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.

#Geometry

Easy Math Editor

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Comments

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 - 5 years, 5 months ago

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

Department 8 - 5 years, 5 months ago

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

Akshat Sharda - 5 years, 5 months ago

No problem brother. (y)

Aditya Agarwal - 5 years, 5 months ago

You would have fun solving Follow Up Problem. :)

Nihar Mahajan - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$