HMMT 二月 2016 · 几何 · 第 9 题

HMMT February 2016 — Geometry — Problem 9

In cyclic quadrilateral ABCD with AB = AD = 49 and AC = 73, let I and J denote the incenters of triangles ABD and CBD . If diagonal BD bisects IJ , find the length of IJ .

解析

In cyclic quadrilateral ABCD with AB = AD = 49 and AC = 73, let I and J denote the incenters of triangles ABD and CBD . If diagonal BD bisects IJ , find the length of IJ . Proposed by: Evan Chen √ 28 Answer: 69 5 2 2 Let O be circumcenter, R the circumradius and r the common inradius. We have IO = JO = R ( R − 2 r ) by a result of Euler; denote x for the common value of IO and JO . Additionally, we know AJ = AB = AD = 49 (angle chase to find that ∠ BJA = ∠ JBA ). Since A is the midpoint of the arc ̂ BD not containing C , both J and A lie on the angle bisector of angle ∠ BCD , so C, J, A are collinear. So by Power of a Point we have 2 2 R − x = 2 Rr = AJ · JC = 49 · 24 . Next, observe that the angle bisector of angle BAD contains both I and O , so A, I, O are collinear. Let M be the midpoint of IJ , lying on BD . Let K be the intersection of IO and BD . Observing that 2 2 the right triangles 4 IM O and 4 IKM are similar, we find IM = IK · IO = rx , so IJ = 4 rx . Now apply Stewart’s Theorem to 4 AOJ to derive 2 2 R ( x ( R − x ) + 4 rx ) = 49 x + x ( R − x ) . Eliminating the common factor of x and rearranging gives 2 2 49 − ( R − x ) = 4 Rr = 48 · 49 49 · 24 49 · 24 168 so R − x = 7. Hence R + x = = 168, and thus 2 R = 175, 2 x = 161. Thus r = = . 7 175 25 √ √ √ 84 · 161 28 69 Finally, IJ = 2 rx = 2 = . 25 5