2016COMC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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2016 COMC真題答案免費(fèi)下載

共計(jì)2.5小時(shí)考試時(shí)間

此套試卷由三部分題目組成

4題簡(jiǎn)答題，每題4分

4題挑戰(zhàn)題，每題6分

4題解答題，每題10分

共計(jì)12題，滿分80分

不可使用任何計(jì)算器

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部分真題預(yù)覽：

Part A Introductory Questions' Solutions:

A2)The area of a circle with radius r is πr²: Since this circle has area 16π, we have r² = 16, and r = 4: So the circle has radius 4 and diameter 8.
Let t be the side length of the square. By the Pythagorean Theorem, t² +t² = 8² so t² = 32; which is the area of the square.

Part B Challenging Questions' Solutions:

B1) Two Solutions:

1. Solution 1: If the equation has roots x₁; x₂; x₃; then x₁+x₂+x₃ = 10; x₁x₂+x₂x₃+x₃x₁ = P and x₁x₂x₃ = 30: Suppose x₁ = 1 is a root. Then x₂ +x₃ = 9 and x₂x₃ = 30. For any pair of integers which multiply to 30, their sum is at least 11, so this is not possible, and 1 is not a root. The only other possibility is that the roots are 2, 3, 5; which gives P = 2 ? 3+3 ? 5+5 ? 2 = 31.
2. Solution 2: The roots of the equation are integers and have a product of 30. The possible sets of roots are (1, 1, 30); (1, 2, 15); (1, 3, 10); (1, 5, 6); (2, 3, 5): Of these, only (2, 3, 5) sums to
   10, so it must be our set of roots. To find P, we substitute x = 2 into the equation to get 8 - 40 + 2P - 30 = 0; which yields P = 31:

Part C Long-form Proof Problems' Solutions:

C3)

1. Reflect the point B in the horizontal x-axis to get the point B'. Then PB = PB'. By the triangle inequality, PA + PB'≥ AB': The line AB' will pass through the segment OC; so taking P along AB' is where PA + PB' will attain its minimum.
   By the Pythagorean Theorem, the length of AB' is X = √[c² + (a + b)²].
2. Observe that at the minimum point attained in (a), POA and PCB are similar triangles. Let a + b = n. We have X² = 144 + n²,or (X - n)(X + n) = 144. Since X and n are both integers, we have X - n and X + n are the same parity. Since their product is a multiple of 2, each must also be a multiple of 2. We can factor 144 in the following ways, where both factors are even: (72, 2); (36, 4); (18, 8); (12, 12); (24, 6): These yield (X, n) pairs of (37, 35); (20, 16); (13, 5); (12, 0); (15, 9): We can immediately eliminate (12, 0); as n must be positive.
   Notice that by similar triangles, p/12 = a/n; so p = 12a/n: If n = 5, 35; there is no value of a for which b will also be positive.
   When n = 9; we have (a, b) = (3, 6); (6, 3) and when n = 16; we have (a, b) = (4, 12); (8, 8); (12, 4).
3. Let q = c - p: By similar triangles, p = ac=(a + b) and q = bc=(a + b).
   Let k be the smallest integer such that kc is divisible by a + b: Then since p and q are integers, k must divide a and b:
   If k = 1, then c = m(a + b) for some integer m, and X =√(1 + m²) × (a + b): This is only an integer when m = 0, which it cannot be, since then c = 0. This k > 1:
   If k = 2; then 2c = m(a + b) for some integer m; and X = [ (a+b)/2 ]√(4 + m²). Again, this is only an integer when m = 0; so k > 2; as required.

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