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, 2011

Geometry Difficulty 8.3 Shortlist Prove it Baltic Way

Let $A$ and $B$ be two circles, external to each other. Let $\ell$ be a line not meeting the circles. For any point $X$ on $\ell$ , let $E$ be a point of contact of a tangent to $A$ through $X$ , and $F$ a point of contact of a tangent to $B$ through $X$ . Find the position of $X$ on $\ell$ such that $EX + FX$ is minimized.

Solution

Denote the centres of $A$ and $B$ by $A$ and $B$ , respectively. Let $C$ and $D$ be the feet of the perpendiculars from $A$ and $B$ to $\ell$ . Let $G$ be a point of contact of $A$ and a tangent to $A$ from $C$ . Define $H$ similarly on $B$ .

Now, by Pythagoras,
$XE^2 = AX^2 - AE^2 = XC^2 + CA^2 - AE^2 = XC^2 + AG^2 + GC^2 - AE^2 = XC^2 + CG^2.$
Now choose a point $M$ on $CA$ such that $CM = CG$ . ( $M$ is a point of intersection of $CA$ and the circle with center $C$ through $G$ ). Then $XE = XM$ . Similarly, if $N$ is the point on $DB$ such that $DN = DH$ and $M$ , $N$ lie on different sides of $\ell$ , then $XF = XN$ . So minimizing $EX + FX$ is equivalent to minimizing $MX + XN$ . Clearly, if $P$ is the point of intersection of the lines $\ell$ and $MN$ , then $P$ solves the problem.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.