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

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For a smooth closed curve C with a convex interior and a point P on C, it is possible to find points Q and R on C such that triangle PQR is equilateral. The suggested solution involves utilizing the properties of convexity and the continuous nature of the curve. By analyzing the angles and distances from point P to the curve, one can demonstrate the existence of such points Q and R. The geometric properties of the triangle and the curve facilitate this construction. Therefore, the conclusion is that the configuration of points Q and R can indeed form an equilateral triangle with point P.
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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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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