2005 Canadian MO Problems/Problem 4

Incomplete 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,

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

Solution Outline

^hahahaha... you can probably use Ravi Sub. to finish the above.

OR

Use the formula $K=\dfrac{abc}{4R}$ to get $KP/R^3=\dfrac{abc(a+b+c)}{4R^4}$. Then use the extended sine law to get something in terms of sines, and use AM-GM and Jensen's to finish. (Jensen's is used for $\sin A+\sin B+\sin C \le \dfrac{3\sqrt3}{2}$.

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