Difference between revisions of "2001 JBMO Problems/Problem 4"
m (→Solution) |
m |
||
| Line 13: | Line 13: | ||
==See also== | ==See also== | ||
{{JBMO box|year=2001|num-b=2|after=Last Question|five=}} | {{JBMO box|year=2001|num-b=2|after=Last Question|five=}} | ||
| + | |||
| + | [[Category:Intermediate Geometry Problems]] | ||
Latest revision as of 23:59, 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.
Solution
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.
See also
| 2001 JBMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
