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. | ||
| − | == | + | ==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. | ||
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] |
| + | |||
| + | ==See also== | ||
| + | {{JBMO box|year=2001|num-b=2|after=Last Question|five=}} | ||
| + | |||
| + | [[Category:Intermediate Geometry Problems]] | ||
Latest revision as of 00:59, 9 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 | ||
