Take any triangle and trisect its internal angles.

No solution was found in the collected postings.

This is Morley's Theorem, and not easy to prove geometrically.  I
think it was written up in the A.M.M. 15 or 20 years ago; you could look
it up.

-- Col. G. L. Sicherman <colonel@buffalo-cs>

[ Editor: further material on this puzzle, edited separately from other collected postings (batch 10); it may repeat some of the above. ]

[ Editor: problem as stated there: ]
Prove Morley's Trisector Theorem: the three points where adjacent angle
trisectors of any triangle meet form an equilateral triangle.

[ Editor: problem statement reconstructed from the postings. ]


I have the undisputedly simplest proof of Morley's Trisector Theorem.

Here it is:

Let your triangle have angles 3a,3b,3c and let x* mean x + pi/3, so
that a + b + c = 0*.  Then triangles with angles

                         0*,0*,0*

          a,b*,c*        a*,b,c*           a*,b*,c

          a**,b,c        a,b**,c           a,b,c**

exist abstractly, since in every case the angle-sum is pi.

Build them on a scale defined as follows:

0*,0*,0*  -  this is equilateral - make it have edge 1

a,b*,c*   -  make the edge joining the angles b* and c* have length 1

          -  similarly for a*,b,c* and a*,b*,c

a,b**,c  (and the other two like it) - let me draw this one:


                           X


              A          Y   Z                C

Let the angles at A,X,C be a,b**,c, and draw lines from X cutting AC at
angle b* in the two senses, so forming an equilateral triangle XYZ.
Choose the scale so that XY and XZ are both 1.

Now just fit all these 7 triangles together!  They'll form a figure
like:-


                             B



                          P     Q


                             X


               A           Y    Z          C

(in which the points X,Y should really be omitted).  The points Y,Z are
what I meant.

To make it a bit more clear, let me say that the angles of APX are
a (at A), b* (at P), c* (at X).

Why do they all fit together?  Well, at each internal vertex, the angles
add up to 2pi, as you'll easily check.  And two coincident edges have
either both been declared to have length 1, or are like the common edge
AX of triangles APX and AXC.

But APX is congruent to the subtriangle AZX of AXC, since PX = ZX = 1,
PAX = ZAX = a, and APX = AZX = b*.

So the figure formed by these 7 triangles is similar to the one you get
by trisecting the angles of your given triangle, and therefore in that
triangle the middle subtriangle must also be equilateral.

-- John Conway <conway@math.Princeton.EDU>

[ Editor: in the original, the sentence about coincident edges reads
  "... or are like the common edge AXZ of sorry - AX of triangles APX
  and AXC", Conway correcting himself in mid-sentence; the correction is
  applied above. ]