Two Symmetric Triangles Are Directly Equidecomposable IV

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

My first copy of the book was perfectly defective. In particular, it did not contain pages with the solution to that problem. So I could not compare mine to the book's. Perhaps due to the disappointment of getting a defective book, I have misread the problem. My attempts, a correct solution and another interesting construction are documented elsewhere. Below is the solution from the book. (MAA dutifully has replaced the book with another copy. Annoyingly, the first swap did not help as the second copy was as much damaged as the first one. After the second swap I am now able to report the original solution.)

This does not produce a new solution but rather a new explanation to the dissection problem.

References

1. J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996, #9
2. V. Klee, S. Wagon, Old and New Unsolved Problems, MAA, 1991

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 Rectangle
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