HMMT 二月 2018 · 几何 · 第 10 题

HMMT February 2018 — Geometry — Problem 10

Let ABC be a triangle such that AB = 6 , BC = 5 , AC = 7. Let the tangents to the circumcircle of ABC at B and C meet at X . Let Z be a point on the circumcircle of ABC . Let Y be the foot of the perpendicular from X to CZ . Let K be the intersection of the circumcircle of BCY with line AB . Given that Y is on the interior of segment CZ and Y Z = 3 CY , compute AK .

解析

Let ABC be a triangle such that AB = 6 , BC = 5 , AC = 7. Let the tangents to the circumcircle of ABC at B and C meet at X . Let Z be a point on the circumcircle of ABC . Let Y be the foot of the perpendicular from X to CZ . Let K be the intersection of the circumcircle of BCY with line AB . Given that Y is on the interior of segment CZ and Y Z = 3 CY , compute AK . Proposed by: Allen Liu 147 Answer: 10 Let ω denote the circumcircle of ABC and ω denote the circle centered at X through B and C . Let 1 2 ′ ′ ω intersect AB, AC again at B , C . The (signed) power of Y with respect to ω is − CY · Y Z . The 2 1 2 2 2 power of Y with respect to ω is XY − CX = − CY . Thus the ratio of the powers of Y with respect 2 to the two circles is 3 : 1. The circumcircle of BCY passes through the intersection points of ω and ω 1 2 ( B and C ) and thus contains exactly the set of points such that the ratio of their powers with respect to ω and ω is 3 : 1 (this fact can be verified in a variety of ways). We conclude that K must be the 1 2 ′ ′ ′ KB ′ ′ AC · B C point on line AB such that = 3. It now suffices to compute AB . Note AB = by similar KA BC ′ ′ triangles. Also an angle chase gives that B XC are collinear. We compute BC 5 7 BX = = = 5 2 cos A 2 2 · 7 49 3 147 ′ ′ ′ ′ and thus B C = 7 so AB = and AK = AB = . 5 2 10