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.

Proof #1

In all likelihood, this proof first appeared in A.Letac, Solution (Morley's triangle), Problem No. 490, Sphinx, 9(1939) 46. I have found 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.

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^o. Also, assuming that the radius of the circle circumscribed around ABC equals 1, we get AB = 2sin(3c), BC = 2sin(3a), AC = 2sin(3b).

Consider now BPC. By the Law of Sines,

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

sin(3a) = 3sin(a) - 4sin3(a) = 4sin(a)[(3/2)2 - sin2(a)] = 4sin(a)[sin2(60o) - sin2(a)] = 4sin(a)(sin(60o) + sin(a))(sin(60o) - sin(a)) = 4sin(a) 2sin[(60o + a)/2]cos[(60o - a)/2] 2sin[(60o - a)/2]cos[(60o + a)/2] = 4sin(a)sin(60o + a)sin(60o - a)

Reaping the fruits of this effort,

BP = 8sin(a)sin(c)sin(60o + a)

Similarly,

BR = 8sin(c)sin(a)sin(60o + c)

Proceeding to the second step and the Law of Cosines,

from where

Note, however, that (60^o + a) + (60^o + c) + b = 180^o. Thus, there exists a triangle with angles (60^o + a), (60^o + 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. Another proof
5. Nikos Dergiades' proof
6. G. Zsolt Kiss' proof
7. M. T. Naraniengar's proof
8. Doodling and Miracles
9. Morley's Pursuit of Incidence
10. Lines, Circles and Beyond
11. On Motivation and Understanding
12. Bankoff's Conundrum
13. Morley's Redux and More, Alain Connes' proof
14. An Unexpected Variant

Copyright © 1996-2008 Alexander Bogomolny