HMMT 二月 2014 · 几何 · 第 10 题

HMMT February 2014 — Geometry — Problem 10

Let ABC be a triangle with AB = 13 , BC = 14, and CA = 15. Let be the circumcircle of ABC , let d O be its circumcenter, and let M be the midpoint of minor arc BC . Circle ! is internally tangent to 1 at A , and circle ! , centered at M , is externally tangent to ! at a point T . Ray AT meets segment 2 1 BC at point S , such that BS CS = 4 / 15. Find the radius of ! . 2 HMMT 2014 Saturday 22 February 2014 Geometry PUT LABEL HERE Name Team ID# Organization Team

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解析

Let ABC be a triangle with AB = 13 , BC = 14, and CA = 15. Let Γ be the circumcircle of ABC , let ̂ O be its circumcenter, and let M be the midpoint of minor arc BC . Circle ω is internally tangent to 1 Γ at A , and circle ω , centered at M , is externally tangent to ω at a point T . Ray AT meets segment 2 1 BC at point S , such that BS − CS = 4 / 15. Find the radius of ω . 2 4 Answer: 1235/108 Let N be the midpoint of BC . Notice that BS − CS = means that 15 2 N S = . let lines M N and AS meet at P , and let D be the foot of the altitude from A to BC . 15 2 / 15 32 SN 3 Then BD = 5 and AD = 12, so DN = 2 and DS = . Thus N P = AD = 12 = . Now 15 SD 32 / 15 4 √ √ ( ) 2 (13)(14)(15) abc 65 65 33 27 2 2 2 OB = R = = = , so ON = OB − BN = − 7 = . Thus OP = 4 A 4(84) 8 8 8 8 19 and P M = OM − OP = . By Monge’s theorem, the exsimilicenter of ω and Γ (which is A ), the 1 4 ′ insimilicenter of ω and ω (which is T ), and the insimilicenter of ω and Γ (call this P ) are collinear. 1 2 2 ′ But notice that this means P = OM ∩ AT = P . From this we get radius of ω M P 38 2 = = . R OP 27 65 38 1235 Thus the radius of ω is · = . 2 8 27 108