2012COMC加拿大数学公开赛真题答案免费下载

历年 Canadian Open Mathematics Challenge加拿大数学公开赛

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2012 COMC真题答案免费下载

共计2.5小时考试时间

此套试卷由三部分题目组成

4题简答题，每题4分

4题挑战题，每题6分

4题解答题，每题10分

共计12题，满分80分

不可使用任何计算器

完整版下载链接见文末

部分真题预览：

Part A Introductory Questions' Solutions:

A3)The answer is r = 3.
Divide the hexagon into six regions as shown, with the centre point denoted by P.

This is possible since the hexagon is regular. By symmetry, note that PA = PB = PC = PD = PE = PF and the six interior angles about P are equal. Then since the sum of the six interior angles about P sum to 360°, ∠APB =∠BPC = ∠CPD = ∠DPE = ∠EPF = ∠FPA = 60°.
Therefore, the six triangles ΔPAB,ΔPBC,ΔPCD,ΔPDE,ΔPEF,ΔPFA are all equilateral and have the same area. Let K be the area of any one of these triangles. Therefore, the hexagon has area 6K.
Note that the area of ΔACD is equal to the area of ΔPCD plus the area of ΔPAC. Since ΔPAB,ΔPBC are both equilateral, PA = AB and PC = CB. Therefore, triangles ΔBAC and ΔPAC are congruent and hence have the same area. Note that the area of ΔPAC plus the area of ΔBAC is the sum of the areas of the equilateral triangles ΔPAB and ΔPBC, which is 2K. Therefore, ΔPAC has area K. We already noted that the area of ΔACD is equal to the area of ΔPCD plus that of ΔPAC. This quantity is equal to K + K = 2K. Hence, the area of ABCDEF is 6K/2K = 3 times the area of ΔACD. The answer is 3.

Part B Challenging Questions' Solutions:

B1) The answer is t = 59 or t = 61.
Since Alice ran exactly 30 laps, Bob meets Alice at where Alice started. Since Bob started diametrically across from Alice, Bob ran n + 1/2 laps for some positive integer n. Since Alice and Bob meet only the first time they meet, the number of laps that Alice ran and the number of laps Bob ran cannot differ by more than 1. Therefore, Bob ran either 29.5 laps or 30.5 laps.
Note that Alice and Bob ran for the same amount of time and the number of seconds each person ran is the number of laps he/she ran times the number of seconds it takes he/she to
complete a lap.
If Bob ran 29.5 laps, then 30t = 29.5 × 60. Hence, t = 29.5 × 2 = 59.
If Bob ran 30.5 laps, then similarly, 30t = 30.5 × 60. Hence, t = 30.5 × 2 = 61.
Therefore, t = 59 or t = 61.

Part C Long-form Proof Problems' Solutions:

C1)

1. (a) The answers are f(2) = 4 and g(f(2)) = 4.
   Substituting x = 2 into f(x) yields f(2) = 2² = 4.
   Substituting x = 2 into g(f(x)) and noting that f(2) = 4 yields g(f(2)) = g(4) =
   3 · 4 − 8 = 4. 
2. The answers are x = 2 and x = 6.
   Note that f(g(x)) = f(3x − 8) = (3x − 8)² = 9x² − 48x + 64 and g(f(x)) = g(x²) = 3x² − 8. Therefore, we are solving 9x² − 48x + 64 = 3x² − 8.
   Rearranging this into a quadratic equation yields 6x² − 48x + 72 = 0 ⇒ 6(x² − 8x + 12) = 0.
   This factors into 6(x − 6)(x − 2) = 0. Hence, x = 2 or x = 6. We now verify these are indeed solutions.
   If x = 2, then f(g(2)) = f(3(2) − 8) = f(−2) = (−2)² = 4 and g(f(2)) = 4 by part(a). Hence, f(g(2)) = g(f(2)). Therefore, x = 2 is a solution.
   If x = 6, then f(g(6)) = f(3 · 6−8) = f(10) = 10² = 100 and g(f(6)) = g(6²) = g(36) = 3·36−8 = 108−8 = 100. Hence, f(g(6)) = g(f(6)). Therefore, x = 6 is also a solution.
3. The answers are r = 3 and r = 8.
   We first calculate f(h(2)) and h(f(2)) in terms of r. f(h(2)) = f(3 · 2 − r) = f(6 − r) = (6 − r)² and h(f(2)) = h(2²) = h(4) = 3 · 4 − r = 12 − r.
   Therefore, (6 − r)² = 12 − r⇒ r² − 12r + 36 = 12 − r. Re-arranging this yields r² − 11r + 24 = 0, which factors as (r − 8)(r − 3) = 0.
   Hence, r = 3 or r = 8. We will now verify that both of these are indeed solutions.
   If r = 3, then h(x) = 3x − 3. Then f(h(2)) = f(3 · 2 − 3) = f(3) = 9 and h(f(2)) =h(2²) = h(4) = 3 · 4 − 3 = 9. Therefore, f(h(2)) = h(f(2)). Consequently, r = 3 is a solution. From the result of part (b), we also verified that r = 8 is a solution.

 

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