PUMaC 2022 · 几何（B 组） · 第 2 题
PUMaC 2022 — Geometry (Division B) — Problem 2
题目详情
- Three spheres are all externally tangent to a plane and to each other. Suppose that the radii of these spheres are 6 , 8 , and, 10 . The tangency points of these spheres with the plane form the vertices of a triangle. Determine the largest integer that is smaller than the perimeter of this triangle. √
解析
- Three spheres are all externally tangent to a plane and to each other. Suppose that the radii of these spheres are 6 , 8 , and, 10 . The tangency points of these spheres with the plane form the vertices of a triangle. Determine the largest integer that is smaller than the perimeter of this triangle. Proposed by Frank Lu Answer: 47 Suppose that the centers of the spheres are O , O , O , with the respective tangency points to 1 2 3 the planes being A , A , A , so that the sphere centered at O has radius 6 , the sphere centered 1 2 3 1 at O has radius 8 , and the sphere centered at O has radius 10 . Our goal will to be to compute 2 3 A A , A A , A A . We go through the computation for A A , and the computation for the 1 2 2 3 3 1 1 2 other two sides follows similarly. Notice that, considering the plane containing A , A , O , O , draw the line perpendicular to 1 2 1 2 O A that goes through O . Say that this line intersects O A at X . Then, O X = A A , 2 2 1 2 2 1 1 1 1 2 and we know that O X = O A − O A = 1 . Therefore, it follows that, by the Pythagorean 2 1 2 2 1 1 √ 2 2 theorem, as O O = 14 , we have that the length of A A is 14 − 2 . 1 2 1 2 √ √ 2 2 2 2 Similarly, we have that A A = 18 − 2 and A A = 16 − 4 . Notice that the sum of 2 3 3 1 these values is less than 14 + 18 + 16 = 48 . We argue that this sum is larger than 47 . To see 2 2 2 this, observe that ( n − 1 / 3) = n − 2 n/ 3 + 1 / 9 < n − 4 , for n ≥ 14 . Therefore, it follows that √ 2 2 14 − 2 > 14 − 1 / 3 , and similarly for the other two sides. Summing up these inequalities yields that the perimeter is larger than 47 . This is our desired floor. √