Question

Let $A$ be a point inside a regular polygon of 10 sides. Let $p_1,p_2,\dots \dots \dots \dots \dots .\dots p_{10}$ 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 $\frac{1}{p_1}+\frac{1}{p_2}+\dots \dots \dots \dots \dots \dots \frac{1}{p_{10}}$.

Answer 1

Answer: 4

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