2001 JBMO Problems/Problem 4


Let $N$ be a convex polygon with 1415 vertices and perimeter 2001. Prove that we can find 3 vertices of N which form a triangle of area smaller than 1.


The largest side has length at least $\frac{2001}{1415}$. Therefore, the sum of the other $1414$ sides is $\frac{2001\cdot1414}{1415}$. Divide these sides into $707$ pairs of adjacent sides, and there exist one pair of sides $a,b$ such that $a+b \le \frac{2001\cdot1414}{1415\cdot707}=\frac{2001\cdot2}{1415}$. Blah blah blah... we know how to prove $\frac{2001\cdot2}{1415} < 2\sqrt2$ except why would a problem want you to do that.... no idea.

Well, then $2\sqrt{ab} < a+b (AM-GM) < 2\sqrt2 \Rightarrow ab < 2$ and the area of the triangle with sides $a,b$ and the angle $C$ between them has area $\frac12ab\sin C \le \frac{1}{2}ab\sin \frac{\pi}{2} <\frac12\cdot2\cdot1 = 1$ as desired.

~ AwesomeToad

See also

2001 JBMO (ProblemsResources)
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Problem 2
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