HMMT 二月 2008 · 几何 · 第 10 题

HMMT February 2008 — Geometry — Problem 10

[ 7 ] Let ABC be a triangle with BC = 2007, CA = 2008, AB = 2009. Let ω be an excircle of ABC that touches the line segment BC at D , and touches extensions of lines AC and AB at E and F , respectively (so that C lies on segment AE and B lies on segment AF ). Let O be the center of ω . Let ` be the line through O perpendicular to AD . Let ` meet line EF at G . Compute the length DG . 1

解析

[ 7 ] Let ABC be a triangle with BC = 2007, CA = 2008, AB = 2009. Let ω be an excircle of ABC that touches the line segment BC at D , and touches extensions of lines AC and AB at E and F , respectively (so that C lies on segment AE and B lies on segment AF ). Let O be the center of ω . Let ` be the line through O perpendicular to AD . Let ` meet line EF at G . Compute the length DG . Answer: 2014024 Let line AD meet ω again at H . Since AF and AE are tangents to ω and ADH is a secant, we see that DEHF is a harmonic quadrilateral. This implies that the pole of AD with respect to ω lies on EF . Since ` ⊥ AD , the pole of AD lies on ` . It follows that the pole of AD is G . 4 A G B D C F E O H Thus, G must lie on the tangent to ω at D , so C, D, B, G are collinear. Furthermore, since the pencil of lines ( AE, AF ; AD, AG ) is harmonic, by intersecting it with the line BC , we see that ( C, B ; D, G ) is harmonic as well. This means that BD CG · = − 1 . DC GB 1 (where the lengths are directed.) The semiperimeter of ABC is s = (2007 + 2008 + 2009) = 3012. So 2 BD = s − 2009 = 1003 and CD = s − 2008 = 1004. Let x = DG , then the above equations gives 1003 x + 1004 · = 1 . 1004 x − 1003 Solving gives x = 2014024. Remark: If you are interested to learn about projective geometry, check out the last chapter of Geometry Revisited by Coxeter and Greitzer or Geometric Transformations III by Yaglom. 5