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 2015-03-04 16:25:43