < z < y < x.
If x > 0, then x+ x2+1>1 . So
0 ) 1
log(x+ x2+ > , which implies y > x
> 0. Similarly, we get z > y > 0 and x > z >
0, yielding the contradiction x > z > y > x.
If x = 0, then x = y = z = 0 is the only
solution.
Other commended solvers: AU Cheuk Yin (Ming Kei College, Form 5). CHEUNG Kwok Koon (HKUST), CHING Wai Hung (STFA Leung Kau
Kui College, Form 6), HO Chung Yu (Ming Kei College, Form 6), KEE Wing
Tao Wilton (PLK Centenary Li Shiu
Chung Memorial College, Form 6), KU
Wah Kwan (Heep Woh College, Form 7), KWOK Chi Hang (Valtorta College,
Form 6), LAM Yee (Valtorta College, Form 6), LAW Ka Ho (Queen Elizabeth School, Form 5), Gary NG Ka Wing (STFA Leung Kau Kui College, Form 5),
TAM Siu Lung (Queen Elizabeth School,
Form 5), WONG Chi Man (Valtoria College, Form 3) and WONG Hau Lun (STFA Leung Kau Kui College, Form 6).
Problem 72. Is it possible to write the
numbers 1,2,…,121 in an 11x11 table so that any two consecutive numbers be written in cells with a common side and all perfect squares lie in a single column? (Source: 1995 Russian Math Olympiad)
Solution: Gary NG Ka Wing (STFA
Leung Kau Kui College, Form 5). Suppose such a table exists. The table would be divided into 2 parts by the single column of perfect squares, with one side 11n (0≤ n≤5) cells and the other side 110 - 11n cells. Note that
numbers between 2 successive perfect squares, say a2, (a+1)2, lie on one side since they cannot cross over the perfect
square column, and those between
2
) 1
(a+ , (a+2)2 lie on opposite side. Now the number of integers (strictly) between 1, 4, 9, 16, K, 100, 121 is 2, 4, 6, 8, K, 20, respectively. So one side has 2 + 6 + 10 + 14 + 18 = 50 numbers while the other side has 4 + 8 + 12 + 16 + 20 = 60 numbers. Both 50 and 60 are not multiple of 11, a contradiction.
Other commended solvers: CHEUNG Kwok Koon (HKUST), HO Chung Yu
(Ming Kei College, Form 6), LAI Chi
Fung Brian (Queen Elizabeth School,
Form 4), LAW Ka Ho (Queen Elizabeth School, Form 5), TAM Siu Lung (Queen Elizabeth School, Form 5),
WONG Hau Lun (STFA Leung Kau
Kui College, Form 6) and WONG Shu
Fai (Valtorta College, Form 6).
Problem 73. Prove that if a and b are
rational numbers satisfying the equation
2 2 5 5 2a b b a + = , then 1−ab is the
square of a rational number. (Source: 26th British Math Olympiad)
Solution: CHAN Wing Sum (City U). If b=0, then 1− ab=12. If b≠0, then 2 3 5 6 2a b ab a + = . So a6−2a3b2+b4 ) 1 ( 4 5 4 ab b ab b − = − = . Therefore, 1−ab 4 4 2 3 6 ) 2 (a − a b +b b = is the square of
the rational number (a3−b2) b2.
Other recommended solvers: CHING Wai Hung (STFA Leung Kau Kui College, Form 6), CHOI Fun Ieng (Pooi To Middle School (Macau), Form 5), KU
Wah Kwan (Heep Woh College, Form 7) and Gary NG Ka Wing (STFA Leung Kau Kui College, Form 5).
Problem 74. Points A2, B2, C2 are the
midpoints of the altitudes AA1, BB1, CC1 of
acute triangle ABC, respectively. Find the sum of ∠B2A1C2 , ∠C2B1A2 and
2 1 2C B
A
∠ . (Source: 1995 Russian Math
Olympiad)
Solution: LAM Po Leung (Ming Kei College, Form 5)
Let A3, B3, C3be the midpoints of BC,
CA, AB, respectively, and H be the
orthocenter of ABC∆ . Since C3A3 is
parallel to AC, so ∠HB2A3 =90° 3 1A HA ∠ = , which implies H, B2, A3, A1 are concyclic. So ∠B2A1H =∠B2A3H.
Since B3A3 is parallel to AB, so
（續第二頁） 8 4 ) ( 4 2 2 ab b a z y x+ + = π +π − 。 這就是所求陰影部分的面積。 綜上所述可知：一般的陰影圖形大多 是由多種規則圖形組成的，所以利用 方程式組解決這類問題時，首先要根 據圖形的特徵（尤其是對稱性）把圖 形分成幾類，用字母表示各類圖形的 面積；其次要仔細觀察圖形的組成， 分析圖形中各部分之間及各部分與整 體圖形的關係，通過規則圖形面積公 式列出方程組；最後解方程求出陰影 面積。 附練習題 １﹒如右圖，已知一塊正方形的地瓷 磚邊長為 a，瓷磚上 的 圖 案 是 以 各 邊 為 直 徑 在 正 方 形 內 畫 圓所圍成的（陰影部 分），那麼陰影部分 的 面 積 是 多 少 ？ （1997 年寮夏回族自治區中考題） ２﹒如右圖，已知圓形 O 的半徑為 R，求圖中 陰 影 部 分 的 面 積 。 （1998 年泰州市中考 模擬題） ３﹒如右圖，正方形的邊長為 a，分別 以 正 方 形 的 四 個 頂點為圓心，邊長 為半徑，在正方形 內畫弧，那麼這四 條 弧 所 圍 成 的 陰 影 部 分 的 面 積 是 多少？（1994 年安 徽省中考題） ４﹒ 如右圖，圓 O 內切於邊長為 a 的 正 三 角 形，分別以 三 角 形 的 三 頂 點 為 圓心，2a為 半 徑 畫 弧 相交成圓中所示的陰影，求陰影部分 的面積。（1996 年泰州市中考模擬題） 參考資料 1. 《陰影部分面積的幾種解法》 《初中生數學園地》 安義人(華南師大主辦) 1997 年 3 月 2. 《列一次方程組解陰影面積題》 《中小學數學》 于志洪(中國教育學會主辦)1997年11 月 練習題答案: 1. (2π−1)a2 2. 2πR2−3 3R2 3. (1− 3+3π)a2 4. 5π−246 3a2 3 1 3 2A 90 HAA HC = °=∠ ∠ , which
implies H, C2, A3, A1 are concyclic. So
H A C H A C2 1 =∠ 2 3 ∠ . Then ∠B2A1C2 H A C H A B H A C H A B2 1 +∠ 2 1 =∠ 2 3 +∠ 2 3 ∠ = BAC B A C =∠ ∠ = 3 3 3 (because 3 3 3BC A ∆ is similar to ∆ABC ).
Similarly, ∠B2C1A2 =∠BCA and
ABC C
B
A =∠
∠ 2 1 2 . Therefore, the sum
of ∠B2A1C2 , ∠C2B1A2 , ∠A2B1C2 is
180o.
Other commended solvers: HO Chung Yu (Ming Kei College, Form 6).
Problem 75. Let P(x) be any polynomial with integer coefficients such that P(21) = 17, P(32) = -247,
P(37) = 33. Prove that if P(N) = N + 51
for some integer N, then N = 26. (Source: 23rd British Math Olympiad)
Solutions: HO Chung Yu (Ming Kei College, Form 6).
If P(N) = N + 51 for some integer N, then P(x) – x – 51 = (x - N)Q(x) for some polynomial Q(x) by the factor theorem. Note Q(x) has integer coefficients because P(x) – x – 51 = P(x) – P(N) – (x – N) is a sum of ai(xi−Ni) terms (with ai’s integer). Since Q(21) and
Q(37) are integers, P(21) – 21 – 51 =
-55 is divisible by 21 – N and P(3) – 37 – 51 = -55 is divisible by 37 – N is 16, we must have N = 26 or 32. However, if
N = 32, then we get -247 = P(32) = 32 +
51, a contradiction. Therefore N = 26.
Other commended solvers: CHEUNG Kwok Koon (HKUST), KU Wah Kwan (Heep Who College, Form 7), TAM Siu
Lung (Queen Elizabeth School, Form 5) and WONG Shu Fai (Valtorta College, Form 6).
Olympiad Corner
(continued from page 1)
Problem 5. Determine the largest of all integers n with the property that n is divisible by all positive integers that are less than 3n.
(Hong Kong team to IMO 98: (from left to right) Lau Wai Tong (Deputy Leader), Law Ka Ho, Chan Kin Hang, Choi Ming Cheung, Lau Lap Ming, Cheung Pok Man, Leung Wing Chung, Liu Kam Moon (Leader).)
