Mathematical and Physical Journal
for High Schools
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Problem A. 380. (October 2005)

A. 380. The convex n-sided polygon K lies in the interior of a unit square. Show that it is possible to select three vertices of the polygon that form a triangle of smaller area than \frac{80}{n^3} units.

(5 pont)

Deadline expired on November 15, 2005.

Solution. We prove that there exist two adjacent sides of K which form a sufficiently small triangle.

K is convex and is a subset of the unit square. Therefore its perimeter p is at most 4.

Denote the length of the ith side of K by di and let the angle between the ith and (i+1)th sides be \pi-\varphii. Since


there exists an index i for which d_i+d_{i+1}+{2\over\pi}\varphi_i\le{12\over n}.

Consider the triangle determnined by the ith and (i+1)th sides; its area is

{1\over2}d_id_{i+1}\sin\varphi_i < 
{1\over2}d_id_{i+1}\varphi_i =
  d_{i+1}\cdot{2\over\pi}\varphi_i\right) \le

{\pi\over4}\left({d_i+d_{i+1}+{2\over\pi}\varphi_i\over3}\right)^3 \le
{\pi\over4}\left({4\over n}\right)^3 = {16\pi\over n^3} < {51\over n^3}.


15 students sent a solution.
5 points:Dücső Márton, Erdélyi Márton, Gyenizse Gergő, Hujter Bálint, Jankó Zsuzsanna, Kisfaludi-Bak Sándor, Kónya 495 Gábor, Nagy 224 Csaba, Paulin Roland, Tomon István.
4 points:Bogár 560 Péter.
1 point:3 students.
0 point:1 student.

Problems in Mathematics of KöMaL, October 2005