HMMT 十一月 2012 · 团队赛 · 第 10 题

HMMT November 2012 — Team Round — Problem 10

[ 6 ] Triangle ABC has AB = 4, BC = 5, and CA = 6. Points A , B , C are such that B C is tangent ′ ′ ′ ′ to the circumcircle of △ ABC at A , C A is tangent to the circumcircle at B , and A B is tangent to ′ ′ the circumcircle at C . Find the length B C .

解析

[ 6 ] Triangle ABC has AB = 4, BC = 5, and CA = 6. Points A , B , C are such that B C is tangent ′ ′ ′ ′ to the circumcircle of △ ABC at A , C A is tangent to the circumcircle at B , and A B is tangent to ′ ′ the circumcircle at C . Find the length B C . 80 ′ ′ ′ ′ ′ ′ Answer: Note that by equal tangents, B A = B C , C A = C B , and A B = A C . Moreover, 3 ′ ′ ′ ′ ′ ′ since the line segments A B , B C , and C A are tangent to the circumcircle of ABC at C , A , and ′ ′ ′ ′ ′ B respectively, we have that ∠ A BC = ∠ A CB = ∠ A , ∠ B AC = ∠ B CA = ∠ B , and ∠ C BA = ′ ′ ′ ∠ C AB = ∠ C . By drawing the altitudes of the isosceles triangles BC A and AC B , we therefore have ′ ′ that C A = 2 / cos C and B A = 3 / cos B . Now, by the Law of Cosines, we have that 2 2 2 a + c − b 25 + 16 − 36 1 cos B = = = 2 ac 2(5)(4) 8 2 2 2 a + b − c 25 + 36 − 16 3 cos C = = = . 2 ab 2(5)(6) 4 Therefore, ( ) 4 80 ′ ′ ′ ′ B C = C A + B A = 2 + 3(8) = . 3 3 Team Round