HMMT 二月 2016 · 几何 · 第 5 题

HMMT February 2016 — Geometry — Problem 5

Nine pairwise noncongruent circles are drawn in the plane such that any two circles intersect twice. ( ) 9 For each pair of circles, we draw the line through these two points, for a total of = 36 lines. Assume 2 that all 36 lines drawn are distinct. What is the maximum possible number of points which lie on at least two of the drawn lines?

解析

Nine pairwise noncongruent circles are drawn in the plane such that any two circles intersect twice. ( ) 9 For each pair of circles, we draw the line through these two points, for a total of = 36 lines. Assume 2 that all 36 lines drawn are distinct. What is the maximum possible number of points which lie on at least two of the drawn lines? Proposed by: Evan Chen Answer: 462 The lines in question are the radical axes of the 9 circles. Three circles with noncollinear centers have a radical center where their three pairwise radical axes concur, but all other intersections between two ( ) 9 of the lines can be made to be distinct. So the answer is 2 ) (( ) ( ) 9 9 2 − 2 = 462 2 3 by just counting pairs of lines, and then subtracting off double counts due to radical centers (each counted three times).