Consider a right triangle ABC and construct a point P inside the triangle such that the angles PBC, PCA and PAB are equal.
*** Adapted from a problem which appeared in a Hungarian Math Contest..
(In reply to
re: Solution Expanded by Bractals)
Just
spotted your additional line. I was intending to approach
this in the opposite direction as follows:
Chasing angles round the three inner triangles and using the
sine rule gives:
PC AP BP sin(A – z) * cos(z) * cos(A + z)
  
= 1 = 
AP BP PC sin(z) * sin(z) *
sin(z)
where z = /PAB. Then, using compound angle formulae,
this reduces to: sin(2A) = 2 tan(z) =
2 tan /PAB
(which is effectively what your additional line says).
Since finding Q is the same procedure as finding P, except
that the labels A and C are interchanged, it follows that
sin(2C) = 2 tan /QCB = 2 tan /QBA
and therefore that /PAB = /QBA,
since sin(2A) = sin(2C).
Much fun.

Posted by Harry
on 20150304 16:25:43 