Difference between revisions of "2005 Canadian MO Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | + | 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. | |
| − | {{solution}} | + | {{incomplete|solution}} |
==See also== | ==See also== | ||
Revision as of 21:33, 17 November 2007
Problem
Let 
be a triangle with circumradius 
, perimeter 
and area 
. Determine the maximum value of 
.
Solution
It would be convenient if this were an equilateral triangle, so we try an equilateral triangle first:
Now we just need to prove that that is the maximum.
See also
| 2005 Canadian MO (Problems) | ||
| Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 | Followed by Problem 5 |
