B.stat -2005 – geometry problem

$B.stat-2005$– (Solution)
[by Neelanjan Mondal] At first use the $AM-HM$ inequality ,
$\frac{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}}{3}$ $\geq$ $\frac{3}{\alpha+\beta+\gamma}$, note that $\alpha $ be equal to $180^\circ$$P$,
$\beta$ must be equal to $90^\circ$+$P/2$ and $\gamma$ must be equal to $2P$. , that is for any acute angled triangle $PQR$ , put the values of $\alpha ,\beta, \gamma$ , so we obtain ,
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$ $\geq$ $\frac{9}{\alpha+\beta+\gamma}$;
=$\frac{9}{270^\circ + \frac{P}{2}+P}$ =$\frac{9}{270^\circ + \frac{3P}{2}}$
= $\frac{18}{540+3P}$.
It is given that $\angle P$ $<$ $90^\circ$ that is $3P + 540^\circ$ $<$ $270^\circ +540^\circ$ =$810^\circ$
that is $\frac{18}{3P + 540^\circ}$ $>$ $\frac{18}{810}$ = $\frac{1}{45}$.
So ,it is easy to say ,$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$ $>$$\frac{1}{45}$.
So , we are done .
[Please note that , $1 = \alpha,2 = \beta , 3 = \gamma $]http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMS81LzIwZWM4YWMyNjE4ODI2N2FiMmZmNGRlYzBkNDBkNGQwMjIwM2EyLnBuZw==&rn=VW50aXRsZWQucG5nNy5wbmc=

Here , we have used a lemma (about the incircle , incenter…etc.) that is …

Note that $A+B+C$ = $\pi$ for any acute angled triangle $\Delta ABC$ , dividing by $2$ to the both sides , we get $\frac{A+B+C}{2} = \frac{\pi}{2}$ .
Now , $\angle BIC + \frac{B}{2} + \frac{C}{2}$ = $\pi$ for the triangle $\Delta BIC$.
Now we will express $\angle BIC$ by the terms of $A$.
so ,$\angle BIC$ = $\pi$ – ($\frac{B+C}{2})$ = $\pi - (\frac{\pi}{2} - \frac{A}{2})$.That is equal to $\frac{\pi}{2} +\frac{A}{2}$ = $90^\circ + \frac{A}{2}$.

http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi82LzY2MjRhMzc4M2ExODQyZTJjNmQzY2JhNDVjYjVlYzQzZTJjY2IxLnBuZw==&rn=VW50aXRsZWQucG5nOC5wbmc=

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