HMMT 十一月 2017 · 冲刺赛 · 第 33 题

HMMT November 2017 — Guts Round — Problem 33

[ 24 ] Tetrahedron ABCD with volume 1 is inscribed in circumsphere ω such that AB = AC = AD = 2 and BC · CD · DB = 16. Find the radius of ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2017, November 11, 2017 — GUTS ROUND Organization Team Team ID# D´ ej` a vu??

解析

[ 24 ] Tetrahedron ABCD with volume 1 is inscribed in circumsphere ω such that AB = AC = AD = 2 and BC · CD · DB = 16. Find the radius of ω . Proposed by: Caleb He 5 Answer: 3 Let X be the foot of the perpendicular from A to ∆ BCD . Since AB = AC = AD , it follows that X is the circumcenter of ∆ BCD . Denote XB = XC = XD = r . By the Pythagorean Theorem on ∆ ABX , √ 2 we have AX = 4 − r . Now, from the extended law of sines on ∆ BCD , we have the well-known identity BC · CD · DB = [ BCD ] , 4 r where [ BCD ] denotes the area of ∆ BCD . However, we have 1 V = AX · [ BCD ] , 3 where V is the volume of ABCD , which yields the expression 3 √ [ BCD ] = . 2 4 − r Now, given that BC · CD · DB = 16, we have 4 3 = √ . 2 r 4 − r 8 Solving, we get r = . Now, let O be the center of ω . Since OB = OC = OD , it follows that the foot 5 of the perpendicular from O to ∆ BCD must also be the circumcenter of ∆ BCD , which is X . Thus, A, X, O are collinear. Let R be the radius of ω . Then we have R = OA = OX + XA √ √ 2 2 2 = R − r + 4 − r √ 64 6 2 = R − + . 25 5 5 Solving, we get R = . (Note: solving for R from OA = OX − XA gives a negative value for R .) 3