MHB Show that there are points Q,R on C such that the triangle PQR is equilateral.

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Let $C$ be a smooth closed curve (no corners) in the plane with a convex interior,
and $P$ a given point on $C$. Show that there are points $Q,R$ on $C$ such that the
triangle $PQR$ is equilateral.
 
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Suggested solution:

Let an angle of $60^{\circ}$ revolve counter-clockwise about $P$, with initial position of one of the arms tangent to $C$ at $P$. The intercepts of the two arms are initially $0$ and some $q > 0$. Turn the angle until the other arm becomes tangent to $C$, and the intercepts are now some $r > 0$ and $0$. Hence the difference of the intercepts changes from: $0-q < 0$ to $r-0 > 0$. By continuity there is a position of the two arms $\overline{PQ}, \overline{PR}$ where $|PQ| = |PR|$, hence the triangle $PQR$ is equilateral.
 
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