I challenge you to bash this

For those who have yet to study advanced Euclidean geometry, $Q$ is called the isogonal conjugate of $P$ with respect to $ABC$ ; it also satisfies $\angle BAQ=\angle CAP=15^{\circ}$ . The locus of points $P$ such that $PQ$ intersects a fixed point is an interesting topic of study in modern Euclidean geometry. In particular, I have written a paper(a collaberation) on a ratio perserving property this locus possesses. Anyway, the fixed point that $PQ$ crosses in this problem is the circumcenter, and I will prove below that such $P$ satisfies $\angle CAP+\angle PCB+\angle ABP=90^{\circ}$ . Since $\angle QBC=\angle ABP$ , the answer to this question is $90-15=\boxed {75^{\circ}}$ . (As one can see, the value of $\angle A$ here is irrelevant, I added it in case bashers want to give it a go.)

Before attempting to prove $\angle CAP+\angle PCB+\angle ABP=90^{\circ}$ , I shall mention a few general properties about isogonal conjugates(try to prove these on your own):

First, the pedal triangles of $P,Q$ with respect to $ABC$ (triangle formed by projecting the points onto the side of $ABC$ ) share the same circumcircle; the circumcenter clearly lies on the midpoint of $QP$ .

Next, let $DEF$ denote $P$ 's pedal triangle( $D,E,F$ lie on $BC,AC,AB$ respectively), $Q'$ is the isogonal conjugate of $P$ with respect to $DEF$ . Then $PQ\parallel QO$ .

Finally, let $LMN$ denote of circumcevian triangle of $P$ (triangle form of the intesection of $AP,BP,CP$ with the circumcircle of $ABC$ . i.e. $L=AP\cap \odot ABC$ and similarly for $M,N$ ). Then $\triangle DEF\sim \triangle LMN$ . Moreover, $Q',P$ are corresponding points in the two similar triangles respectively.

With these general properties about $P,Q$ in mind, we now look at what happens with $O$ lying on $PQ$ . From the second property, we immediately deduce that $Q'$ lies on $PQ$ . Let $O'$ denote the circumcenter of $DEF$ , which lies on $PQ$ from the first property. By the third property, we know that $Q'O', PO$ are corresponding elements in the two similar triangles $DEF, LMN$ . Since they are parallel(they all lie on the same line), we can deduce that the triangles $DEF, LMN$ are homothetic. Again, we have established that the pedal and cevian triangle of $P$ are homothetic, a property that is only true with $O$ lies on $PQ$ . We are now ready to prove our angle equality(just angle chasing from now on)

It is a general fact about isogonals that $FE\perp AQ$ (easy to prove, like the ones above), since we've proven $MN\parallel EF$ , therefore $MN\perp AQ$ $\implies 90^{\circ}=\angle NAQ+\angle ANM=\angle BAQ+\angle BAN+\angle ABP=\angle CAP+\angle PCB+\angle ABP$ . We are done.

Feel free to ask me anything about isogonal conjugates.