HMMT 二月 2007 · 冲刺赛 · 第 26 题

HMMT February 2007 — Guts Round — Problem 26

[ 12 ] ABCD is a cyclic quadrilateral in which AB = 4 , BC = 3 , CD = 2, and AD = 5. Diagonals AC and BD intersect at X . A circle ω passes through A and is tangent to BD at X . ω intersects AB and AD at Y and Z respectively. Compute Y Z/BD .

解析

[ 12 ] ABCD is a cyclic quadrilateral in which AB = 4 , BC = 3 , CD = 2, and AD = 5. Diagonals AC and BD intersect at X . A circle ω passes through A and is tangent to BD at X . ω intersects AB and AD at Y and Z respectively. Compute Y Z/BD . 115 Answer: . Denote the lengths AB, BC, CD , and DA by a, b, c , and d respectively. Because 143 AX BX a ABCD is cyclic, 4 ABX ∼ 4 DCX and 4 ADX ∼ 4 BCX . It follows that = = and DX CX c AX DX d = = . Therefore we may write AX = adk, BX = abk, CX = bck , and DX = cdk for some k . BX CX b Now, ∠ XDC = ∠ BAX = ∠ Y XB and ∠ DCX = ∠ XBY , so 4 BXY ∼ 4 CDX . Thus, XY = BX abk 2 2 DX · = cdk · = abdk . Analogously, XY = acdk . Note that XY /XZ = CB/CD . Since CD c 2 ∠ Y XZ = π − ∠ ZAY = ∠ BCD , we have that 4 XY Z ∼ 4 CBD . Thus, Y Z/BD = XY /CB = adk . Finally, Ptolemy’s theorem applied to ABCD gives ( ad + bc ) k · ( ab + cd ) k = ac + bd It follows that the answer is ad ( ac + bd ) 20 · 23 115 = = ( ab + cd )( ad + bc ) 22 · 26 143