HMMT 二月 2022 · 几何 · 第 4 题
HMMT February 2022 — Geometry — Problem 4
题目详情
- Parallel lines ℓ , ℓ , ℓ , ℓ are evenly spaced in the plane, in that order. Square ABCD has the property 1 2 3 4 that A lies on ℓ and C lies on ℓ . Let P be a uniformly random point in the interior of ABCD and 1 4 let Q be a uniformly random point on the perimeter of ABCD . Given that the probability that P lies 53 a between ℓ and ℓ is , the probability that Q lies between ℓ and ℓ can be expressed as , where 2 3 2 3 100 b a and b are relatively prime positive integers. Compute 100 a + b .
解析
- Parallel lines ℓ , ℓ , ℓ , ℓ are evenly spaced in the plane, in that order. Square ABCD has the property 1 2 3 4 that A lies on ℓ and C lies on ℓ . Let P be a uniformly random point in the interior of ABCD and 1 4 let Q be a uniformly random point on the perimeter of ABCD . Given that the probability that P lies 53 a between ℓ and ℓ is , the probability that Q lies between ℓ and ℓ can be expressed as , where 2 3 2 3 100 b a and b are relatively prime positive integers. Compute 100 a + b . Proposed by: Daniel Zhu Answer: 6100 Solution: The first thing to note is that the area of ABCD does not matter in this problem, so for the sake of convenience, introduce coordinates so that A = (0 , 0), B = (1 , 0), and C = (0 , 1). Suppose A and B lie on the same side of ℓ . Then, by symmetry, C and D lie on the same side of ℓ . Now 2 3 suppose BC intersects ℓ and ℓ at X and Y , respectively, and that DA intersects ℓ and ℓ at U and V , 2 3 2 3 respectively. Note that XY V U is a parallelogram. Since BC = BX + XY + Y C = BX +2 XY > 2 XY, we have that XY is less than half the side length of the square, so the area of XY V U is at most half 1 of the area of square ABCD . However, since 0 . 53 > , this can’t happen. Similar reasoning applies if 2 B and C lie on the same side of ℓ . Therefore, points B and D lie between ℓ and ℓ . 3 2 3 Let AB and AD intersect ℓ at points M and N , respectively. Let r = AM and s = AN . By symmetry, 2 y x [ AM N ] = 0 . 235, so rs = 0 . 47. Additionally, in coordinates line ℓ is just + = 1. Therefore line ℓ 2 4 r s y x 1 1 is given by + = 3. Since C = (1 , 1) lies on this line, + = 3. r s r s The answer that we want is 2 r + 2 s r + s 1 − = 1 − . 4 2 1 1 On the other hand, the condition + = 3 rearranges to 3 rs = r + s , so r + s = 1 . 41. Thus the answer r s