A Solution to Crux Mathematicorum Problem 5109

The original problem statement is intentionally not reproduced here. This article follows its notation and presents only the independently prepared solution published here, its diagram, and a related generalization. The complete statement is available in the official issue of Crux Mathematicorum.

Solution

The condition $BA=BC$ is unnecessary.

For a nondegenerate position with $B<M<C$ , let $D$ be a point such that $ABCD$ is a rectangle. Let

S=PQ\cap BC, \qquad R=PQ\cap AD,

Rectangle ABCD with M between B and C, N beyond C, perpendicular feet P and Q from B to AM and AN, and line PQ meeting BC at S and line AD at R. — Figure 1. The auxiliary rectangle and intersections used in the solution.Select the diagram to enlarge it.

Proof

Since $\triangle RAP\sim\triangle SMP$ ,

RA:SM=AP:MP=AB^2:MB^2.

Hence,

SM=\frac{RA\cdot MB^2}{AB^2}.

Similarly,

SN=\frac{RA\cdot NB^2}{AB^2}.

Here $S,M,N$ occur in this order. Therefore,

\begin{aligned} NM &=SN-SM\\ &=\frac{RA(NB+MB)(NB-MB)}{AB^2}\\ &=\frac{RA\cdot 2BC\cdot NM}{AB^2}. \end{aligned}

It follows that

RA=\frac{AB^2}{2BC}.

Therefore, the line $PQ$ passes through the fixed point $R$ .

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A projective generalization

More generally, the following proposition holds.

Proposition

Let $\Gamma$ be a nonsingular conic, and let $A$ and $B$ be distinct fixed points on $\Gamma$ . Let $\ell$ be a fixed line passing through $B$ , with $A\notin\ell$ . For a variable point $M\in\ell$ , let $N\in\ell$ be its conjugate under a fixed nonidentity projective involution on $\ell$ .

Let $P$ and $Q$ be the second intersections of the lines $AM$ and $AN$ with $\Gamma$ , respectively, counted with multiplicity. Then, as $M$ varies along $\ell$ , the line $PQ$ passes through a fixed point. At a fixed point of the involution, $PQ$ is interpreted as the tangent to $\Gamma$ .

The original problem is the special case in which $\Gamma$ is the circle with diameter $AB$ , $\ell$ is the line through $B$ perpendicular to $AB$ , and the involution on $\ell$ is the reflection in $C$ .

Proof

Define a projectivity $\phi:\ell\to\Gamma$ by letting $\phi(X)$ be the second intersection of the line $AX$ with $\Gamma$ .

The given involution on $\ell$ induces an involution on $\Gamma$ . Thus, if $M$ and $N$ are conjugate points on $\ell$ , then the corresponding points $P=\phi(M)$ and $Q=\phi(N)$ are conjugate points of a fixed involution on $\Gamma$ .

A classical theorem in projective geometry states that the lines determined by conjugate pairs of an involution on a conic are concurrent. Hence the line $PQ$ passes through a fixed point.

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