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# 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
2 years, 1 month 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.

- 2 years, 1 month 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.

- 2 years, 1 month ago

- 2 years, 1 month 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 $PA_1+PA_2+...+PA_{2n}=2nr=k$

- 2 years, 1 month ago

Comment deleted Jan 08, 2016

What do you mean?

- 2 years, 1 month ago

Oh wait , I got your point. Can you think of some other way to prove it?

- 2 years, 1 month ago

Is there an even simpler one? I believe my method wasn't the solution you had in mind?

- 2 years, 1 month ago

I had solved this question about an year ago.I don't have solution ready , but I know the basic theme

- 2 years, 1 month ago

Try proving $$\displaystyle\sum_{i=1}^{n} PA_{2i-1} =\sum_{j=1}^{n} PA_{2j} =nr$$ .

- 2 years, 1 month ago

Comment deleted Jan 08, 2016

See the edit. You misunderstood my solution.

- 2 years, 1 month ago

Can we directly prove the last thing?

- 2 years, 1 month ago

Yes. Then the main problem becomes obvious from there.

- 2 years, 1 month ago