HMMT 二月 2021 · 几何 · 第 4 题

HMMT February 2021 — Geometry — Problem 4

Let ABCD be a trapezoid with AB ‖ CD , AB = 5, BC = 9, CD = 10, and DA = 7. Lines BC and DA intersect at point E . Let M be the midpoint of CD , and let N be the intersection of the a 2 circumcircles of 4 BM C and 4 DM A (other than M ). If EN = for relatively prime positive integers b a and b , compute 100 a + b .

解析

Let ABCD be a trapezoid with AB ‖ CD , AB = 5, BC = 9, CD = 10, and DA = 7. Lines BC and DA intersect at point E . Let M be the midpoint of CD , and let N be the intersection of the 2 a circumcircles of 4 BM C and 4 DM A (other than M ). If EN = for relatively prime positive integers b a and b , compute 100 a + b . Proposed by: Milan Haiman Answer: 90011 Solution: From 4 EAB ∼ 4 EDC with length ratio 1 : 2, we have EA = 7 and EB = 9. This means ′ that A, B, M are the midpoints of the sides of 4 ECD . Let N be the circumcenter of 4 ECD . Since ′ ′ ′ ◦ ′ N is on the perpendicular bisectors of EC and CD , we have ∠ N M C = ∠ N BC = 90 . Thus N is ′ ′ on the circumcircle of 4 BM C . Similarly, N is on the circumcircle of 4 DM A . As N 6 = M , we must ′ 2 have N = N . So it suffices to compute R , where R is the circumradius of 4 ECD . √ We can compute K = [ 4 ECD ] to be 21 11 from Heron’s formula, giving 10 · 14 · 18 30 √ R = = . 4 K 11 900 2 So R = , and the final answer is 90011. 11