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

Revision as of 22:33, 17 November 2007

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

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.

@@ Line 5: / Line 5: @@
 ==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>
-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==

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