PUMaC 2014 · 几何（B 组） · 第 8 题

PUMaC 2014 — Geometry (Division B) — 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 . 2 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 ) p = ) 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