2005COMC加拿大數(shù)學(xué)公開賽真題答案免費(fèi)下載

歷年 Canadian Open Mathematics Challenge加拿大數(shù)學(xué)公開賽

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2005 COMC真題答案免費(fèi)下載

共計(jì)2.5小時(shí)考試時(shí)間

此套試卷由兩部分題目組成

Part A共8題，每題5分

Part B共4題，每題10分

共計(jì)12題，滿分80分

不可使用任何計(jì)算器

完整版下載鏈接見文末

部分真題預(yù)覽：

Part A Solutions:

A5)Answer: 14 . Two Solutions:

1. Solution 1: Let G be the number of gold storeys in the top half of the building.
   Then there are 25 ? G black storeys in the top half of the building.
   Since there are 25 black storeys in total, then the number of black storeys in the bottom half of the building is 25 ? (25 ? G) = G.
   Since the sum of the number of gold storeys in the top half of the building and the number of black storeys in the bottom half of the building is 28, then G + G = 28, or G = 14.
   Thus, there are 14 gold storeys in the top half of the building.
2. Solution 2: Let G and g be the number of gold storeys in the top and bottom halfs of the building, and B and b the number of black storeys in the top and bottom halfs of the building.
   Then G + B = 25 and g + b = 25, looking at the top and bottom halfs of the building.
   Also, G + g = 25 and B + b = 25, since 25 of the storeys are painted in each colour.
   Also, G + b = 28 from the given information, or b = 28 ? G.
   Since B + b = 25, then B + 28 ? G = 25, so B = G ? 3.
   Since G + B = 25, then G + G ? 3 = 25 or 2G = 28 or G = 14.
   Thus, there are 14 gold storeys in the top half of the building.

Part B Solutions:

B2)

1. Three solutions:
   1. Solution 1: Factoring the left side of the second equation, we get 2a2 + ab ? 3b2 = (a ? b)(2a + 3b).
      Since a ? b = 1, we get (1)(2a + 3b) = 22 or 2a + 3b = 22.
      So we now have a ? b = 1 and 2a + 3b = 22.
      Adding 3 times the first equation to the second equation, we get 5a = 25 or a = 5.
      Substituting back into the first equation, we get b = 4.
      Thus, the only solution is (a, b) = (5, 4).
   2. Solution 2: From the first equation, a = b + 1.
      Substituting into the second equation, we obtain
      2(b + 1)2 + (b + 1)(b) ? 3b2 = 22
      (2b2 + 4b + 2) + (b2 + b) ? 3b2 = 22
      5b = 20
      b = 4
      Substituting back into the first equation, we get a = 5, so the only solution is (a, b) = (5, 4).
2. Two Solutions:
   1. Solution 1: If we add the second equation to the third equation, we obtain
      y² ? zx + xy + yz + z² ? xy + zx + yz = ?18 + 18
      y² + 2yz + z² = 0
      (y + z)² = 0
      y + z = 0
      z = ?y
      Substituting back into the three equations, we obtain
      x² + y² = 82
      2xy = ?18
      ?2xy = 18
      Thus, x² + y² = 82 and xy = ?9.
      Therefore, (x + y)² = x² + 2xy + y² = 82 + (?18) = 64, so x + y = ±8.
      If x+y = 8, then y = 8?x and so since xy = ?9, then x(8?x) = ?9 or x² ?8x?9 = 0 or (x ? 9)(x + 1) = 0 so x = 9 or x = ?1.
      Since x + y = 8, then if x = 9, we have y = ?1 and z = ?y = 1.
      Since x + y = 8, then if x = ?1, we have y = 9 and z = ?y = ?9.
      If x+y = ?8, then y = ?8?x and so since xy = ?9, then x(?8?x) = ?9 or x²+8x?9 = 0 or (x + 9)(x ? 1) = 0 so x = ?9 or x = 1.
      Since x + y = ?8, then if x = ?9, we have y = 1 and z = ?y = ?1.
      Since x + y = ?8, then if x = 1, we have y = ?9 and z = ?y = 9.
      Therefore, the four solutions are (x, y, z) = (9,?1, 1), (?1, 9,?9), (?9, 1,?1), (1,?9, 9).

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