Given a convex angle and a line segment of length k, determine the locus of those points... Got this problem last week on a geometry test, but didn't manage to solve. I have tried reflecting the points, but I didn't get anywhere, I am in 10th class, and I am familiar whith basic geometry tools.
Given a convex angle and a line segment of length k, determine the locus of those points inside the angle through which there exists a line cutting off a triangle of perimeter k from the angle.
 A: Let $O$ be the apex and $OA$, $OB$ be the two rays bounding the convex angle.
For any $P \in OA$, $Q \in OB$ such that $\triangle OPQ$ has perimeter $k$. Draw a circle tangent to $OA$, $OB$ and $PQ$ (within the convex angle, on different side of $PQ$ with respect to $O$). Let $A'$, $B'$ and $C'$ be the points on the circle tangent to rays $OA$, $OB$ and line segment $PQ$ respectively. Notice
$$PC' = PA'\quad\text{and}\quad QC' = QB'$$
We have
$$\begin{align} OA' + OB' &= (OP + PA') + (OQ + QB') = OP + OQ + (PC' + C'Q)\\ &= OP + OQ + PQ = k\end{align}$$
Since $OA' = OB'$, this implies $OA' = OB' = \frac{k}{2}$. This means the two tangent points $A', B'$ and hence the circle through $A',B',C'$ is independent of the choice of $P$, $Q$.
From this, it is easy to see the set of point $X$ where you
can find a line segment $PQ \ni X$ with $\triangle OPQ$ having perimeter $k$ is a curved triangular region.
Two sides of it lies on the rays $OA$, $OB$ with length $\frac{k}{2}$
(ie. $OA'$ and $OB'$). The remaining side is the circular arc $A'B'$ (the one facing $O$) on the circle mentioned above.
