Two sets A and B are said to be equidecomposable if you can partition set A into a finite number of subsets and reassemble them (by rigid motions only) to form set B.
Let A be a unit circle, and let B be a unit circle with one point X missing (called a “deleted circle”). Are sets A and B equidecomposable?
Believe it or not, yes! In fact you can do it using just 2 subsets. Can you figure out how?
Presentation Suggestions:
Some students will be tempted to “push together” the ends of the deleted circle, but this is not a rigid motion, and because of the openness of the endpoints, the ends will never “meet” unless they intersect.
The Math Behind the Fact:
Consider set B and let U be the subset consisting of all points that are a positive integer number of radians clockwise from X along the circle. This is a countably infinite set (the irrationality of Pi prevents two such points from coinciding). Let set V be everything else.
If you pick set U up and rotate it counterclockwise by one radian, something very interesting happens. The deleted hole at X gets filled by the point 1 radian away, and the point at the (n-1)-th radian gets filled by the point at the n-th radian. Every point vacated gets filled, and in addition, the empty point at X gets filled too!
Thus, B may be decomposed into sets U and V, which after this reassembling, form set A, a complete circle!
This elementary example forms the beginnings of the idea of how to accomplish the Banach-Tarski paradox.
How to Cite this Page:
Su, Francis E., et al. “Equidecomposability.” Math Fun Facts. <https://www.math.hmc.edu/funfacts>.
Fun Fact suggested by:
Francis Su