HMMT 二月 2010 · 几何 · 第 9 题
HMMT February 2010 — Geometry — Problem 9
题目详情
- [ 7 ] Let ABCD be a quadrilateral with an inscribed circle centered at I . Let CI intersect AB at E . If ◦ ◦ ◦ ∠ IDE = 35 , ∠ ABC = 70 , and ∠ BCD = 60 , then what are all possible measures of ∠ CDA ?
解析
- [ 7 ] Let ABCD be a quadrilateral with an inscribed circle centered at I . Let CI intersect AB at E . If ◦ ◦ ◦ ∠ IDE = 35 , ∠ ABC = 70 , and ∠ BCD = 60 , then what are all possible measures of ∠ CDA ? ◦ ◦ Answer: 70 and 160 Arbitrarily defining B and C determines I and E up to reflections across BC . D lies on both the circle ◦ determined by ∠ EDI = 35 and the line through C tangent to the circle (and on the opposite side of B ); since the intersection of a line and a circle has at most two points, there are only two cases for ABCD . The diagram below on the left shows the construction made in this solution, containing both cases. The diagram below on the right shows only the degenerate case. ′ B D A D E A I I ◦ ◦ ◦ ◦ 70 70 60 60 B C B C Geometry Subject Test ′ ′ Reflect B across EC to B ; then CB = CB . Since BA and BC are tangent to the circle centered ′ ◦ ′ at I , IB is the angle bisector of ∠ ABC . Then ∠ IBE = ∠ IB E = 35 . If B = D , then ∠ ADC = ′ ◦ ′ ◦ ′ ∠ EB C = 70 . Otherwise, since ∠ IB E = 35 = ∠ IDE (given), EB DI is a cyclic quadrilateral. ′ ◦ ◦ Then ∠ IED = ∠ IB D = 35 and ∠ BCI = ∠ ECD = 30 , so 4 CED ∼ 4 CBI . ◦ Since ∠ CID is exterior to 4 DIE , ∠ CID = ∠ IDE + ∠ DEI = 70 . Then 4 CDI ∼ 4 CEB . Because ′ ′ ◦ ◦ ◦ ◦ ◦ EB DI is cyclic, ∠ IDC = ∠ IEB = ∠ IEB = 180 − 70 − 30 = 80 . Then ∠ ADC = 2 ∠ IDC = 160 . ◦ ◦ Thus, the two possible measures are 70 and 160 .