Let $\text{Pol}_n$ be the set of all convex polygons on a plane with $n$ sides. For $P\in \text{Pol}_n$ denote by $\text{Tr}(P)$ the set of all triangles whose vertices are some vertices of $P$. I want to find an explicit formula for the function $$ \Phi(n)=\inf\limits_{P\in \text{Pol}_n}\max\limits_{T\in \text{Tr}(P)}\frac{\text{area}(T)}{\text{area}(P)} $$ It is not hard to prove that $\Phi(3)=1$, $\Phi(4)=1/2$. For $n\geq 5$ we have an estimation $\Phi(n)\geq 1/(n-2)$.
Minmax problem for polygons
17
$\begingroup$
geometry
optimization
-
3Suppose $\triangle ABC$ has maximal area among sub-triangles of $P$. Among vertices of $P$, $A$ is at maximal distance from line $BC$: that is, $P$ lies between the line through $A$ parallel to $BC$ and the reflection of that line in $BC$. Likewise, $B$ (resp $C$) is at maximal distance from line $CA$ (resp $AB$). The parallel-line boundaries determine $\triangle A'B'C'$ for which $\triangle ABC$ is the (similar) midpoint triangle. Thus, $|P| \le |\triangle A'B'C'| = 4 \; |\triangle ABC|$, so that $\frac{1}{4} \le \; \frac{|\triangle ABC|}{|P|} = \max_T \frac{|T|}{|P|}$, regardless of $n$. – 2011-12-12
-
0By the way: For $\frac{|\triangle ABC|}{|P|}$ to attain the minimum $\frac{1}{4}$, we'd need that $P$ covers the entirety of $\triangle A'B'C'$. This would require each of $A'$, $B'$, and $C'$ to be vertices of $P$; but then $\triangle A'B'C'$ is itself a sub-triangle of $P$, contradicting $\triangle ABC$'s maximality. Thus, the maximal ratio of areas is *strictly* greater than $\frac{1}{4}$. Further, it seems the infemum is also strictly greater than $\frac{1}{4}$, since $P$ can't even have three vertices "close" to $A'$, $B'$, and $C'$ (or even just two vertices close to two of them). – 2011-12-12
-
0On the other hand, for a regular $n$-gon with $n$ large the maximal triangle has area approximately $3 \sqrt{3}/(4 \pi)$ times the area of the $n$-gon. So $\limsup_{n \to \infty} \Phi(n) \le \frac{3 \sqrt{3}}{4 \pi}$. – 2011-12-19
-
1Just to acknowledge the cases explicitly: Clearly, if $\Phi(3)=1$. If $n=4$, we minimize the ratio by taking the fourth vertex to be one of $A'$, $B'$, or $C'$, so that $\Phi(4) = 1/2$. Indeed, if, for any $n$, $P$ contains any one of those vertices, say $A'$, then its boundary must exactly match that of parallelogram $A'BAC$ (with sub-divided edges), giving ratio $1/2$. Thus, we can concentrate on situations where $P$ contains none of those vertices; and where $P$ has vertices in at least two of $\triangle A'BC$, $\triangle AB'C$, and $\triangle ABC'$. – 2011-12-20
-
0It seems that the formula would be the same if we take vertices of triangle from boundary of P, or points bounded by P. Would this help the computation? – 2011-12-23