For any regular \(n\) sided polygon, if we choose any point inside the polygon (not on side) and drop one perpendicular to every side from the point prove that the sum of all the heights remain constant, irrespective where the points lies inside the polygon.

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TopNewestLet the side of each polygon be \(b\). And join the vertices with the point.

Now area of each triangle (formed by two adjacent vertices of the polygon and that point), will be \[\frac12bh_i\] where \(h_i\) is one of those heights.

Now, the sum of all the triangles would equal the area of the polygon, let the total area of polygon be \(A\).

So we get \[\frac12b(h_1+h_2+...+h_n)=A\] And thus,

\[\sum h_i=\frac{2A}{b}\]

This shows it will always be constant until the point is in the polygon. – Aditya Agarwal · 8 months, 3 weeks ago

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– Akshat Sharda · 8 months, 3 weeks ago

Oops !! I didn't saw your solution and you beated me while writing the solution.Log in to reply

– Aditya Agarwal · 8 months, 3 weeks ago

No problem brother. (y)Log in to reply

– Lakshya Sinha · 8 months, 3 weeks ago

I have edited out your solution. Nice (+1)Log in to reply

You would have fun solving Follow Up Problem. :) – Nihar Mahajan · 8 months, 3 weeks ago

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– Akshat Sharda · 8 months, 3 weeks ago

Should I delete mine or let it be ?Log in to reply

– Lakshya Sinha · 8 months, 3 weeks ago

I believe you should delete but if you have another different solution then upload that.Log in to reply

– Akshat Sharda · 8 months, 3 weeks ago

I didn't saw that....maybe he wrote at the same time as me but a bit faster.Log in to reply