PUMaC 2015 · 几何（A 组） · 第 8 题
PUMaC 2015 — Geometry (Division A) — Problem 8
题目详情
- [ 8 ] The incircle of acute triangle ABC touches BC , AC , and AB at points D , E , and F , respectively. Let P be the second intersection of line AD and the incircle. The line through P tangent to the incircle intersects AB and AC at points M and N , respectively. Given that a AB = 8, AC = 10, and AN = 4, let AM = where a and b are positive coprime integers. b What is a + b ? 1
解析
- [ 8 ] The incircle of acute triangle ABC touches BC , AC , and AB at points D , E , and F , respectively. Let P be the second intersection of line AD and the incircle. The line through P tangent to the incircle intersects AB and AC at points M and N , respectively. Given that a AB = 8, AC = 10, and AN = 4, let AM = where a and b are positive coprime integers. b What is a + b ? Solution: Lemma. M C , N B , P D and EF form a harmonic pencil of lines. Proof. By Newton’s Theorem, M C , N B , P D and EF have a common point X . So AM BXCN is a complete quadrilateral, and according to the property of complete quadrilaterals, the diagonals AX and M N harmonically divide the third diagonal BC . Let G be the intersection of lines M N and BC , then CDBG is a harmonic range of points. Similarly, by Ceva’s Theorem, AD , BE and CF has a common point Y , and similar to the discussion above, AF BY CE is a complete quadrilateral and the diagonals AY and EF ′ harmonically divide the third diagonal BC . Let G be the intersection of lines EF and BC , ′ then CDBG is a harmonic range of points. ′ Therefore G = G , and G , E , F and P are collinear. So the fact that CDBG is a harmonic range of points implies that P C , P D , P B and P G , or equivalently, M C , N B , P D and EF , form a harmonic pencil of lines. 5 Now back to the original problem. The lemma implies that AM F B is a harmonic range of points, because these points are the intersections of the harmonic pencil of lines and the line AB . So according to the property of harmonic range of points, we have the equation 1 1 2
- = AM AB AF Similarly we have 1 1 2
- = AN AC AE The lengths of the two tangential segments AE and AF are obviously equal. 1 1 1 1 ∴ + = + AM AB AN AC 40 And this gives AM = , so our answer is 49 . 9 6