Consider a natural number n≥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 8n2.
Problem
Source: Moldova TST 2024 P5
Tags: combinatorial geometry
09.06.2024 21:22
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 x1,x2,…,xn. Since (x1+x2)+(x2+x3)+⋯+(xn+x1)≤8 there must exist a index i for which xi+xi+1≤8/n. Now we can just choose the triangle with side lengths xi and xi+1 that has area xixi+1sin(θ)/2≤8/n2.
10.06.2024 11:54
I will show that there always is a triangle with area <16πn3. Assume not, let the vertices of the polygon be Ai, and consider li to be the length of the side connecting Ai and Ai+1.Let αi=∠Ai−1AiAi+1 Considering a triangle formed by 3 adjacent vertices, it's area is equal to li−1lisinαi2. Taking the product we get ∏li−1lisinαi2≥(16πn3)n, or regrouping that ∏li2∏sinαi≥(32πn3)n, now knowing that αi≤π we get sinαi=sin(π−αi)≤π−αi, thus ∏sinαi≤∏(π−αi), and by AM−GM that ∏(π−αi)≤(∑ni=1(π−αi)n)n=(2πn)n, and substituting this back we get that ∏li≥(4n)n, but by AM−GM and ∑li≤4 we get that ∏li≤(4n)n however equality, can't hold everywhere. This is enough to show the inequality for n≥7, and for other n just use ∏sinαi≤∏sin2πn, and now instead of the sinx<x idea just substitute the actual values. Looks like asymptotically, n3 is the best power we can get, however the constant seems to not be the best one, as equality never holds.