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Tiling a Triangulated Hexagon

A hexagon is cut out of a triangular grid. You have a supply of rombi the size of two adjacent grid triangles with which to tile the hexogon. The rombi come in three varieties each associated with one of the three directions of the common side of their constituent grid triangles. Prove that if the rombi tile the whole hexagon, there is exactly the same number of rombi of each variety.

Instead of tiling the hexagon with rombi, we may think of spliting it into pairs of adjacent triangles. An easy way to do that is to join the centers of any two paired triangles. In the applet below, click on one of the triangles and then move (not drag) the cursor to the other one and click there too. The lines joining the centers may have one of three directions. When no single triangle is left over, there are as many lines in one direction as in any other.

<hr> <h3> This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet. </h3> <hr>

This problem was posted by Guy David and Carlos Tomei in the Proofs Without Words section of the American Mathematical Monthly (v. 96, no. 5, May 1989, pp. 429-430). It was reprinted in [Proofs Without Words, p. 142] and in a concise form in [Winkler, p. 46]. An extended discussion can be found in [The Art of Mathematics, #24].

The problem has been motivated by the form of a French candy and the shape of the box it is usually sold in:

A calisson is a French sweet that looks like two equilateral triangles meeting along an edge. Calissons could come in a box shaped like a regular hexagon, and their packing would suggest an intersiting combinatorial problem. Suppose a box with side length n is filled with sweets of sides 1. Calissons could be placed in three different directions. Prove that a filled out box contains as many calissons in one direction as in any other.


  1. B. Bollobás, The Art of Mathematics: Coffee Time in Memphis, Cambridge University Press, 2006
  2. R. B. Nelson, Proofs Without Words, MAA, 1993
  3. P. Winkler, Mathematical Puzzles: A Connoisseur's Collection, A K Peters, 2004

Copyright © 1996-2008 Alexander Bogomolny