# 1415-gon and triangle

1. Feb 23, 2013

### Numeriprimi

Hello. I have an example for you. I'm curious how. Yesterday I was on the mathematical competition. One example I can not solve. I want to know how. Can you help me, please?

Consider a convex polygon of 1415 sides, which circumference is 2001 cm. Prove that between its peaks, there are 3 such that vertices form a triangle with an area less than 1 cm2.

So... We know it is 1415-gon. The sum of its interior angles is π(n-2) rad (where n is 1415). We can calculate the average length of a side: 2001/1415 cm...
And the last what I think: there is Pigeonhole principle but i don't know how to do it.

Thanks very much for your ideas and sorry for my bad English.

2. Feb 23, 2013

### Staff: Mentor

This is so extremely close to sqrt(2) that I don't think it is a coincidence. As the corresponding problem for n=4 sides is exact, I would expect that induction is possible.

3. Feb 23, 2013

### Numeriprimi

Hmmm... Where is sqrt(2)?

4. Feb 23, 2013

### Staff: Mentor

2001/1415 = 1.414134...
sqrt(2) = 1.414214...
and 2001/1415 < sqrt(2) < 2002/1415

Edit: Forget induction, there are polygons which would not be covered there.

I found a direct proof, and there is a good safety margin - it is possible to prove a better bound (smaller than 0.1cm^2).

Last edited: Feb 23, 2013
5. Feb 24, 2013

### Numeriprimi

Well, still do not know how to do it ... I don't understand why.

6. Feb 24, 2013

### Staff: Mentor

It is probably not the intended way to solve it (as it does not use this sqrt(2)-relation), but here is a possible approach:

For any corner, can you calculate the corresponding area as function of the lengths of the adjacent sides and the exterior angle at this corner? This relation is true for all corners and you know something about the sum of those exterior angles.