# How to Solve this BMO 2005 round 2 problem?

Let $$ABC$$ be a triangle with $$AC > AB$$. The point $$X$$ lies on the side $$BA$$ extended through $$A$$, and the point $$Y$$ lies on the side $$CA$$ in such a way that $$BX = CA$$ and $$CY = BA$$. The line $$XY$$ meets the perpendicular bisector of side $$BC$$ at $$P$$. Show that $$\angle BPC + \angle BAC = 180^{\circ}$$.

Let $$l$$ denote the angle bisector of $$\angle{BAC}$$ and $$m$$ the parallel to $$l$$ passing through the midpoint $$M$$ of $$BC$$. It suffices to prove that $$XY$$ is reflection of $$l$$ over $$m$$. Indeed, since midpoint of arc $$BC$$ of the circumcircle of $$ABC$$ which doesn't contain $$A$$ lies on $$l$$ and $$M$$ lies on $$m$$, the reflection of the first one with respect to the second one lies on $$XY$$ which happens to be a point $$P$$. Since the midpoint of arc satisfies the equality of angles in question and reflection with respect to $$M$$ preserves this property we get the result. Now, we will prove fact mentioned at the beginning it in three different ways:
1) If we denote the intersection of $$l$$ and $$BC$$ by $$L$$ as well as the intersection of $$XY$$ and $$BC$$ by $$Z$$, by the angle bisector and Thales theorem (as previously mentioned $$AL || XY$$) we have: $$\frac{CZ}{CY}=\frac{CL}{CA}=\frac{BL}{BA}$$ and since $$CY=BA$$ we have $$BL=CZ$$ i.e. $$Z$$ and $$L$$ are symmetric with respect to $$M$$ which yields the result.
2) If we consider glide reflection with respect to $$m$$ such that the image of $$AB$$ is $$AC$$ we notice that $$X$$ goes to $$A$$ (because $$B$$ goes to $$C$$ in this reflection as the midpoint of $$BC$$ stays fixed). We can consider analogous reflection taking $$AC$$ to $$AB$$. This proves that the distances of $$A$$, $$X$$ and $$Y$$ to $$m$$ are the same which finishes the proof.
3) One can also consider $$X'$$ and $$Y'$$ to be the midpoints of $$AX$$ and $$AY$$ and make use of the fact that $$AX'=\frac{1}{2}AX=\frac{BX-AB}{2}=\frac{b-c}{2}$$ and similarly for $$AY'$$ by applying Menelaus theorem to $$X'$$, $$Y'$$ and $$M$$ to see that they are collinear.
The line through $$A$$ parallel to $$XY$$ is the bisector of angle $$\angle BAC$$. It meets again the circle through $$ABC$$ at the midpoint of arc $$BC$$, i.e. on the perpendicular bisector of $$BC$$.
A few things to note: $$PB=PC$$, since $$P$$ is on the perp. bisector of $$BC$$. Also, we have $$AC=BX$$ and $$CY=AB$$ so $$AY=AX$$. Also, do you see that the claim is proven if you can prove that $$\angle BPC=\angle XAY$$?