HMMT 二月 2009 · GEN1 赛 · 第 10 题

HMMT February 2009 — GEN1 Round — Problem 10

[ 6 ] A kite is a quadrilateral whose diagonals are perpendicular. Let kite ABCD be such that ∠ B = ◦ ∠ D = 90 . Let M and N be the points of tangency of the incircle of ABCD to AB and BC respectively. ′ ′ ′ Let ω be the circle centered at C and tangent to AB and AD . Construct another kite AB C D that is ′ ′ ′ ′ similar to ABCD and whose incircle is ω . Let N be the point of tangency of B C to ω . If M N ‖ AC , then what is the ratio of AB : BC ?

解析

[ 6 ] A kite is a quadrilateral whose diagonals are perpendicular. Let kite ABCD be such that ∠ B = ◦ ∠ D = 90 . Let M and N be the points of tangency of the incircle of ABCD to AB and BC respectively. ′ ′ ′ Let ω be the circle centered at C and tangent to AB and AD . Construct another kite AB C D that is ′ ′ ′ ′ similar to ABCD and whose incircle is ω . Let N be the point of tangency of B C to ω . If M N ‖ AC , then what is the ratio of AB : BC ? 2 √ 1+ 5 Answer: 2 Solution: Let’s focus on the right triangle ABC and the semicircle inscribed in it since the situation is symmetric about AC . First we find the radius a of circle O . Let AB = x and BC = y . Drawing the radii OM and ON , we see that AM = x − a and 4 AM O ∼ 4 ABC . In other words, AM AB = M O BC x − a x = a y xy a = . x + y Now we notice that the situation is homothetic about A . In particular, ′ ′ 4 AM O ∼ 4 ON C ∼ 4 CN C . ′ ′ ′ Also, CB and CN are both radii of circle C . Thus, when M N ‖ AC , we have ′ AM = CN = CB x − a = y xy a = = x − y x + y 2 2 x − xy − y = 0 √ 2 y y 2 x = ± + y 2 4 √ AB x 1 + 5 = = . BC y 2 3