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Difference between revisions of "2001 JBMO Problems/Problem 4"
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Let <math>N</math> 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.
 
Let <math>N</math> 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.
  
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==Soultion==
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==Solution==
 
The largest side has length at least <math>\frac{2001}{1415}</math>. Therefore, the sum of the other <math>1414</math> sides is <math>\frac{2001\cdot1414}{1415}</math>. Divide these sides into <math>707</math> pairs of adjacent sides, and there exist one pair of sides <math>a,b</math> such that <math>a+b \le \frac{2001\cdot1414}{1415\cdot707}=\frac{2001\cdot2}{1415}</math>. Blah blah blah... we know how to prove <math>\frac{2001\cdot2}{1415} < 2\sqrt2</math> except why would a problem want you to do that.... no idea.
 
The largest side has length at least <math>\frac{2001}{1415}</math>. Therefore, the sum of the other <math>1414</math> sides is <math>\frac{2001\cdot1414}{1415}</math>. Divide these sides into <math>707</math> pairs of adjacent sides, and there exist one pair of sides <math>a,b</math> such that <math>a+b \le \frac{2001\cdot1414}{1415\cdot707}=\frac{2001\cdot2}{1415}</math>. Blah blah blah... we know how to prove <math>\frac{2001\cdot2}{1415} < 2\sqrt2</math> except why would a problem want you to do that.... no idea.
  
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~ [https://artofproblemsolving.com/community/user/61542 AwesomeToad]
 
~ [https://artofproblemsolving.com/community/user/61542 AwesomeToad]
  
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{{JBMO box|year=2001|num-b=2|after=Last Question}}
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{{JBMO box|year=2001|num-b=2|after=Last Question|five=}}

Revision as of 23:55, 8 January 2023

Problem

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.

Solution

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


2001 JBMO (Problems • Resources)
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