Two Symmetric Triangles Are Directly Equidecomposable III

The problem #9 in a delightful collection Which Way Did the Bicycle Go? reads:

There is a complete solution of two cuts into 3 pieces. While trying to solve the problem we also found two 3-cuts decompositions: one into 3, the other into 4 pieces. The former - valid only for acute triangles - was obtained by joining the circumcenter of the triangle with its vertices. An additional dissection [Frederickson, p. 24] is determined by the incenter of the triangle. It splits the triangle into quadrilateral and works for any triangle with no restriction. It can be naturally extended to inscriptible polygons with more than 3 vertices.

(Note in passing that each of the quadrilaterals is a kite, i.e. they all have two pairs of adjacent equal sides. Each then can be cut into two isosceles triangles. It follows that every triangle is decomposable into a union of 6 isosceles triangles. Another solution to the decomposition problem at hand shows that the feat can be acomplished with just four isosceles triangles. An acute triangle can be split into three isosceles.)

References

1. G. N. Frederickson, Dissections: Plane & Fency, Cambridge University Press, 1997
2. J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996, #9

Equidecomposition by Dissection

1. Carpet With a Hole
2. Equidecomposition of a Rectangle and a Square
3. Equidecomposition of Two Parallelograms
4. Equidecomposition of Two Rectangles
5. Equidecomposition of a Triangle and a Rectangles
6. Equidecomposition of a Triangle and a Rectangle II
7. Perigal's Proof of the Pythagorean Theorem
8. Two Symmetric Triangles Are Directly Equidecomposable
9. Wallace-Bolyai-Gerwien Theorem

Copyright © 1996-2008 Alexander Bogomolny