Equal Areas in Regular 2n-gons
The applet below illustrates the following theorem:
Let P be a point inside a regular 2n-gon. The lines joining P to the vertices of the polygon split the letter into 2n triangles. Then the sum of the areas of odd-numbered triangles equals that of the even-numbered triangles.
| What if applet does not run? |
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Copyright © 1996-2018 Alexander Bogomolny
Let P be a point inside a regular 2n-gon. The lines joining P to the vertices of the polygon split the letter into 2n triangles. Then the sum of the areas of odd-numbered triangles equals that of the even-numbered triangles.
| What if applet does not run? |
The claim is a consequence of a generalization of the Viviani's theorem that claims that in either equilateral or equiangular polygons the sum of distances from an arbitrary point in the plane of the polygon to its sidelines is constant.
Given a regular 2n-gon. Form two regular n-gons by extending the sides of the former. In each of the latter the distances from a point to the sides add up to a constant; in addition, the two constants are equal, and the sums in each of the n-gons add up to the sum of the altitudes of either odd-numbered or even-numbered triangles in the 2n-gon.
For the 4n-gon, one need not invoke the Viviani's theorem. As Matt Hendersen has observed, for the regular 4n-gons, the required result follows from the fact that the distance between any pair of opposite sides is exactly the same. (This is of course true for any regular 2n-gons, but for the 4n-gons the fact acquires extra significance. Do you see why?)
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
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