# 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**

- T. Andreescu, B. Enescu,
*Mathematical Olympiad Treasures*, Birkhäuser, 2004

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