Morley's Miracle

In 1899, almost a hundred years ago, Frank Morley, then professor of Mathematics at Haverford College, came across a result so surprising that it entered mathematical folklore under the name of Morley's Miracle. Morley's marvelous theorem states that

The applet below serves to demonstrate that, indeed, whatever the shape of the given triangle, the Morley triangle is always equilateral.

Morley's original proof stemmed from his results on algebraic curves tangent to a given number of lines. As usual in mathematics, numerous attempts have been made to find a simple, elementary proof that could match the level of knowledge and proficiency required to grasp the statement of the theorem. The simplest proofs proceed backwards starting with an equilateral triangle. They differ in subsequent steps. Most such proofs highlight some additional features of the configuration but complicate matters unnecessarily as a few most trivial proofs convincingly demonstrate.

Before giving several backward proofs, here is a direct one which, while logically absolutely transparent, requires some high school trigonometry.

In all likelihood, this proof first appeared in A.Letac, Solution (Morley's triangle), Problem No. 490, Sphinx, 9(1939) 46. I came across it in a Russian book D.O.Shklyarsky, N.N.Chentsov, Y.M.Yaglom, Selected Problems and Theorems of Elementary Mathematics, v. 2, problem 97, Moscow, 1952 and also in The Art of Mathematics by B. Bollobás (Cambridge University Press, 2006, p. 126-127.)

The idea of the proof is fairly straightforward.

1. In triangles ARB, BPC, CQA, we know the bases - AB, BC, and AC - and the adjacent angles. The Law of Sines then yields the segments AR, BR, BP, CP, CQ, and AQ.
2. Next we apply the Law of Cosines to triangles AQR, BPR, and CPQ to determine (and compare) the segments QR, PR, and PQ. The fact that they come out equal proves the theorem.

For simplicity, let (angles) A = 3a, B = 3b, and C = 3c. This implies that a + b + c = 60°. Also, assuming that the radius of the circle circumscribed around ΔABC equals 1, we get AB = 2sin(3c), BC = 2sin(3a), AC = 2sin(3b).

Consider now ΔBPC. By the Law of Sines,

Therefore, BP = 2sin(3a)sin(c)/sin(60° - a). To simplify the expression note that

Proceeding to the second step and the Law of Cosines,

from where

Note, however, that (60° + a) + (60° + c) + b = 180°. Thus, there exists a triangle with angles (60° + a), (60° + c), and b. Indeed, there is a whole family of similar triangles with those angles. Out of this family, choose the one with the circumscribed radius equal to 1 (then, as above, by the Law of Sines, its sides have a very simple form.) In that triangle, apply the Law of Cosines:

Which gives

an expression which is symmetric in a, b, and c. QR and PQ are similarly found to be equal to the same expression. Therefore, PR = PQ = QR.

Morley's Miracle

1. J.Conway's proof
   • Remarks on J. Conway's proof
2. D. J. Newman's proof
3. Bankoff's proof
4. B. Bollobás' proof
5. Another proof
6. Nikos Dergiades' proof
7. G. Zsolt Kiss' proof
8. M. T. Naraniengar's proof
9. Doodling and Miracles
10. Morley's Pursuit of Incidence
11. Lines, Circles and Beyond
12. On Motivation and Understanding
13. Bankoff's conundrum
14. Of Looking and Seeing
15. Morley's Redux and More, Alain Connes' proof
16. An Unexpected Variant
17. Proof by B. Stonebridge and B. Millar
18. Proof by B. Stonebridge
19. Proof by Nolan L Aljaddou
20. Proof by Roger Smyth
21. Proof by H. D. Grossman
22. Proof by R. J. Webster

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	Equilateral and 3-4-5 Triangles
	Rusty Compass Construction of Equilateral Triangle
	Equilateral Triangle on Parallel Lines
	When a Triangle is Equilateral?
	Viviani's Theorem
	Viviani's Theorem (PWW)
	Triangle Classification
	Napoleon's Theorem

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