1st CIS 1992 Problem 21 An equilateral triangle side 10 is divided into 100 equilateral triangles of side 1 by lines parallel to its sides. There are m equilateral tiles of 4 unit triangles and 25 - m straight tiles of 4 unit triangles (as shown below). For which values of m can they be used to tile the original triangle. [The straight tiles may be turned over.]
m = 5, 7, 9, ... or 25. Solution There are 45 upside down triangles (yellow in the diagram) and 55 right way up. A straight tile always covers two of each. A triangular tile covers 3 of one and 1 of the other. So suppose there are t1 triangular tiles placed the right way up and t2 triangular tiles placed upside down. Then we have 3t 1 + t 2 + 2(25-t 1 -t 2 ) = 55, so t 1 - t 2 = 5. So t 1 and t 2 have opposite parity and hence m = t 1 + t 2 must be odd. Also m ≥ t 1 ≥ 5. Now each rhombus shown may be filled with two triangular tiles or two straight tiles. So any tiling with m odd and ≥ 5 is possible. 1st CIS 1992 © John Scholes jscholes@kalva.demon.co.uk 5 December 2003 Last corrected/updated 5 Dec 03