Difference between revisions of "2005 Canadian MO Problems/Problem 4"

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==Solution==
 
==Solution==
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Since equilateral triangles are awesome, we try an equilateral triangle first:
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<math>\dfrac{KP}{R^3}=\dfrac{27}{4}</math>
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now we just need to prove that that is the maximum.
  
 
{{solution}}
 
{{solution}}

Revision as of 11:12, 8 October 2007

Problem

Let $ABC$ be a triangle with circumradius $R$, perimeter $P$ and area $K$. Determine the maximum value of $KP/R^3$.

Solution

Since equilateral triangles are awesome, we try an equilateral triangle first:

$\dfrac{KP}{R^3}=\dfrac{27}{4}$

now we just need to prove that that is the maximum.

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

2005 Canadian MO (Problems)
Preceded by
Problem 3
1 2 3 4 5 Followed by
Problem 5