PDF《对数表的构造》

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for all x. Therefore, the equation has no solution. Other commended solvers: Antonio LEI (Colchester Royal Grammar School, UK, Year 12), LEUNG Wai Ying (Queen Elizabeth School, Form 7), OR Kin (HKUST, Year 1) and SIU Tsz Hang (STFA Leung Kau Kui College, Form 6). Problem 144. (Proposed by José Luis Díaz-Barrero, Universitat Politècnica de Catalunya, Barcelona, Spain) Find all (non-degenerate) triangles ABC with consecutive integer sides a, b, c and such that . 2A C = Solution. CHAO Khek Lun Harold (St. Paul'

s College, Form 7), CHUNG Tat Chi (Queen Elizabeth School, Form 5), KWOK Tik Chun (STFA Leung Kau Kui College, Form 4), LAM Wai Pui Billy (STFA Leung Mathematical Excalibur Mathematical Excalibur Mathematical Excalibur Mathematical Excalibur, Vol. 7, No. 1, Mar 02- Apr

02 Page

4 Kau Kui College, Form 4), Antonio LEI (Colchester Royal Grammar School, UK, Year 12), LEUNG Wai Ying (Queen Elizabeth School, Form 7), POON Ming Fung (STFA Leung Kau Kui College, Form 4), WONG Chun Ho (STFA Leung Kau Kui College, Form 7), WONG Tsz Wai (Hong Kong Chinese Women'

s Club College, Form 6) and YEUNG Wing Fung (STFA Leung Kau Kui College). Let a=BC, b=CA, c=AB. By sine and cosine laws, . cos

2 sin sin

2 2

2 bc a c b A A C a c ? + = = = This gives .

3 2

2 2 a ac ab bc ? + = Factoring, we get ? ? ?

2 2 )( ( a c b a ab) = 0. Since the sides are consecutive integers and A C >

implies , a c >

we have (a, b, c) = (n, n C 1, n + 1), (n C 1, n + 1, n) or (n C 1, n, n + 1) for some positive integer .

1 >

n Putting these into ,

0 2

2 = ? ? ab a c the first case leads to ,

0 1

3 2 = + + ? n n which has no integer solution. The second case leads to ,

0 2

2 = ? n n which yields a degenerate triangle with sides 1, 2, 3. The last case leads to ,

0 5

2 = ? n n which gives (a, b, c) = (4, 5, 6). Other commended solvers: CHENG Ka Wai (STFALeung Kau Kui College, Form 4), Clark CHONG Fan Fei (Queen'

s College, Form 5), SIU Tsz Hang (STFA Leung Kau Kui College, Form6), WONG Chun Ho (STFA Leung Kau Kui College, Form 7) and WONG Wing Hong (La Salle College, Form 4). Problem 145. Determine all natural numbers k >

1 such that, for some distinct natural numbers m and n, the numbers m k +

1 and + n k

1 can be obtained from each other by reversing the order of the digits in their decimal representations. (Source:

1992 CIS Math Olympiad) Solution. CHAO Khek Lun Harold (St. Paul'

s College, Form 7), LEUNG Wai Ying (Queen Elizabeth School, Form 7), Ricky TANG (La Salle College, Form 4) and WONG Tsz Wai (Hong Kong Chinese Women'

s Club College, Form 6). Without loss of generality, suppose such numbers exist and . m n >

By the required property, both numbers are not power of 10. So n k and m k have the same number of digits. Then

10 >

. k k k k m n m n ≥ = ? Since every number and the sum of its digits are congruent (mod 9), we get

1 1 + ≡ + m n k k (mod 9). Then )

1 ( ? = ? ?m n m m n k k k k is divisible by 9. Since the two factors are relatively prime, k >

10 and m n k ? >

9 C 1, we can only have k = 3,

6 or 9. Now

28 1

33 = + and

82 1

34 = + show k =

3 is an answer. The case k =

6 cannot work as numbers of the form

1 6 + i end in

7 so that both

1 + m k and

1 + n k would begin and end with 7, which makes k k k m n ≥ / impossible. Finally, the case k =

9 also cannot work as numbers of the form

1 9 + i end in

0 or

2 so that both numbers would begin and end with 2, which again makes k k k m n ≥ / impossible. Other commended solvers: SIU Tsz Hang (STFA Leung Kau Kui College, Form 6). Olympiad Corner (continued from page 1) Problem 4. Determine all real valued functions f(x) in one real variable for which y x xf y f x f f + = + ) ( )) ( ) ( (