HMMT 二月 2018 · 几何 · 第 8 题

HMMT February 2018 — Geometry — Problem 8

Let ABC be an equilateral triangle with side length 8. Let X be on side AB so that AX = 5 and Y be on side AC so that AY = 3. Let Z be on side BC so that AZ, BY, CX are concurrent. Let ZX, ZY intersect the circumcircle of AXY again at P, Q respectively. Let XQ and Y P intersect at K . Compute KX · KQ .

解析

Let ABC be an equilateral triangle with side length 8. Let X be on side AB so that AX = 5 and Y be on side AC so that AY = 3. Let Z be on side BC so that AZ, BY, CX are concurrent. Let ZX, ZY intersect the circumcircle of AXY again at P, Q respectively. Let XQ and Y P intersect at K . Compute KX · KQ . Proposed by: Allen Liu Answer: 304 ∼ Let BY and CX meet at O . O is on the circumcircle of AXY , since 4 AXC 4 CY B . = We claim that KA and KO are tangent to the circumcircle of AXY . Let XY and BC meet at L . Then, LBZC is harmonic. A perspectivity at X gives AY OP is harmonic. Similarly, a perspectivity at Y gives AXOQ is harmonic. Thus, K is the pole of chord AO . Now we compute. Denote r as the radius and θ as ∠ AXO . Then, √ √ 2 2 XY 5 + 3 − 3 · 5 19 r = √ = √ = ; 3 3 3 √ √ AC 3 8 4 ◦ sin θ = sin 60 · = · √ = 3; 2 2 XC 2 7 5 + 8 − 5 · 8 √ √ 19 2 2 2 KX · KQ = KA = ( r · tan θ ) = ( · 4 3) = 304 . 3