CG with convex polygon inside a square and weak bound

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CG with convex polygon inside a square and weak bound

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Source: Moldova TST 2024 P5

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Marius_Avion_De_Vanatoare

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Marius_Avion_De_Vanatoare

#1 h Jun 9, 2024, 5:33 PM

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Consider a natural number $n \ge 3$ . A convex polygon with $n$ sides is entirely placed inside a square with side length 1. Prove that we can always find three vertices of this polygon, the triangle formed by which has area not greater than $\frac{8}{n^2}$ .

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#2 h Jun 9, 2024, 6:22 PM • 1 Y

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Notice that the polygon has perimeter at most $4$ as if one convex polygon lies in another it has smaller perimeter. Let the sides of the polygon be $x_1,x_2,\dots,x_n$ . Since $(x_1+x_2)+(x_2+x_3)+\dots+(x_n+x_1)\leq 8$ there must exist a index $i$ for which $x_i+x_{i+1} \leq 8/n$ . Now we can just choose the triangle with side lengths $x_i$ and $x_{i+1}$ that has area $x_ix_{i+1}\sin(\theta)/2\leq 8/n^2$ .

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Marius_Avion_De_Vanatoare

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Marius_Avion_De_Vanatoare

#4 h Jun 10, 2024, 8:54 AM

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I will show that there always is a triangle with area $< \frac{16\pi}{n^3}$ . Assume not, let the vertices of the polygon be $A_i$ , and consider $l_i$ to be the length of the side connecting $A_i$ and $A_{i+1}$ .Let $\alpha_i=\angle A_{i-1}A_iA_{i+1}$ Considering a triangle formed by 3 adjacent vertices, it's area is equal to $\frac{l_{i-1}l_{i}\sin{\alpha_i}}{2}$ . Taking the product we get $\prod \frac{l_{i-1}l_{i}\sin{\alpha_i}}{2} \ge (\frac{16\pi}{n^3})^n$ , or regrouping that $\prod{l_i}^2\prod{\sin{\alpha_i}} \ge (\frac{32\pi}{n^3})^n$ , now knowing that $\alpha_i \le \pi$ we get $\sin{\alpha_i}=\sin{(\pi-\alpha_i)} \le \pi-\alpha_i$ , thus $\prod{\sin{\alpha_i}} \le \prod{(\pi-\alpha_i)}$ , and by $AM-GM$ that $\prod{(\pi-\alpha_i)} \le (\frac{\sum_{i=1}^{n} {(\pi-\alpha_i})}{n})^n=(\frac{2\pi}{n})^n$ , and substituting this back we get that $\prod {l_i} \ge (\frac{4}{n})^n$ , but by $AM-GM$ and $\sum{l_i} \le 4$ we get that $\prod {l_i} \le (\frac{4}{n})^n$ however equality, can't hold everywhere.
This is enough to show the inequality for $n \ge 7$ , and for other $n$ just use $\prod{\sin{\alpha_i}} \le \prod{\sin{\frac{2\pi}{n}}}$ , and now instead of the $\sin{x}<x$ idea just substitute the actual values. Looks like asymptotically, $n^3$ is the best power we can get, however the constant seems to not be the best one, as equality never holds.

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