HMMT 二月 2016 · 几何 · 第 5 题
HMMT February 2016 — Geometry — Problem 5
题目详情
英文原题
- Nine pairwise noncongruent circles are drawn in the plane such that any two circles intersect twice.
( )
For each pair of circles, we draw the line through these two points, for a total of = 36 lines. Assume 9
that all 36 lines drawn are distinct. What is the maximum possible number of points which lie on at 2
least two of the drawn lines?
解析
英文解析
- Nine pairwise noncongruent circles are drawn in the plane such that any two circles intersect twice.
( )
For each pair of circles, we draw the line through these two points, for a total of = 36 lines. Assume 9
that all 36 lines drawn are distinct. What is the maximum possible number of points which lie on at 2
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
( )
of the lines can be made to be distinct. So the answer is 9
)2
(( ) ( ) − 2 = 462299
2 3
by just counting pairs of lines, and then subtracting off double counts due to radical centers (eachcounted three times).