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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?

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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.

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  • $\begingroup$ What if it isn´t a triangle? A pentagon for example. The pentagon with least possible area would have consecutive vertices? $\endgroup$ – Vmimi Oct 14 '18 at 3:21
  • $\begingroup$ Unfortunately, the proof in my answer works only for triangles, because every triangle has a circumscribed circle that touches all of its vertices. If you take 5 consecutive vertices of a regular convex $n$-polygon ($n>5$) to construct a pentagon that pentagon will not be regular and may be that some of the pentagon's vertices do not lie on its circumscribed circle. $\endgroup$ – Alice Schwarze Oct 14 '18 at 3:35
  • $\begingroup$ Actually, the reasoning in my first comment was incorrect. The vertices of course still lie on a circle. But I am not aware of a relation that links the area of such an irregular polygon with its angles. So a straightforward extension of this proof from triangles to shapes with more vertices does not seem possible. Perhaps one could triangulate the irregular polygon and show that all but one of those triangles need to be constructed on consecutive vertices to get a minimal area for the irregular polygon. $\endgroup$ – Alice Schwarze Oct 14 '18 at 18:08

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