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2005 Canadian MO Problems/Problem 4

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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}

This solution is incomplete. You can help us out by completing it.

See also

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