Solving a 1415-gon Triangle Problem with the Pigeonhole Principle

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Discussion Overview

The discussion revolves around a mathematical problem involving a convex polygon with 1415 sides and a circumference of 2001 cm. Participants are exploring how to prove that among its vertices, there exist three that form a triangle with an area less than 1 cm². The conversation includes various mathematical approaches and reasoning related to the problem.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the problem and mentions the Pigeonhole Principle as a potential method for solving it.
  • Another participant notes a numerical proximity to sqrt(2) and suggests that induction might be applicable, although they later retract this idea, indicating that not all polygons would fit that approach.
  • A calculation is provided comparing the average side length of the polygon to sqrt(2), with a suggestion that a better bound could be established for the area of the triangle.
  • A different approach is proposed, focusing on calculating the area of a triangle based on the lengths of adjacent sides and the exterior angle at a vertex, while acknowledging that this may not align with the sqrt(2) relation.

Areas of Agreement / Disagreement

Participants express uncertainty about the best method to solve the problem, with no consensus on a definitive approach or solution. Multiple viewpoints and methods are presented, indicating an ongoing debate.

Contextual Notes

Some assumptions regarding the properties of the polygon and the implications of the Pigeonhole Principle are not fully explored. The discussion also reflects varying levels of understanding and familiarity with the mathematical concepts involved.

Who May Find This Useful

Individuals interested in advanced geometry, mathematical problem-solving techniques, and the application of principles like the Pigeonhole Principle in complex scenarios may find this discussion relevant.

Numeriprimi
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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.
 
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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.
 
Hmmm... Where is sqrt(2)?
 
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:
Well, still do not know how to do it ... I don't understand why.
 
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.
 

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