Question

Let $A$ be a point inside a regular polygon of 10 sides. Let $p _1, p _2, \ldots \ldots \ldots \ldots \ldots . \ldots p _{10}$ be the distances of A from the sides of the polygon. If each side is of length 2 then find the minimum possible integral value f $\frac{1}{p_1}+\frac{1}{p_2}+\ldots \ldots \ldots \ldots \ldots \ldots \frac{1}{ p _{10}}$.

Answer 1

$\Theta$ Area of polygon $=10\left(\frac{1}{2} \times 2 \times \cot \frac{\pi}{10}\right)=\frac{1}{2} \times 2 \times p_1+\frac{1}{2} \times 2 \times p_2+\ldots . . \frac{1}{2} \times 2 \times p_{10}$
$\Rightarrow \therefore p _1+ p _2 \ldots \ldots \ldots \ldots \ldots p _{10}=10 \cot \frac{\pi}{10} $
$\Theta AM \geq HM $
$\Rightarrow \frac{ p _1+ p _2 \ldots \ldots \ldots p _{10}}{10} \geq \frac{10}{\frac{1}{ p _1}+\frac{1}{ p _2}+\ldots \ldots . \frac{1}{ p _{10}}}$
$\Rightarrow \frac{1}{ p _1}+\frac{1}{ p _2}+\ldots \ldots \ldots \ldots \ldots \ldots \frac{1}{ p _{10}} \geq 10 \tan \frac{\pi}{10}$
$ \Theta \tan \theta>\theta \forall \theta \in\left(0, \frac{\pi}{2}\right)$
$\therefore \frac{1}{ p _1}+\frac{1}{ p _2}+\ldots \ldots \ldots \ldots . . \frac{1}{ p _{10}}>\pi$
$\therefore \text { Minimum positive integral value }=4$