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 \(\displaystyle\sum_{i=1}^{n} PA_{2i-1} =\sum_{j=1}^{n} PA_{2j} =nr\)

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TopNewestThe 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. – Adarsh Kumar · 11 months, 1 week ago

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– Nihar Mahajan · 11 months, 1 week 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.Log in to reply

– Adarsh Kumar · 11 months, 1 week ago

Thanx!Sorry for the late reply.Log in to reply

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 \[PA_1+PA_2+...+PA_{2n}=2nr=k\] – Aditya Agarwal · 11 months, 1 week ago

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– Aditya Agarwal · 11 months, 1 week ago

What do you mean?Log in to reply

– Nihar Mahajan · 11 months, 1 week ago

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

– Aditya Agarwal · 11 months, 1 week ago

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

– Nihar Mahajan · 11 months, 1 week ago

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

– Nihar Mahajan · 11 months, 1 week ago

Try proving \(\displaystyle\sum_{i=1}^{n} PA_{2i-1} =\sum_{j=1}^{n} PA_{2j} =nr\) .Log in to reply

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– Aditya Agarwal · 11 months, 1 week ago

See the edit. You misunderstood my solution.Log in to reply

Can we directly prove the last thing? – Aditya Agarwal · 11 months, 1 week ago

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– Nihar Mahajan · 11 months, 1 week ago

Yes. Then the main problem becomes obvious from there.Log in to reply