# Least area of maximal triangle inside convex $n$-gon.

A previous question Maximum area of triangle inside a convex polygon asks to show that for any convex polygon $P_s$ of $s$ sides having area $1$ there is a triangle contained in that polygon with area at least $c$, where $c = \frac38$. It also asks how much larger one can make $c$ and still have a true statement. The second part of the question is somewhat difficult; I speculated that the largest $c$ for such a statement is $\frac{3\sqrt{3}}{4\pi}$, which is the area ratio for an equilateral triangle inscribed in a circle.

If that conjecture holds, then defining $c_s$ as the largest $c$ such that any convex polygon of unit area and $s$ sides contains a triangle of area at least $c_s$, then for all $s$, $c_s > \frac{3\sqrt{3}}{4\pi}$, and for any $\epsilon > 0$ there exists some $s$ such that $c_s < \frac{3\sqrt{3}}{4\pi}+\epsilon$. As I said, proving this seems to be hard.

This question, which should be easier, restricts the previous question to specifically convex polygons of $5, 6$, or $7$ sides. Here I will show how to setlle the matter for $4$ sides:

For $s=4$, it is easy to show that $c_4$ is at most $\frac12$: Consider a unit square. The largest inscribed triangles have width $1$ along a side, and the third vertex touching the opposite side; that is area $\frac12$. It is also easy to show that $c_4$ is at least $\frac12$: For convex quadrilateral $ABCD$ draw line $AC$ dividing the quadrilateral into to triangles. One of these triangles must have area greater than or equal to $\frac12$. So $c_4 = \frac12$.

For a regular hexagon, the largest inscribed triangle is that with verticies on alternating hexagon vertices, and this has area $\frac12$, so $c_6 \leq \frac12$. But I have not been able to find a convex hexagon of unit area that does not allow an included triangle of area at least $\frac12$, and this leads me to weakly suspect that $c_6 = \frac12$. (If that is so, then $c_5 = \frac12$ as well, since $c_s$ is a non-increasing function of $s$.)

Eventually for large enough $s$, $c_s$ will be less than $\frac12$. It looks like that should happen at relatively small $s$.

So my problem is, to determine $c_5$ and $c_6$, and to demonstrate that $c_7 < \frac12$ or $c_8 < \frac12$ by constructing a convex septagon or octagon of unit area that does not include any triangle of area at least $\frac12$.

• Your claim seems to be a known theorem due to Blashcke: books.google.it/… – Jack D'Aurizio Jul 20 '16 at 20:44
• We may notice that the largest inscribed triangle $ABC$ in our convex body $K$ has to have the following property: $C\in\partial K$ and the parallel to $AB$ through $C$ meets $K$ only at $C$. By a duality principle, the problem is equivalent to finding the circumscribed triangle with the least area. We may also notice that a large inscribed triangle is given by the largest triangle belonging to the inellipse of $K$. – Jack D'Aurizio Jul 20 '16 at 21:09
• – Jack D'Aurizio Jul 20 '16 at 21:18
• And an almost-trivial observation, useful if $n$ is large: the inscribed ellipse with maximum area (John-Loewner inellipse $E^-$) is unique. If it has area $\Delta$, the largest inscribed triangle in our convex polygon $P$ has at least area $\frac{3\sqrt{3}}{4\pi}\Delta$, since that is the area of the largest triangle inscribed in $E^-\subset P$. – Jack D'Aurizio Jul 20 '16 at 21:35
• @MarkFischler, Fleischer, R., Mehlhorn, K., Rote, G., Welzl, E., & Yap, C.-K. (1992). Simultaneous Inner and Outer Approximation of Shapes. Algorithmica, 8, 365-389. There is an online copy of this and in the appendix, there is a proof that $c_6 = 4/9$. – achille hui Jul 26 '16 at 20:30

This is not really an answer but a summary of what I can find:

• $c_5 = \frac{1}{\sqrt{5}}$, achieved by affine images of regular pentagon${}^{\color{blue}{[1]}}$.
• $c_6 = \frac{4}{9}{}^{\color{blue}{[1]}}$, achieved by affine images of a cyclic hexagon with vertices at $$\begin{cases} p(\pm\theta) &= \frac{1}{2\sqrt{2}}(\sqrt{5}, \pm \sqrt{3}),\\ p(\pm 3\theta) &= \frac{1}{4\sqrt{2}}(-\sqrt{5}, \pm 3\sqrt{3}),\\ p(\mp 5\theta) &= \frac{1}{8\sqrt{2}}(-5\sqrt{5}, \pm \sqrt{3}) \end{cases} \quad\text{ where }\quad \begin{cases} p(t) &= (\cos(t),\sin(t))\\ \theta &= \frac12\cos^{-1}\left(\frac14\right) \end{cases}$$
• For $n > 6$, we have${}^{\color{blue}{[2]}}$ $$c_n \le r(n) \stackrel{def}{=} \frac{ \sin\left(\frac{2\pi}{n}\left\lfloor \frac{n}{3}\right\rfloor\right) + \sin\left(\frac{2\pi}{n}\left\lfloor \frac{n+1}{3}\right\rfloor\right) + \sin\left(\frac{2\pi}{n}\left\lfloor \frac{n+2}{3}\right\rfloor\right) }{n\sin\left(\frac{2\pi}{n}\right)}\tag{*1}$$ where $r(n)$ is the area of the largest inscribed triangle for a regular $n$-gon of unit area.

It can show that for $k \ge 2$, we have in general $$r(3k-1) \color{red}{<} r(3k) > r(3k+1) > r(3k+2)$$ Since $c_n$ is non-increasing in $n$, we can improve $(*1)$ a little bit to $$c_n \le \begin{cases} r(n),& n \not\equiv 0 \pmod 3\\r(n-1),& n \equiv 0 \pmod 3\end{cases}\quad\text{ for }n > 6$$

In particular, we have

• $c_7 \le r(7) = \frac{\sin\left( \frac{6 \pi}{7}\right) +2 \sin\left( \frac{4 \pi}{7}\right) }{7 \sin\left( \frac{2 \pi}{7}\right)} \approx 0.4355596199317579$

• $c_8, c_9 \le r(8) = \frac{2+\sqrt{2}}{8} \approx 0.4267766952966369$

Notes/References

• $\color{blue}{[1]}$ - Fleischer, R., Mehlhorn, K., Rote, G., Welzl, E., & Yap, C.-K. (1992).
Simultaneous Inner and Outer Approximation of Shapes.
Algorithmica, 8, 365-389

• $\color{blue}{[2]}$ - Kiyoshi Hosono, Ferran Hurtado, Masatsugu Urabe, Jorge Urrutia (2003).
On a Triangle with the Maximum Area in a Planar Point Set.
Proceeding IJCCGGT'03 Proceedings of the 2003 Indonesia-Japan joint conference on Combinatorial Geometry and Graph Theory. Pages 102-107