Optimisation Problem about convex curve

Question

A smooth closed curve C is said to be convex if it lies wholly to one side of each tangent . Show that for the triangle of minimum area circumscribed about C that each side is tangent to C at its midpoint.

Answer 1

A geometric lemma first: Let $P$ be a point in the interior of angle $CAB$, and let $XPY$ be a line segment with $X$ on ray $AB$ and $Y$ on ray $AC$, and further suppose $P$ bisects the segment $XY$. Then any other triangle formed by passing a line through $P$ and intersecting the line with the sides of angle $CAB$ has larger area than the area of triangle $AXY.$ This lemma is fairly easy to show, though I could not locate a reference to it. A proof can be based on the triangles formed by two nearby chords through $P$ and the formula $(1/2)uv\sin\theta$ for the area of a triangle; I can include some details if needed.

Now to apply this to the case at hand. Assume given an angle at $A$ for which the closed curve $\gamma$ is tangent to the sides of angle $A$ at points $M$ and $N$. We wish to find where on the rest of $\gamma$ to place the third vertex. If we suppose we have a minimal area triangle superscribed about $\gamma,$ and this triangle cuts the ray $AM$ at $B$, and ray $AN$ at $C$, our claim is that the point $P$ at which the curve $\gamma$ is tangent to segment $BC$ must bisect that segment. For assume not. Then a copy of line $BC$ can be rolled along $\gamma$ in such a way that it ends up tangent to $\gamma$ at a point $Q$, and this new tangent line at $Q$ is cut by the sides of angle $A$ into a segment such that $Q$ now bisects this segment. It is clear that $Q$ is interior to the triangle $ABC$. By the above geometric fact, we may now rotate the line through $Q$ about $Q$, which increases its length, until it is parallel to side $BC$. But then clearly the area formed by the sides of the angle $A$ and the tangent line through $Q$ is less than the area of triangle $ABC$, which was supposed to be minimal.

The same argument may be applied to the other sides, so that indeed all three sides of the minimal triangle are bisected by the contact points of $\gamma.$