Art of Problem Solving

2005 Canadian MO Problems/Problem 4: Difference between revisions

1=2 (talk | contribs)
No edit summary
Temperal (talk | contribs)
incomplete
Line 5: Line 5:
==Solution==
==Solution==


Since equilateral triangles are awesome, we try an equilateral triangle first:
It would be convenient if this were an equilateral triangle, so we try an equilateral triangle first:


<math>\dfrac{KP}{R^3}=\dfrac{27}{4}</math>
<math>\dfrac{KP}{R^3}=\dfrac{27}{4}</math>


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


{{solution}}
{{incomplete|solution}}


==See also==
==See also==

Revision as of 21:33, 17 November 2007

Problem

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

Solution

It would be convenient if this were an equilateral triangle, so we try an equilateral triangle first:

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

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

Template:Incomplete

See also

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