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Prove that within any convex polygon of area $A$, there exists a triangle with area at least $cA$, where $c=\tfrac{3}{8}$. Are there any better constants $c$?

I'm not sure how to approach this problem. It is easily proven that such a triangle should have its vertices on the perimeter of the polygon, but I don't know how to proceed from here.

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  • $\begingroup$ Perhaps a fruitful line of investigation is to proceed by the number of sides of the convex polygon. Clearly $c=1$ when $A$ itself is a triangle. For a quadrilateral, $c=1/2$. We are aided in this analysis by the fact that affine transformations do not affect the ratio of area of a triangle inscribed in a convex polygon. $\endgroup$ – hardmath Jul 20 '16 at 23:26
  • $\begingroup$ In addition to showing that the maximum area triangle has vertices on the perimeter, it can be shown that the vertices of the triangle may be chosen to be vertices of the convex polygon, without loss of generality. $\endgroup$ – hardmath Jul 21 '16 at 4:40
  • $\begingroup$ @hardmath: Are you sure about the vertex comment? Consider a regular pentagon. A vertex-only triangle invoIves two adjacent vertices. It looks like sliding one of those up the side away from the other might increase the area of the triangle. I have not checked this, though. $\endgroup$ – Mark Fischler Jul 21 '16 at 13:44
  • $\begingroup$ @MarkFischler : Let $c$ be a vertex of a maximum area triangle inscribed in a convex polygon. If $c$ lies on the interior of an edge of the polygon, consider the slope of the edge in relation to that of the triangle side (base) opposite to $c$. $\endgroup$ – hardmath Jul 21 '16 at 14:32
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As first pointed out by Mark Fischler in his comment, one can take $c = \frac{3\sqrt{3}}{4\pi}$.

In addition to Steiner symmetrization mentioned in Jack D'Aurizio's answer$\color{blue}{{}^{[1],[2]}}$, there is another elegant analytic approach which can be generalized to inscribed $n$-gon.

E. Sas (1939) - For any $n \ge 3$, let $c_n = \frac{n}{2\pi}\sin\left(\frac{2\pi}{n}\right)$. For any convex body $B$ in the plane, there exists a $n$-gon $P$ inside $B$ such that $$\verb/Area/(P) \ge c_n \verb/Area/(B) \tag{*1}$$

When $n = 3$ and $B$ is a convex polygon, the claim we can take $c = c_3 = \frac{3\sqrt{3}}{4\pi}$ follows immediately.

Following argument is based on a paper by E. Sas in German$\color{blue}{{}^{[3]}}$. All mistakes are mine and in case anything looks fishy. Please refer to E. Sas paper for correct statement.


Since I am lazy, I will assume $B$ is a convex body whose boundary $\partial B$ is a smooth Jordan curve. This avoids all sort of potential pathologies and save me from justifying a lot of stuff.

Let $2\ell$ be diameter of $B$. Let $L$, $R$ be two points on $\partial B$ at a distance $2\ell$ apart. Choose a coordinate system such that $L,R$ is located at $(-\ell,0)$ and $(\ell,0)$ respectively. Under such a coordinate system, $\partial B$ has a parametrization $\gamma$ of the form:

$$[0,2\pi] \ni t \quad\mapsto\quad \gamma(t) = (\ell\cos t, e(t)\sin t ) \in \partial B$$

where $e(t) > 0$ is some smooth function. Extend $\gamma(t)$ and $e(t)$ to smooth periodic functions over $\mathbb{R}$ with period $2\pi$.

For any fixed $t \in [0,2\pi]$ and $k \in \mathbb{Z}$, let $t_k = t + \frac{2k\pi}{n}$. It is clear $t_{k+n} = t_k$.

Let $P(t)$ be the $n$-gon with vertices $\gamma(t_0), \gamma(t_1), \gamma(t_2), \ldots, \gamma(t_{n-1})$. Since $B$ is convex and $\gamma(t_k) \in B$, $P$ lies inside $B$.

Let $f(t)$ be the area of $P(t)$. It is easy to work out

$$f(t) = \frac{\ell}{2} \sum_{k=1}^n e(t_k)\sin t_k(\cos(t_{k-1}) - \cos(t_{k+1})) = \ell\sin\left(\frac{2\pi}{n}\right)\sum_{k=1}^n e(t_k)\sin^2(t_k)$$

Now treat $t$ as a variable and average $f(t)$ over $[0,2\pi]$, one get

$$\frac{1}{2\pi}\int_0^{2\pi} f(t) dt = \frac{n}{2\pi}\sin\left(\frac{2\pi}{n}\right)\times \ell\int_0^{2\pi} e(t)\sin^2(t)dt = c_n \verb/Area/(B)$$

This implies there exists a $t_{*} \in [0,2\pi]$ such that $f(t_{*}) \ge c_n \verb/Area/(B)$. In other words, there exists a poylgon $P = P(t_{*}) \subset B$ whose area is at least $c_n$ of that of $B$.

Back to the problem at hand for convex polygon.

For any polygon $P$, let $|P|$ be its number of sides. Given any convex polygon $Q$ and any $0 < c < c_n$, approximate $Q$ by a convex body $B$ with smooth boundary whose area is at least $\frac{c}{c_n}\verb/Area/(Q)$. By $(*1)$, there is a $n$-gon $P \subset B \subset Q$ with $\verb/Area/(P) \ge c_n\verb/Area/(B) \ge c\verb/Area/(Q)$. Since the set of polygon $P \subset Q$ with $|P| \le n$ is compact under the topology induced from $\mathbb{R}^2$ and $\verb/Area/(\cdot)$ is continuous with respect to this topology, there is a polygon $P_{*} \subset Q$ with $|P_{*}| \le n$ and $\verb/Area/(P_*) \ge c_n \verb/Area/(Q)$. If $|P_{*}| < n$, we can turn $P_*$ to a $n$-gon by adding some extra vertices along its edges.

Fix $n$ to $3$, this means there is a triangle inside $Q$ whose area at least $c_3 = \frac{3\sqrt{3}}{4\pi}$ of that of $Q$.

Notes

  • $\color{blue}{[1]}$ - For a proof of $(*1)$ when $n = 3$, see this answer by Christian Blatter.

  • $\color{blue}{[2]}$ - Christian Blatter's answer uses Steiner symmetrization. For more detail, please refer to
    Wilhelm Blaschke, Über affine Geometrie III: Eine Minimumeigenschaft der Ellipse. Leipziger Berichte 69 (1917), pages 3–12.

  • $\color{blue}{[3]}$ - E. Sas, Über eine Extremaleigenschaft der Ellipsen, Compositio Math. 6 (1939), 468–470.

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  • $\begingroup$ Indeed from your proof I don't where you used that $\partial D$ is smooth? $\endgroup$ – user99914 Jul 26 '16 at 9:47
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    $\begingroup$ @ArcticChar This is because I am lazy. When the boundary is smooth, it avoid all sort of possible pathologies and save me the trouble to justify a lot of stuff. $\endgroup$ – achille hui Jul 26 '16 at 10:19
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There is not-so-painful way to prove a weaker inequality, i.e. that for any convex polygon $P$ with unit area there is a triangle $T\subset P$ with area $\geq \color{purple}{\frac{4}{21}}$. The approach is simple: let, counter-clockwise, $P_1,P_2,\ldots,$ $P_n=P_0,$ $ P_{n+1}=P_1$ be the vertices of the polygon, and let $P_{a-1}P_{a} P_{a+1}$ be the triangle with the smallest area among all the triangles of such a form. We remove the vertex $P_a$ and re-index the vertices, then continue the removal procedure until our polygon is a triangle. That is our $T$. It is not difficult to check that if the area of $P$ is $1$, the area of $T$ is at least $$ \prod_{n\geq 4}\left(1-\frac{4\sin^2\frac{\pi}{n}}{n}\right)\geq \frac{4}{21}. $$ Some comments of mine to your revised question may be useful to improve the above inequality/approach: for instance, there is for sure a triangle $T\subset P$ with area $$ \frac{3\sqrt{3}}{4\pi} \Delta(E^-) $$ where $E^-$ is the John-Loewner inellipse (the inscribed ellipse in $P$ with maximal area), and we have very accurate bounds when the vertices of $P$ all lie on a ellipse (so for any $n\leq 5$).

The original Blaschke argument is outlined in this answer by Christian Blatter.
Blaschke's approach is extremely slick: there is a Steiner symmetrization procedure that is area-preserving, so we may assume without loss of generality that the original polygon is a cyclic polygon. But that gives a way easier optimization problem, just related with the distribution of $n$ points on a circle.

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The statement is true using any $c < \frac{3\sqrt{3}}{4\pi} \approx 0.413$, which is more than $ \frac38$. But I have tried and failed to find an easy proof.

EDITED A DAY LATER

If you impose a restriction on the number $s$ of sides in the convex polygon, the $c$ for that set of polygons becomes greater. For $s=3$, $c=1$ of course. For $s=4$ it is easy to show that $c=\frac12$: $c$ is at least $\frac12$, as demonstrated by a line connecting two non-adjacent vertices (one of the triangles thus formed must have at least half the area); and $c$ is at least $\frac12$, as demonstrated by a square.

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  • $\begingroup$ Heh, I had not seen your latest edit before making a similar observation as a comment on the Question. $\endgroup$ – hardmath Jul 20 '16 at 23:28
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Let's turn the problem on its head. Let triangle $\Delta abc$ be a maximum area triangle contained in convex polygon $\mathscr{P}$. We are asked to show that the area of the triangle is at least $C$ times the area $A$ of $\mathscr{P}$, where $C=3/8$. We offer a simple proof that $C \ge 1/4$ can be achieved (and improved upon).

As noted in the Question, the points $a,b,c$ must lie on the perimeter of $\mathscr{P}$. Further these points may be chosen, without loss of generality, to be vertices of the polygon $\mathscr{P}$. To wit: If (say) vertex $a$ is in the interior of an edge $e$ of the polygon, then sliding $a$ one way or the other would increase the area of $\triangle abc$ unless the edge $e$ is parallel to the side $bc$ of the triangle opposite to $a$ (by the familiar formula for area of a triangle being one-half the height of the triangle with respect to base $bc$).

Now any affine transformation applied to $\mathscr{P}$ preserves the ratio of its area to that of a triangle inscribed within it. Therefore we may apply some affine transformation that maps $\triangle abc$ to an equilateral triangle, and the ratio of areas remains the same. It also preserves convexity of $\mathscr{P}$, and the identification of $a,b,c$ with vertices of $\mathscr{P}$. Henceforth we will simply refer to the transformed (equilateral) triangle as $\triangle abc$ and the transformed polygon as $\mathscr{P}$.

A picture may help the Reader to follow our argument to its next conclusion, as we draw lines through the vertices of an equilateral triangle parallel to their respective opposing sides:

equilateral triangle with extended lines through vertices Fig. 1 Equilateral triangle $\triangle abc$ with lines through vertices

Here $\triangle abc$ is the central equilateral triangle. No point $p$ of $\mathscr{P}$ can lie outside the circumscribing equilateral triangle, because to do so would induce a triangle with area strictly greater than $\triangle abc$. For if (say) point $p$ were on the outside of the outer line through $a$, opposite from side $bc$, then $\triangle pbc$ would have area strictly greater than $\triangle abc$.

Therefore $\mathscr{P}$ is constrained to have area no greater than four times the area of $\triangle abc$. This proves that $C \ge 1/4$ can be achieved, that the maximum area of an inscribed triangle is at least one-fourth the area of convex polygon $\mathscr{P}$.

I know a somewhat tedious but elementary way to improve this to $C = 3/11$, still a good ways short of the desired $C = 3/8$, but suggesting that improvements on this argument might be worth pursuing.

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  • $\begingroup$ +1 I essentially have the same idea. a simple way to improve from $1/4$ to $1/3$ is consider the family of triangles inside the big triangle obtainable by expanding $\triangle ABC$ by a small factor $1+\epsilon$, reflect wrt origin and then parallel translate... $\endgroup$ – achille hui Jul 22 '16 at 14:08

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