HMMT 二月 2005 · 冲刺赛 · 第 14 题

HMMT February 2005 — Guts Round — Problem 14

个人笔记

[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 perpendicularto ℓ with the following property: the three circles with centers at the given points andtangent to the line all concur at some point. Find all possible values of n .
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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 perpendicularto ` with the following property: the three circles with centers at the given points andtangent 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 parabolacontaining the three points. Three noncollinear points in the coordinate plane determine 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 aparabola, unless two of the points lie on a line perpendicular to the directrix. Thiscase is ruled out by the given condition, so the answer is 1.
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