Morley's Miracle |
| (1) | BR = AR·sin(a)/sin(b). |
Construct ΔBPR with BR given above. Then
| (2) | BP = BR·sin(a*)/sin(c*). |
Keep constructing triangles so that their sides match. By the Law of Sines, we'll sequentially get
| (3) | CP = BP·sin(b)/sin(c) |
| (4) | CQ = CP·sin(b*)/sin(a*) |
| (5) | AQ = CQ·sin(c)/sin(a) |
| (6) | AR = AQ·sin(c*)/sin(b*) |
The last one should actually read "AR' = AQ·sin(c*)/sin(b*)". For we do not know in advance whether it has the same length as AR. To see that it does, multiply (1)-(6): after reducing like factors we end up with AR' = AR. Therefore, putting together the six triangles fills a triangle with a triangular hole. Counting angles we see that the hole is equiangular. Which is what was needed.
One should appreciate the simplicity of Conway's proof that not only avoids the trigonometry but makes it also unnecessary to handle all 6 triangles separately. D.J.Newman's proof is a simplified trigonometric variant where only 3 (out of 6) triangles are handled with the Law of Sines.
Morley's Miracle
- J.Conway's proof
- Remarks on J. Conway's proof
- D. J. Newman's proof
- Bankoff's proof
- B. Bollobás' proof
- Another proof
- Nikos Dergiades' proof
- G. Zsolt Kiss' proof
- M. T. Naraniengar's proof
- Doodling and Miracles
- Morley's Pursuit of Incidence
- Lines, Circles and Beyond
- On Motivation and Understanding
- Bankoff's conundrum
- Of Looking and Seeing
- Morley's Redux and More, Alain Connes' proof
- An Unexpected Variant
- Proof by B. Stonebridge and B. Millar
- Proof by B. Stonebridge
- Proof by Nolan L Aljaddou
- Proof by Roger Smyth
- Proof by H. D. Grossman
- Proof by R. J. Webster
- Proof by H. Shutrick
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Copyright © 1996-2012 Alexander Bogomolny
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