Two points $A$ and $B$ lie inside an acute angle. Construct an isosceles triangle $PQR$ whose base $PR$ lies on one side of the angle, whose third vertex $Q$ lies on the other side of the angle, and whose two lateral sides $PQ$ and $QR$ pass through $A$ and $B$, respectively.
I was thinking of reflecting $A$ and $B$ in one of the arms and then figuring out some line which will intersect the angle to give the points, but cant figure out any useful reflection. Please help. Thanks a lot.
On
Suppose line $AB$ is not perpendicular to side $s$ of the angle. Let $a$ and $b$ be the distances of $A$ and $B$ from $s$. Let $h_A$ and $h_B$ be the lengths of the chords through $A$ and $B$, perpendicular to $s$. We want to find on $S$ the midpoint $H$ of $PR$.
If $h$ is the altitude of $PQR$, then by similar triangles we have $$ {AH\over BH}={1-a/h\over 1-b/h}={h_A-h\over h-h_B}. $$ This equation can be solved for $h$ to get $$ h={1\over4}\left( h_A+h_B+a+b+\sqrt{(h_A+h_B+a+b)^2-8(ah_B+bh_A)} \right). $$ Once $h$ is found, completing the construction is easy.
Hint: Let's assume that the problem is solved and $PQR$ is the required triangle. Reflect $A$ in $OQ$ to $A'$.($O$ is the vertex of the acute angle and $OY$ is the arm having $Q$)
$\angle A'QB=\angle A'QA+\angle PQR$ and $\angle A'QA=2\angle OQP=2(\angle QPR-\angle QOR)$. Thus $\angle A'QB=180^o-2\angle QOR$. Thus $\angle A'QA $ is fixed and we can find $Q$ as the point of intersection of $OY$ an the circle with arc $A'B$, its not difficult to see that this circle will be unique.