Two Symmetric Triangles Are Directly Equidecomposable II
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 attempts at solving this problem led me a little astray. I was put back on the right track with the help from Nathan Bowler. To all intents and purposes this is indeed the intended solution, but the manner in which it's arrived at is different.
References
- J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996, #9
Equidecomposition by Dissiection
- Carpet With a Hole
- Equidecomposition of a Rectangle and a Square
- Equidecomposition of Two Parallelograms
- Equidecomposition of Two Rectangles
- Equidecomposition of a Triangle and a Rectangle
- Equidecomposition of a Triangle and a Rectangle II
- Two Symmetric Triangles Are Directly Equidecomposable
- Two Symmetric Triangles Are Directly Equidecomposable II
- Two Symmetric Triangles Are Directly Equidecomposable III
- Two Symmetric Triangles Are Directly Equidecomposable IV
- Wallace-Bolyai-Gerwien Theorem
- Perigal's Proof of the Pythagorean Theorem
- A Proof Perigal and All Others After Him Missed
- Dissection of a Vase
- Curvy Dissection
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Copyright © 1996-2018 Alexander Bogomolny
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.
Solution
| What if applet does not run? |
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
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