2010COMC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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2010 COMC真题答案免费下载

共计2.5小时考试时间

此套试卷由两部分题目组成

Part A共8题，每题5分

Part B共4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 Solutions:

A3)Answer: 18π.Two solutions:

1. Solution 1:We call the ring between the middle and inner circles the "inner ring". We reflect the shaded portion of the inner ring across line segment CD. The area of the shaded region does not change when we do this.
   The shaded region is now the entire semi-circle to the right of CD.
   Thus, the area of the shaded region is half of the area of the outer circle.
   Since OC = 6, then the outer circle has radius 6 and so has area π6² = 36π.
   Therefore, the area of the shaded region is 1/2(36π) = 18π.
2. Solution 2: We call the ring between the outer and middle circles the "outer ring", and the ring between the middle and inner circles the "inner ring".
   Since OC = 6, then the outer circle has radius 6 and so has area π6² = 36π.
   Since OB = 4, then the middle circle has radius 4 and so has area π4² = 16π.
   Since OA = 2, then the inner circle has radius 2 and so has area π2² = 4π.
   Since the outer circle has area 36π and the middle circle has area 16π, then the area of the outer ring is 36π - 16π = 20π.
   Since the diameter CD divides each ring into two parts of equal area, then the shaded region of the outer ring has area 1/2(20π) = 10π.
   Since the middle circle has area 16π and the inner circle has area 4π, then the area of the inner ring is 16π - 4π = 12π.
   Since the diameter CD divides each ring into two parts of equal area, then the shaded region of the inner ring has area 1/2(12π) = 6π.
   Since the inner circle has area 4π and line segment CD passes through the centre of this circle, then the shaded region of the inner circle has area 1/2(4π) = 2π.
   Therefore, the total shaded area is 10π + 6π + 2π = 18π.

Part B Solutions:

B1)

1. Two Solutions:
   1. Solution 1:From the first row, A + A = 50 or A = 25.
      From the second column, A + C = 57. Since A = 25, then C = 57 - 25 = 32.
   2. Solution 2:From the first row, A + A = 50 or A = 25.
      From the rst column, A + B = 37. Since A = 25, then B = 37 - 25 = 12.
      From the second row, B + C = 44. Since B = 12, then C = 44 - 12 = 32.
2. Three solutions:
   1. Solution 1:The sum of the nine entries in the table equals the sum of the column sums, or 50+n+40 = 90 + n. (This is because each entry in the table is part of exactly one column sum.)
      Similarly, the sum of the nine entries in the table also equals the sum of the row sums, or 30 + 55 + 50 = 135. Therefore, 90 + n = 135 or n = 45.
   2. Solution 2: The sum of the nine entries in the table equals the sum of the row sums, or 30+55+50 = 135. (This is because each entry in the table is part of exactly one row sum.)
      Since the entries in the table include three entries equal to each of D, E and F, then the sum of the entries in the table is also 3D + 3E + 3F = 3(D + E + F).
      Therefore, 3(D + E + F) = 135 or D + E + F = 45.
      From the second column, D + E + F = n. Thus, n = 45.
   3. Solution 3: From the first row, D + D + D = 30 or D = 10.
      From the rst column, D + 2F = 50. Since D = 10, then 2F = 50 - 10 and so F = 20.
      From the third column, D + 2E = 40. Since D = 10, then 2E = 40 - 10 and so E = 15.
      Therefore, n = D + E + F = 10 + 15 + 20 = 45.
3. Three solution:
   1. Solution 1: From the third row, 3R + T = 33.
      From the fourth row, R + 3T = 19.
      Adding these equations, we obtain 4R + 4T = 52 or R + T = 13.
      From the rst row, P + Q + R + T = 20.
      Since R + T = 13, then P + Q = 20 - 13 = 7.

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