Equal Segments

A problem from the geometer Jean-Louis Ayme, see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&p=3590641.

Let [math]ABC[/math]be a triangle, with circumcenter [math]O[/math]. Denote the intersection of the internal bisector of [math]\angle BAC[/math] with [math]BC[/math] as [math]M[/math]. Let [math]P[/math] be on [math]CA [/math]such that [math]MP\perp AO[/math]. Prove that [math]AB = AP[/math].

Perpendicular bisector

A surprising result from the 36th Spanish Mathematical Olympiad, see solution http://www.artofproblemsolving.com/Forum/blog.php?u=214539&b=107527.

Let [math]\mathcal{S}_1, \mathcal{S}_2[/math] be two circles of different radii, centers [math]O_1, O_2[/math] respectively, which intersect at [math]A,B[/math]. Let [math]PQ[/math] be a line through [math]B[/math] meeting [math]\mathcal{S}_1, \mathcal{S}_2[/math] at [math]P,Q[/math] respectively. Prove that a) For any choice of [math]\overline{PQ}[/math], the perpendicular bisector of [math]PQ[/math] passes through a fixed point [math]M[/math], and b) [math]AO_1O_2M[/math] is an isosceles trapezium.

APMO 2014 Q5

APMO 2014 Q5, see solution http://www.artofproblemsolving.com/Forum/blog.php?u=214539&b=109581.

Circles [math]\omega[/math] and [math]\Omega[/math] meet at points [math]A[/math] and [math]B[/math]. Let [math]M[/math] be the midpoint of the arc [math]AB[/math] of circle [math]\omega[/math] ([math]M[/math] lies inside [math]\Omega[/math]). A chord [math]MP[/math] of circle [math]\omega[/math] intersects [math]\Omega[/math] at [math]Q[/math] ([math]Q[/math] lies inside [math]\omega[/math]). Let [math]\ell_P[/math] be the tangent line to [math]\omega[/math] at [math]P[/math], and let [math]\ell_Q[/math] be the tangent line to [math]\Omega[/math] at [math]Q[/math]. Prove that the circumcircle of the triangle formed by the lines [math]\ell_P[/math], [math]\ell_Q[/math] and [math]AB[/math] is tangent to [math]\Omega[/math].