Below are the 11 possible triangular pillows, or short *trillows* (up to motions).

They are obtained by replacing in an equilateral triangle some edges by circular arcs curving either inwards or outwards, The numbers indicate their *defect*, which is the difference between the number of convex and concave edges. This defect is additive when you tile larger regions. This means that the sum of the defects of the tiles will be equal to the defect of the entire region.

For instance, suppose we want to tile a straight triangle with the three trillows above. The green one has defect +2, while the red and yellow ones have defect -1. To see how this becomes useful, let’s suppose in such a tiling there are just trillows with defect +2 (A many) and trillows with defect -1 (B many).

A triangle of edge length 3 requires 9 trillows, so A+B=9. On the other hand, the total defect gives the equation 2A-B=0, because a straight edged triangle has defect 0. Solving this shows that A=3 and B=6, as is indeed the case in the left figure. Now go ahead and try to tile a triangle of edge length 4 with the same type of tiles…

I have written about the four purely circular version briefly before. If you want to use them for a periodic tiling of the plane, the total defect of a fundamental piece must be 0. Above and below are examples that use just two trillows.

Every purely circular simply connected region has total defect +6, so if we want to tile a large circular triangle with C cyan, B blue, P purple and G green circular pillows, we have two equations to satisfy: C+B+P+G=22 for the total number of tiles, and3C+B-P-3G=6 for the total defect. Solving this for B and P gives B=14-2C+G and P=8+C-2G. Below you can see the region of values of C and G for which C,B,P,G to be nonnegative. Each dots in this regions represents a pair of integers (G,C) for which there is *maybe* a solution.

The edges correspond to solutions where only three of the tiles are used, and the vertices to those with only two tiles. The blue and purple lines are examples for which G and C are constant. At their intersection we have G=3 and C=4, which determines P=6 and B=9. Below are two different solutions with these numbers.

For instance, if we wanted to tile this triangle without blue, the potential solutions are on the top edge. Below are actual solutions for G=0,2,4,6. Is there one for G=8 or even for G=10?

It would be interesting to find further simple obstructions to the possibility of tiling such shapes.