A Solution to Crux Mathematicorum Problem 5109

Problem 5109

Proposed by Ion Patrascu

Let $ABC$ be an isosceles right-angled triangle with $BA=BC$ . On the segment $BC$ we consider the point $M$ to be mobile and denote by $N$ its symmetry with respect to $C$ . The points $P$ and $Q$ are the projections of $B$ onto $AM$ and $AN$ , respectively. Prove that the line $PQ$ passes through a fixed point.

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.

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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Solution

A projective generalization

Diagrams and complete source