Morley's Miracle: An Unexpected Variant

Larry, thank you. I find it indeed amazing and think that F. Morley himself have not foreseen this variant.

One of the vertices of ABC, say C, is pushed off to infinity so that the two sides AC and BC and the trisector lines at C become parallel. In this case the angles at A and B are supplementary:

(1)	A + B = 180o

As will be demonstrated below, in order that the "Morley triangle" be equilateral, the four lines need be equidistant.

I'll follow D. J. Newman's proof, but use J. Conway's notations.

We backtrack, i.e. start with an equilateral triangle PQR of side 1 and errect two triangles, AQR and BPR, with angles a, and 60^o, b* and, respectively, b, 60^o, and a*. As in the original proof, angle RAB is shown to be A, while angle RBA is shown to be b.

We now draw four parallel lines through the points B, P, Q, A, such that the added angle at vertex A is a and that at B is b. This is possible because of (1). Counting angles around P and Q we immediately get PQ = a* while QP = b*. (It is of course possible to draw lines at the indicated angles and prove them parallel afterwards. This would be closer to Newman's proof.)

Now the only thing that remains to be proven is the fact that the four lines are equidistant.

From the Law of Sines in AQR,

Therefore, if QT AT, then QT = AQ·sin(a). Which gives QT = sin(b*). Similarly, BP = sin(a*)/sin(b) and PS = sin(a*), where PS BS. However, from (1), a* + b* = 180^o. We thus have

Let QU PU. Then QU = sin(b*), too. And all three distances coincide:

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