Difference between revisions of "2001 JBMO Problems/Problem 4"
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Well, then <math>2\sqrt{ab} < a+b (AM-GM) < 2\sqrt2 \Rightarrow ab < 2</math> and the area of the triangle with sides <math>a,b</math> and the angle <math>C</math> between them has area <math>\frac12ab\sin C \le \frac{1}{2}ab\sin \frac{\pi}{2} <\frac12\cdot2\cdot1 = 1</math> as desired. | Well, then <math>2\sqrt{ab} < a+b (AM-GM) < 2\sqrt2 \Rightarrow ab < 2</math> and the area of the triangle with sides <math>a,b</math> and the angle <math>C</math> between them has area <math>\frac12ab\sin C \le \frac{1}{2}ab\sin \frac{\pi}{2} <\frac12\cdot2\cdot1 = 1</math> as desired. | ||
| − | ~ | + | ~ [https://artofproblemsolving.com/community/user/61542 AwesomeToad] |
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| + | {{JBMO box|year=2001|num-b=2|after=Last Question}} | ||
Revision as of 23:53, 8 January 2023
Contents
Problem
Let 
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.
Soultion
The largest side has length at least 
. Therefore, the sum of the other 
sides is 
. Divide these sides into 
pairs of adjacent sides, and there exist one pair of sides 
such that 
. Blah blah blah... we know how to prove 
except why would a problem want you to do that.... no idea.
Well, then 
and the area of the triangle with sides and the angle 
between them has area 
as desired.
| 2001 JBMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 | ||
| All JBMO Problems and Solutions | ||
