HMMT 二月 2020 · 冲刺赛 · 第 29 题

HMMT February 2020 — Guts Round — Problem 29

[18] Let ABCD be a tetrahedron such that its circumscribed sphere of radius R and its inscribed sphere 2 of radius r are concentric. Given that AB = AC = 1 ≤ BC and R = 4 r , find BC .

解析

[18] Let ABCD be a tetrahedron such that its circumscribed sphere of radius R and its inscribed 2 sphere of radius r are concentric. Given that AB = AC = 1 ≤ BC and R = 4 r , find BC . Proposed by: Michael Ren √ 7 Answer: 1 + 15 Solution: Let O be the common center of the two spheres. Projecting O onto each face of the tetrahedron will divide it into three isosceles triangles. Unfolding the tetrahedron into its net, the reflection of any of these triangles about a side of the tetrahedron will coincide with another one of these triangles. Using this property, we can see that each of the faces is broken up into the same three triangles. It follows that the tetrahedron is isosceles, i.e. AB = CD , AC = BD , and AD = BC . Let P be the projection of O onto ABC and x = BC . By the Pythagorean Theorem on triangle P OA , √ √ 2 2 P has distance R − r = r 15 from A , B , and C . Using the area-circumcenter formula, we compute AB · AC · BC x √ [ ABC ] = = . 4 P A 4 r 15 However, by breaking up the volume of the tetrahedron into the four tetrahedra OABC , OABD , V OACD , OBCD , we can write [ ABC ] = , where V = [ ABCD ]. Comparing these two expressions 4 r 3 √ for [ ABC ], we get x = 3 15 V . Using the formula for the volume of an isosceles tetrahedron (or some manual calculations), we can √ 2 1 2 compute V = x (2 − x ). Substituting into the previous equation (and taking the solution which 72 √ 7 2 is ≥ 1), we eventually get x = 1 + . 15