Proof Contest Day 2

For any regular nn 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 Department 8
3 years, 11 months ago

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Let the side of each polygon be bb. 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 12bhi\frac12bh_i where hih_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 AA.

So we get 12b(h1+h2+...+hn)=A\frac12b(h_1+h_2+...+h_n)=A And thus,

hi=2Ab\sum h_i=\frac{2A}{b}

This shows it will always be constant until the point is in the polygon.

Aditya Agarwal - 3 years, 11 months ago

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I have edited out your solution. Nice (+1)

Department 8 - 3 years, 11 months ago

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

Akshat Sharda - 3 years, 11 months ago

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No problem brother. (y)

Aditya Agarwal - 3 years, 11 months ago

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You would have fun solving Follow Up Problem. :)

Nihar Mahajan - 3 years, 11 months ago

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