Carpets in Triangle

What Might This Be About?

1 March 2014, Created with GeoGebra

Problem

Let $M$ be a point in the interior $\Delta ABC.$ Three lines are drawn through $M$ parallel to the sides of the triangle, thereby producing three trapezoids. With a reference to the diagram below, in each of the trapezoids choose a diagonal so that the three have no common points: $AA',$ $BB',$ $CC'.$ These diagonals divide $\Delta ABC$ into seven pieces of which four are triangles.

Prove that the area of the central triangle equals the sum of the areas of the other three. More accurately,

$[XYZ] = [AB'Z] = [BC'X] + [CA'Y],$

where the brackets are used to denote the area of the included shape.

Solution

The three triangles $ABB',$ $BCC',$ $CAA'$ cover a certain area, that excludes $\Delta XYZ,$ but with triangles $AB'Z,$ $BC'X,$ $CA'Y$ covered twice.

Due to Euclid I.37, their total area is exactly $[ABC]:$

$[ABB'] + [BCC'] + [CAA'] = [ABM] + [BCM] + [CAM] = [ABC].$

Motivated by the Carpet Theorem, we observe that the area of their overlap equals the uncovered area.

Acknowledgment

This is one of the problems from the most wonderful book by T. Andreescu and B. Enescu. This is actually where I first came across the Carpet Theorem.

References

1. T. Andreescu, B. Enescu, Mathematical Olympiad Treasures, Birkhäuser, 2004

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Copyright © 1996-2018 Alexander Bogomolny
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