PUMaC 2014 · 几何（A 组） · 第 8 题

PUMaC 2014 — Geometry (Division A) — Problem 8

[ 8 ] ABCD is a cyclic quadrilateral with circumcenter O and circumradius 7. AB intersects p CD at E , DA intersects CB at F . OE = 13, OF = 14. Let cos ∠ F OE = , with p, q coprime. q Find p + q . 1

解析

[ 8 ] ABCD is a cyclic quadrilateral with circumcenter O and circumradius 7. AB intersects p CD at E , DA intersects CB at F . OE = 13, OF = 14. Let cos ∠ F OE = , with p, q coprime. q Find p + q . Solution: Since we’re given EO = 13 and the radius is 7, by power of a point, the power of E with respect to O is P ( E ) = (13 − 7)(13 + 7) = 120. Similarly, the power of point F with respect to O is P ( F ) = (14 − 7)(14 + 7) = 147. Construct point X on EF such that ∠ CXE = ∠ CDF . Note that given ABCD is cyclic, this implies ∠ CBA = 180 − ∠ ADC = ∠ CDF . Thus note that ∠ EBC + ∠ CXE = (180 − ∠ CBA ) + ∠ CDF = 180 . Therefore quadrilateral BEXC is cyclic; and similarly, quadrilateral CXF D is cyclic. Note that the circle about BEXC and ◦ O have a common chord BC . Thus the power of point F with respect to both circles is F C · F B , and so the power of F with respect to the circle about BEXC is P ( F ) = 147. Similarly, the power of E with respect to the circle about CXF D is P ( E ) = 120. Therefore, by power of a point, ( EX + XF )( XF ) = 147 = P ( F ) and ( EX )( EX + XF ) = 120 = P ( E ). Then by adding, 2 267 = 120 + 147 = ( EX + XF )( EX ) + ( EX + XF )( XF ) = ( EX + XF ) √ = ⇒ EF = 267 . Now by Law of Cosine on ∆ EOF , 2 2 2 EF = EO + OF − 2 · EO · OF · cos( ∠ EOF ) 267 = 169 + 196 − 2 · 13 · 14 cos( ∠ EOF ) − 98 7 = ⇒ cos( ∠ EOF ) = = . − 2 · 13 · 14 26 Hence p + q = 33 . 3