What is the envelope formed by a triangle fixed to two points? Take two fixed points in a plane and a triangle of fixed shape. Constrain two sides of the triangle to each touch one of the two points. As the triangle moves under this constraint the third side forms an envelope curve. Is this curve just an elliptical arc or something else?
This question arises in "lost in a forest" type problems.
 A: Not an answer, just an illustration.

               


A: By elementary differential geometry, the locus is an arc of a circle.
A: With thanks to @Joseph O'Rourke and @goleta who gave me the clues I will post a solution to my own question. I will use the nomenclature for the points provided by @Aaron Meyerowitz.

Construct a triangle $OPQ$ with the given side $PQ$ such that it is similar to the reflection of the given triangle $ABC$ i.e. $\angle OPQ$ = $\angle ABC$ and $\angle OQP$ = $\angle ACB$.
Draw the circumscribed circle of triangle $OPQ$. Because $\angle POQ$ = $\angle CAB$ = $\angle PAQ$ it follows that $A$ must also lie on this circle (angles subtended from points on a circle by a chord of that circle are equal)
Also, $\angle OAP$ = $\angle OQP$ = $\angle ACB$, therefore $OA$ is parallel to $CB$. Therefore the distance $OY$ from the point $O$ to the side $BC$ is equal to the distance $r$ from the point $A$ to the side $BC$ and this is the fixed height of the triangle. Therefore the side $BC$ is tangent to the circle with centre $O$ and radius $r$ and the envelope of the edge must be an arc of that circle.
If anyone knows a prior proof of this please comment so that I can cite it.
