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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## Comments

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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.

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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.

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Thanx!Sorry for the late reply.

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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 \[PA_1+PA_2+...+PA_{2n}=2nr=k\]

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Can we directly prove the last thing?

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Yes. Then the main problem becomes obvious from there.

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