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- #1

- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,597

- Thread starter
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- #2

- Feb 14, 2012

- 3,597

$\cos A+\cos B+\cos C=1+\dfrac{r}{R}$ where $r$ and $R$

$\sin A+\sin B+\sin C=\dfrac{s}{R}$ where $s$ represents the triangle's semi-perimeter

$\cos A \cos B \cos C=\dfrac{s^2-(2R+r)^2}{4R^2}$

$\sin A \sin B \sin C=\dfrac{rs}{2R^2}$

$2r\le R$ (Euler's inequality)

$s^2\le 4R^2+4Rr+3r^2$ (Gerretsen inequality)

We try to show

$\dfrac{\cos A+\cos B+\cos C}{\sin A+\sin B+\sin C}\le 3\cot A \cot B \cot C$

$\dfrac{1+\dfrac{r}{R}}{\dfrac{s}{R}}\le 3\left(\dfrac{\dfrac{s^2-(2R+r)^2}{4R^2}}{\dfrac{rs}{2R^2}} \right)$

$5r^2+14rR+12R^2 \le 3s^2 \le 9r^2+12rR+12R^2$ which implies $R\le 2r$

This is impossible but that suggests $R=2r$must be true or $\dfrac{\cos A+\cos B+\cos C}{\sin A+\sin B+\sin C}= 3\cot A \cot B \cot C$. This can happen if and only if $ABC$ is an equilateral triangle.