Morley's Miracle: An Unexpected Variant

Subject:Morley's theorem
From:Larry Hammick


Since CTK has such a good account of Morley's theorem, I thought I would point out this little sidelight on it:


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

One of the vertices of ΔABC, 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 = 180°

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°, b* and, respectively, b, 60°, 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 ΔAQR,

AQ = sin(b*)/sin(a).

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°. We thus have

QT = PS.

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

QT = PS = QU.

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