# Two Symmetric Triangles Are Directly Equidecomposable IV

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

Can you cut an arbitrary triangle into pieces so that the pieces can be rotated and translated (but not flipped) so as to form the mirror image of the given triangle? It can be done in just two cuts.

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 applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

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

Note: the above decomposition shows that every triangle can be decomposed into 4 isosceles triangles. For isosceles triangles, there is a special decomposition. Since any triangle is cyclic, the acute triangles can be decomposed into 3 isosceles triangles by joining the circumcenter to the vertices. Right triangles are decomposed into 2 isosceles triangles by joining the right angled vertex to the midpoint of the hypotenuse.

### 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 Dissiection

Copyright © 1996-2018 Alexander Bogomolny

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