05 November 2020

Morley's Miracle — Yet Another Simplest Proof

(Compare here.) 

We use the rule of sines (if R is the circumradius, the side facing angle a is 2R sin a) and two more elementary facts.

Property 1.

Property 2. If we have the situation of the first triangle below (k some positive constant), then the blue elements are uniquely determined 

and likewise for

Verification of the first case: consider the third triangle, which has angles a,x,y and circumradius R. It is similar to the first triangle, because the angle a is common to both and the adjacent sides are proportional. Hence, the angles in the first triangle are also a,x,y and the rule of sines implies that its third side is k (sin a). Similarly for the second case.

To obtain Morley's trisectors theorem there is nothing more to do than to fill in successive elements in triangles. R is the circumradius of the triangle and we abbreviate 8R sin x sin y sin z to x y z.

In black: the primary data on trisected angles and, by the rule of sines, the sides 2R sin(3a), 2R sin(3b), 2R sin (3c) expressed by Property 1.

In blue: sides implied by the rule of sines. Consider for instance the lowest placed triangle. The black angle and side show that a side is obtained by multiplying the angle by 8R sin a sin a+. Hence the blue sides.

In red: sides and angles implied by Property 2. Consider for instance the highest placed triangle. We have the situation of Property 2, with k=8R sin b sin c and a+b++c+=180°. Hence the red angles and sides, forming Morley's equilateral triangle.

Note that the proof is straightforward, moves from the data to the solution and requires no computation. Also, it results in a geometric configuration that is completely solved (every segment and angle determined). 

The usefulness of this approach is illustrated by considering an outer Morley triangle, obtained by trisecting one interior and two exterior angles.

The only difference with the previous proof is, that the second case of Property 2 also occurs. Such is the case in the highest placed blue-sided triangle, where k= 8R sin a sin b-  and c-+a-+(180°-b+)=180°, and in the lowest triangle right, where k= 8R sin a sin c-  and b-+a-+(180°-c+)=180°.