## Octagon in Parallelogram: Overlapping Regions

In a parallelogram ABCD, the midpoints M, N, P, Q of the sides are joined to the non-adjacent vertices. These lines form an octagon at the center of the parallelogram. Determine the area of the octagon relative to the area of the parallelogram.

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Copyright © 1996-2018 Alexander Bogomolny

This problem has been also discussed on another page, where two solutions were given. Below is an absolutely beautiful solution by Stuart Anderson.

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Each of the eight triangles AMD, ABQ, BCM, etc., constitutes one fourth of the parallelogram ABCD and has 1/4 of the area of the latter. We consider four other regions:

- I, the intersection of triangles ABP and CDM,
- J, the intersection of triangles ADN and BCQ,
- K, the intersection of quadrilaterals BPDQ and DMBN,
- L, the intersection of quadrilaterals ANCP and CQAM.

Assume for simplicity that the area of ABCD is 1. Then the grids displayed by the applet help one surmise that the areas of I and J are 1/4 each, and that of K and L are 1/3 each.

The eight triangles and the four regions I, J, K, L cover the parallelogram with a considerable overlap: every point in the parallelogram is covered three times, i.e. every point, except those that belong to the middle octagon which are covered four times. If the area of the octagon is O, we get an equation:

8·1/4 + 2·1/4 + 2·1/3 = 3 + O, |

from which

O = 2 + 1/2 + 2/3 - 3 = 1/6. |

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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