#### Posers and Puzzles

1.  talzamir
Art, not a Toil
26 May '12 05:56
Is it possible to divide any triangle ABC into three parts by choosing points D and E on the side BC so that the angles CAD, DAE and EAB are all equal, and that the lengths BD = DE = EC = BC / 3 ?
2. 26 May '12 20:46
Originally posted by talzamir
Is it possible to divide any triangle ABC into three parts by choosing points D and E on the side BC so that the angles CAD, DAE and EAB are all equal, and that the lengths BD = DE = EC = BC / 3 ?
No (if I understand the question correctly), the positions D and E can be found by trisecting D&E, which can be done with compass and straight-edge:
http://www41.homepage.villanova.edu/robert.styer/trisecting%20segment/

If the angles were also equal then BAC would have been trisected with compass and straight edge, which is impossible.
http://en.wikipedia.org/wiki/Angle_trisection
3.  SwissGambit
Caninus Interruptus
31 May '12 05:19
Originally posted by talzamir
Is it possible to divide any triangle ABC into three parts by choosing points D and E on the side BC so that the angles CAD, DAE and EAB are all equal, and that the lengths BD = DE = EC = BC / 3 ?
Three possibilities:

1) BD < BE