HMMT 二月 2020 · 几何 · 第 8 题
HMMT February 2020 — Geometry — Problem 8
题目详情
- Let ABC be an acute triangle with circumcircle Γ. Let the internal angle bisector of ∠ BAC intersect ′ BC and Γ at E and N , respectively. Let A be the antipode of A on Γ and let V be the point where ′ ′ ′ AA intersects BC . Given that EV = 6, V A = 7, and A N = 9, compute the radius of Γ.
解析
- Let ABC be an acute triangle with circumcircle Γ. Let the internal angle bisector of ∠ BAC intersect ′ BC and Γ at E and N , respectively. Let A be the antipode of A on Γ and let V be the point where ′ ′ ′ AA intersects BC . Given that EV = 6, V A = 7, and A N = 9, compute the radius of Γ. Proposed by: James Lin 15 Answer: 2 Solution 1: Let H be the foot of the altitude from A to BC . Since AE bisects ∠ H AV , by the angle a a AH AV AN AH ′ a a bisector theorem = . Note that 4 AH E ∼ 4 AN A are similar right triangles, so = . a ′ H E V E N A H E a a √ √ ′ ′ 2 ′ 2 2 Let R be the radius of Γ. We know that AA = 2 R , so AN = AA − N A = 4 R − 81 and ′ ′ AV = AA − V A = 2 R − 7. Therefore √ 2 4 R − 81 AN AH AV 2 R − 7 a = = = = . ′ 9 N A H E V E 6 a The resulting quadratic equation is 2 2 2 0 = 9(2 R − 7) − 4(4 R − 81) = 20 R − 252 R + 765 = (2 R − 15)(10 R − 51) . 15 ′ We are given that ABC is acute so V A < R . Therefore R = . 2 A B C H E V a ′ K A N √ Solution 2: Let Ψ denote inversion about A with radius AB · AC composed with reflection about ′ AE . Note that Ψ swaps the pairs { B, C } , { E, N } , and { H , A } . Let K = Ψ( V ), which is also the a ′ ′ second intersection of AH with Γ. Since AE bisects ∠ KAA , we have N K = N A = 9. By the a inversion distance formula, AB · AC · V E AE · AN · V E AN · V E N K = = = . AE · AV AE · AV AV This leads to the same equation as the previous solution.