Given a triangle ABC, how can you find a point P on AC and a point Q on BC, such that AP=PQ=QB?

N.B. A construction method is sought, and only compass and straightedge are allowed.

(In reply to re: Two solutions for one triangle! by Charlie)

Sorry, I overlooked the significance of Charlie's first comment.  Although I am not sure that there are two solutions for every triangle, Charlie has made it clear that two solutions is commonplace (but not yet clear that there is at least one solution).  Isosceles triangles with the angle at C different from 60 degrees always have two solutions.  But if the triangle is equilateral, there is only one solution,isn't there?

Here's why I think there is always at least one solution.  Draw a line segment P'Q'.  Draw two circles of radius the length of P'Q' centered at P' and Q'.  Let C' be any point different from P' and Q' so that angle P'C'Q' equals angle ACB.  There is a unique circle containing the points P', Q', and C'.  Every point X on the arc of this circle containing C' has angle P'XQ' equal to angle ACB.  Extend C'P' to the point A' on the circumference of the circle centered at P' and extend C'Q' to the point B' on the circumference of the circle centered at Q'.  As C' moves along the circle containing C', P', and Q', the angle C'A'B' varies from 0 to 180 degrees.  Hence we can move C' until the triangle A'B'C' is similar to triangle ABC.  Moreover, P' and Q' form a solution of Kereki's problem for triangle A'B'C'.  Indeed, it would appear, there is always a solution with, in Charlie's language, with x positive.  (My examples came from extending one of C'P' and C'Q' and "shrinking" the other, since these segments usually meet the circumference of the circles in two places)

I imagine Charlie's computation can include the examples I gave by simply allowing his x to be added in one expression and subtracted in another.

But, alas, this discussion gives me no clue towards a straightedge-compass construction.

Posted by McWorter on 2005-05-08 02:48:19