Let $\alpha , \beta ,\gamma$ be the angles of a triangle $ABC$ of perimeter $ 2p $ and $R$ is the radius of its circumscribed circle. $(a)$ Prove that \[\cot^{2}\alpha +\cot^{2}\beta+\cot^{2}\gamma\geq 3\left(9\cdot \frac{R^{2}}{p^{2}} - 1\right).\] $(b)$ When do we have equality?
Problem
Source: Moroccan MO 11th grade 5th exam 2011
Tags: inequalities, geometry, perimeter, trigonometry, inequalities proposed
Goutham
24.04.2011 12:28
This rearranges to $\sum\frac{1}{\sin^2\alpha}\ge\frac{27R^2}{p^2}$ but $\frac{R}{p}=\frac{1}{\sum \sin\alpha}$ and $(\sum\sin\alpha)^2\ge 9\sqrt[3]{\prod\sin^2\alpha}$. So, $\frac{27R^2}{p^2}\le \frac{3}{\sqrt[3]{\prod\sin^2\alpha}}\le \sum\frac{1}{\sin^2\alpha}$ by AM-GM and we are done. Equality occurs when $\alpha=\beta=\gamma$, that is, when the triangle is equilateral.
momo1729
24.04.2011 14:58
Goutham wrote: but $\frac{R}{p}=\frac{1}{\sum \sin\alpha}$ Is this a direct consequence of the law of sines ?
Goutham
24.04.2011 15:53
Moderator Edit: do not quote the whole post above. Yes. \[\frac{p}{R}=\sum\frac{a}{2R}=\sum\sin \alpha\]
Dijkschneier
24.04.2011 15:54
Moderator Edit: do not quote the whole post above. Indeed. Use also the fact that if $\frac{a}{x}=\frac{b}{y}$, then we also have : $\frac{a}{x}=\frac{b}{y}=\frac{a+b}{x+y}$.
Dijkschneier
24.04.2011 15:58
In fact, this problem comes originally from Estonia, as it was mentionned on the exam paper.