2015 Ninth STARS OF MATHEMATICS Competition

Since the lines $AC$, $BD$ and $MN$ are concurrent, $(AM/MB)(BC/CN)(ND/DA)=1$, by Ceva's theorem in trigonometric form along with the sine law in the triangle $ABN$; and since $MA=MB$, it follows that $CN/DN=BC/AD$.

Next, the triangles $ADP$ and $QDA$ are similar, so $A{D}^2=DP\cdot DQ$. Similarly, $B{C}^2=CP\cdot CQ$, so, by the preceding, $(CN/DN{)}^2=(CP/DP)(CQ/DQ)$.



By a well-known theorem of Steiner, the lines $NP$ and $NQ$ are isogonal with respect to the lines $NC$ and $ND$; that is, the angles $CNP$ and $DNQ$ are congruent.

Finally, if $X$ is a point on the tangent $t$ at $N$ to $\gamma$, other than $N$, then a standard angle chase shows the angle $QNX$ congruent to one of the angles $CPN$, $NPQ$, depending on which side of $N$ the point $X$ lies on $t$, so $t$ is also tangent at $N$ to the circle $NPQ$. The conclusion follows.