HMMT 二月 2021 · 几何 · 第 9 题

HMMT February 2021 — Geometry — Problem 9

Let ABCD be a trapezoid with AB ‖ CD and AD = BD . Let M be the midpoint of AB , and let P 6 = C be the second intersection of the circumcircle of 4 BCD and the diagonal AC . Suppose that a BC = 27 , CD = 25 , and AP = 10. If M P = for relatively prime positive integers a and b , compute b 100 a + b .

解析

Let ABCD be a trapezoid with AB ‖ CD and AD = BD . Let M be the midpoint of AB , and let P 6 = C be the second intersection of the circumcircle of 4 BCD and the diagonal AC . Suppose that a BC = 27 , CD = 25 , and AP = 10. If M P = for relatively prime positive integers a and b , compute b 100 a + b . Proposed by: Krit Boonsiriseth Answer: 2705 Solution 1: As ∠ P BD = ∠ P CD = ∠ P AB , DB is tangent to ( ABP ). As DA = DB , DA is also tangent to ( ABP ). Let CB intersect ( ABP ) again at X 6 = B ; it follows that XD is the X -symmedian of 4 AXB . As ∠ AXC = ∠ DAB = ∠ ADC , X also lies on ( ACD ). Therefore ∠ P XB = ∠ P AB = ∠ P CD = ∠ AXD , so XP is the median of 4 AXB , i.e. XP passes through M . Now we have 4 M P A ∼ 4 M BX and 4 ABX ∼ 4 BCD . Therefore M P M B AB BC BC · AP 27 · 10 27 = = = = ⇒ M P = = = . AP XB 2 XB 2 DC 2 DC 2 · 25 5 Solution 2: Let E be the point such that BDCE is a parallelogram. Then ADCE is an isosceles trapezoid, therefore ∠ P AB = ∠ CAB = ∠ BED = ∠ CDE ; this angle is also equal to ∠ P BD = ∠ P CD . Now, ∠ ABP = ∠ ABD − ∠ P BD = ∠ BEC − ∠ BED = ∠ DEC , therefore 4 P AB ∼ 4 CDE . Let F be the midpoint of DE , which lies on BC because BDCE is a parallelogram. It follows that ( 4 P AB, M ) ∼ ( 4 CDE, F ), therefore M P F C BC = = , AP DC 2 DC 27 and again this gives M P = . 5