HMMT 二月 2016 · 几何 · 第 9 题

英文解析

9. In cyclic quadrilateral ABCD with AB = AD = 49 and AC = 73, let I and J denote the incenters oftriangles ABD and CBD . If diagonal BD bisects IJ , find the length of IJ .
   Proposed by: Evan Chen
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   Answer: 6928
   2 25
   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
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   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 . Nowapply 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
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   84 · 161 28 69
   Finally, IJ = 2 rx = 2 = .
   25 5