HMMT 二月 2005 · 冲刺赛 · 第 14 题

HMMT February 2005 — Guts Round — Problem 14

[7] Three noncollinear points and a line ℓ are given in the plane. Suppose no two of the points lie on a line parallel to ℓ (or ℓ itself). There are exactly n lines perpendicular to ℓ with the following property: the three circles with centers at the given points and tangent to the line all concur at some point. Find all possible values of n . 2 2

解析

Three noncollinear points and a line are given in the plane. Suppose no two of the points lie on a line parallel to ` (or ` itself). There are exactly n lines perpendicular to with the following property: the three circles with centers at the given points and tangent to the line all concur at some point. Find all possible values of n . Solution: 1 The condition for the line is that each of the three points lies at an equal distance from the line as from some fixed point; in other words, the line is the directrix of a parabola containing the three points. Three noncollinear points in the coordinate plane deter- mine a quadratic polynomial in x unless two of the points have the same x -coordinate. Therefore, given the direction of the directrix, three noncollinear points determine a parabola, unless two of the points lie on a line perpendicular to the directrix. This case is ruled out by the given condition, so the answer is 1. 2 2