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Posers and Puzzles Forum

Third of a Triangle

Original post by Subscriber talzamir, 26 May '12 05:56
  1. Art, not a Toil
    60.13N / 25.01E
    Joined : 19 Sep '11
    Moves : 40871
    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. Joined : 26 Apr '03
    Moves : 21050
    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. Caninus Interruptus
    2014.05.01
    Joined : 11 Apr '07
    Moves : 92274
    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
    But then CAD > DAE. (DAE is a 'sub-angle' of CAD)

    2) BD > BE
    But then BD > ED (ED is part of the segment BD)

    3) BD = BE
    But then DE = 0

    So, no.
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