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Proof Contest Day 2

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.

Note by Lakshya Sinha
11 months ago

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Let 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 · 11 months ago

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@Aditya Agarwal Oops !! I didn't saw your solution and you beated me while writing the solution. Akshat Sharda · 11 months ago

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@Akshat Sharda No problem brother. (y) Aditya Agarwal · 11 months ago

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@Aditya Agarwal I have edited out your solution. Nice (+1) Lakshya Sinha · 11 months ago

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You would have fun solving Follow Up Problem. :) Nihar Mahajan · 11 months ago

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@Lakshya Sinha Should I delete mine or let it be ? Akshat Sharda · 11 months ago

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@Akshat Sharda I believe you should delete but if you have another different solution then upload that. Lakshya Sinha · 11 months ago

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@Lakshya Sinha I didn't saw that....maybe he wrote at the same time as me but a bit faster. Akshat Sharda · 11 months ago

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