2005 Canadian MO Problems/Problem 4

Revision as of 11:16, 5 May 2008 by 1=2 (talk | contribs) (A better incomplete solution.)

Problem

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

Solution

Let the sides of triangle $ABC$ be $a$, $b$, and $c$. Thus $\dfrac{abc}{4K}=R$, and $a+b+c=P$. We plug these in:

$\dfrac{K(a+b+c)}{\dfrac{a^3b^3c^3}{64K^3}}=\dfrac{64K^4(a+b+c)}{a^3b^3c^3}$.

Now Heron's formula states that $K=\sqrt{(\dfrac{a+b+c}{2})(\dfrac{-a+b+c}{2})(\dfrac{a-b+c}{2})(\dfrac{a+b-c}{2})}$. Thus,

\[\dfrac{KP}{R^3}=\dfrac{(a+b+c)^3(-a+b+c)^2(a-b+c)^2(a+b-c)^2}{4a^3b^3c^3}\]

Template:Incomplete

See also

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