Least possible area of a triangle with vertices on...

Question

Assume you have a regular polygon of $n$ sides and its circumcircle $(n>3)$. Assume that $A,B$ and $C$ are $3$ different vertices of the polygon. Is the triangle $ABC$ with least possible area the one that is formed by $3$ consecutive vertices? Can this be applied to a polygon of any number of sides? What would be the traingle with most area?

Answer 1

Yes, the triangle with least possible area one that is formed by 3 consecutive vertices.

A regular convex polygon $P$ has a circumscribed circle. A triangle whose vertices are a subset of the vertices of $P$ has the same circumscribed circle. The area $a$ of a triangle with angles $\alpha, \beta, \gamma$ and circumscribed circle with radius $r$ is [ref] \begin{equation}a=2r^2 sin(\alpha) \sin(\beta) \sin(\gamma)\,.\end{equation}

The radius $r$ is the same for all triangles whose vertices are a subset of the vertices of $P$. Hence, the triangle with the smallest area is the triangle that minimizes $sin(\alpha) \sin(\beta) \sin(\gamma)$ and the triangle with the largest area is the triangle that maximizes $sin(\alpha) \sin(\beta) \sin(\gamma)$.

Every angle in a triangle is smaller than 180° and the sum of the three angles is 180°. Looking at $\sin(x)$ on the interval [0°,180°], one can see that one minimizes $sin(\alpha) \sin(\beta) \sin(\gamma)$ when choosing one angle to be close to 180° as possible and the other two angles as small as possible. The largest angle formed by a triplet of vertices of a regular polygon is $\angle ABC$, where $A, B, C$ are consecutive vertices. The smallest angle formed by a triplet of vertices of a regular polygon is $\angle BAC$. So a triangle whose vertices are a subset of the vertices of a regular convex polygon has a minimal area if its vertices $A,B,C$ are consecutive vertices of that polygon.