Carpets in Triangle

What Might This Be About?

1 March 2014, Created with GeoGebra


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.

Carpets in Triangle - problem

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.


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.

Carpets in Triangle - solution

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.


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.


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


Carpets Theorem

|Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny


Search by google: