Track / Stage 5 / 344 of 400 #944 of 1964

Problem 944

AIME late

Geometry Difficulty 5.9 Prove it

It is known that a certain point $M$ is equidistant from two intersecting lines $m$ and $n$ . Prove that the orthogonal projection of point $M$ onto the plane of lines $m$ and $n$ lies on the bisector of one of the angles formed by lines $m$ and $n$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $M_1$ be the orthogonal projection of point $M$ onto the plane $\alpha$ passing through the lines $m$ and $n$ ; $A$ be the point of intersection of lines $m$ and $n$ ; $P$ and $Q$ be the feet of the perpendiculars dropped from point $M$ to lines $m$ and $n$ respectively. Since $M_1P$ and $M_1Q$ are the orthogonal projections of the oblique lines $AP$ and $AQ$ onto the plane $\alpha$ , by the theorem of three perpendiculars, $M_1P \perp m$ and $M_1Q \perp n$ . Since $MP = MQ$ , it follows that $M_1P = M_1Q$ , i.e., point $M_1$ is equidistant from the sides of angle $PAQ$ . Therefore, point $M_1$ lies on the bisector of this angle.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.