HMMT 二月 2011 · CALCGEO 赛 · 第 18 题
HMMT February 2011 — CALCGEO Round — Problem 18
题目详情
- Collinear points A , B , and C are given in the Cartesian plane such that A = ( a, 0) lies along the x -axis, B lies along the line y = x , C lies along the line y = 2 x , and AB/BC = 2. If D = ( a, a ), the circumcircle of triangle ADC intersects y = x again at E , and ray AE intersects y = 2 x at F , evaluate AE/EF .
解析
- Collinear points A , B , and C are given in the Cartesian plane such that A = ( a, 0) lies along the x -axis, B lies along the line y = x , C lies along the line y = 2 x , and AB/BC = 2. If D = ( a, a ), the circumcircle of triangle ADC intersects y = x again at E , and ray AE intersects y = 2 x at F , evaluate AE/EF . Answer: 7 Calculus & Geometry Individual Test Q P C D B F E O A Let points O , P , and Q be located at (0 , 0), ( a, 2 a ), and (0 , 2 a ), respectively. Note that BC/AB = 1 / 2 implies [ OCD ] / [ OAD ] = 1 / 2, so since [ OP D ] = [ OAD ], [ OCD ] / [ OP D ] = 1 / 2. It follows that [ OCD ] = [ OP D ]. Hence OC = CP . We may conclude that triangles OCQ and P CA are congruent, so C = ( a/ 2 , a ). It follows that ∠ ADC is right, so the circumcircle of triangle ADC is the midpoint of AC , which is located at (3 a/ 4 , a/ 2). Let (3 a/ 4 , a/ 2) = H , and let E = ( b, b ). Then the power of the point O with respect to the circumcircle of ADC is OD · OE = 2 ab , but it may also be computed as 2 2 OH − HA = 13 a/ 16 − 5 a/ 16 = a/ 2. It follows that b = a/ 4, so E = ( a/ 4 , a/ 4). We may conclude that line AE is x + 3 y = a , which intersects y = 2 x at an x -coordinate of a/ 7. Therefore, AE/EF = ( a − a/ 4) / ( a/ 4 − a/ 7) = (3 a/ 4) / (3 a/ 28) = 7. Remark: The problem may be solved more quickly if one notes from the beginning that lines OA , OD , OP , and OQ form a harmonic pencil because D is the midpoint of AP and lines OQ and AP are parallel. Calculus & Geometry Individual Test