2017COMC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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2017 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:

A2)The area of the intersection of each circle and the triangle is 4π/6 cm². The three circles do not overlap, thus the total area is 2π cm².

Part B Challenging Questions' Solutions:

B1) Solution 1: From their success rates we conclude that each of them must have made a multiple of 15 throws. Specically, from Andrew's success rate, his number of throws must be a multiple of 3. Since the total number of throws (105) is also a multiple of 3, Beatrice's number of throws must be a multiple of 3 too. From Beatrice's success rate, her number of throws must be a multiple of 5, and thus must in fact be a multiple of 15. Similarly, since 105 is a multiple of 5, Andrew's number of throws must be a multiple of 5 and thus a multiple of 15 too.
Since 1/3 < 3/5, to maximize the result we should assume that Andrew made the least possible number of throws, that is 15. Then Beatrice made 90 throws.
Then the number of successful free throws they could have made between them is

The maximum possible number of successful free throws they could have made between them is 59.
Solution 2: Suppose Andrew made a free throws and Beatrice b free throws, then a + b = 105, a > 0, b > 0. Let M be the number of successful free throws. We have

M is maximal when 4a/15 is minimal. That is, a = 15 and so M = 59.
The maximum possible number of successful free throws they could have made between them is 59.

Part C Long-form Proof Problems' Solutions:

C3)

1. Subtracting coordinates of corresponding points we have b = z₁ - y₁ = 4 - 0 = 4, h = x₂ - y₂ = x₂ - z₂ = 4 - 0 = 4. Note that points (1, 2) and (3, 2) lie on sides XY and XZ respectively and together with points (1, 0) and (3, 0) they define a square that satisfies the conditions of the problem. This square has side s = 2. (Alternatively, from similar triangles we have s/b = (h-s)/h that is s/4 = (4-s)/4 , so s = 2.)  The answer is b = 4; h = 4; s = 2 .
2. Since PQ is parallel to YZ, triangle XPQ is similar to XYZ. Because h = 3 and corresponding height of XPQ is 3 - 2 = 1, we conclude that the base of XYZ is 3 × 2 = 6. The answer is b = 6 .
3. Two solutions：
   1. Solution 1: From the similarity of XPQ and XY Z we have s/b =(h - s)/h. Equivalently, s =bh/(b + h). Thus, s² =(bh)² /(b + h)² = 2K × [bh/(b + h)² ]. Here K = bh/2 is the area of XYZ.
      By AM-GM inequality (arithmetic mean is greater or equal than geometric mean), we have 4bh/(b + h)² ≤ 1, thus s² ≤K/2.
      We have 2017 ≤ K/2. Thus, 4034 ≤ K.
      Now, we show that the minimum area is achieved for b = h = 2s = 2√2017. Indeed, if b = h then s = bh/(b + h)=b/2=h/2 and K = 2s² = 4034.
      The minimum value for the area of XY Z is 4034.
      The answer is 4034 .
   2. Solution 2: From the similarity of XPQ and XY Z we have s/b = (h - s)/h. Thus b = sh/(h - s) . The area
      [XYZ] =bh/2=h²s/2(h - s)
      Finding the minimum value of this expression is equivalent to finding the maximum of its reciprocal 2(h - s)/h²s → max.
      Now, note that the reciprocal is a quadratic function in the variable 1/h , that is
      so the maximum is achieved at 1/h=1/2•1/sor equivalently, for h = 2s. Then b = 2s and the area [XY Z] = 2s2 = 4034 .

 

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