# Proof Contest Day 2 (Follow Up problem)

The previous problem dealt with proving that the sum of heights remains constant. However , in this follow up problem we would find the exact value of the sum of heights for an even sided polygon.

PROBLEM:

Let $P$ be any point in the interior of a regular polygon of $2n$ sides. Perpendiculars $PA_1 \ , \ PA_2 \ , \ PA_3 \ , \ \dots \ , \ PA_{2n}$ are drawn to the sides of the polygon. Show that: $\displaystyle\sum_{i=1}^{2n} PA_{i} = 2nr$ where $r$ is the radius of inscribed circle of polygon.

Also show that $\displaystyle\sum_{i=1}^{n} PA_{2i-1} =\sum_{j=1}^{n} PA_{2j} =nr$

###### This problem is not original

Note by Nihar Mahajan
5 years ago

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The first one can be pretty easily proved using the fact that the radius of the inscribed circle is the height of the triangle.The second one can be proved using the fact that since the polygpn has an even number of sides the two perpendiculars,$PA_{i},PA_{i+n}$ form a straight line,from here we easily get that,$PA_{i}+PA_{i+n}=PA_{k}+PA_{k+n}$,now just put values,$1,3,5...(n-1)$ for i and the rest for k and then add to get the result,i am writing this on phone so cant explain very clearly.

- 5 years ago

Exactly! Nice use of the fact that since the polygon has an even number of sides the two perpendiculars, $PA_i , PA_{i+n}$ form a straight line.

- 5 years ago

- 5 years ago

We know that the sum of the distances of the perpendiculars from the interior point is always constant. (As proved earlier). So we know that $PA_1+...+PA_{2n}=k$ a constant.

Now the incentre of the polygon is also such a point so the sum of the perpendiculars from it will be the sum of $2n$ radii. So we get that $PA_1+PA_2+...+PA_{2n}=2nr=k$

- 5 years ago

Can we directly prove the last thing?

- 5 years ago

Yes. Then the main problem becomes obvious from there.

- 5 years ago

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