China Mathematical Competition (Hainan)

The center of a circle passing through points $M$ and $N$ is on the perpendicular bisector $y=3-x$ of $MN$. Denote the center by $S(a,3-a)$, then the equation of the circle $S$ is $$(x-a{)}^2+(y-3+a{)}^2=2(1+a^2).$$ Since for a chord with a fixed length, the angle at the circumference subtended by the corresponding arc will become larger as the radius of the circle becomes smaller. When $\angle MPN$ reaches its maximum value, the circle $S$ through the three points $M$, $N$ and $P$ will be tangent to the $x$-axis at $P$, which means the value $a$ in the equation of $S$ has to satisfy the condition $2(1+a^2)=(a-3{)}^2$. Solve the above equation we have $a=1$ or $a=-7$. Thus the points of contact are $P(1,0)$ and $P^{\prime }(-7,0)$ respectively. But the radius of the circle through points $M$, $N$, and $P^{\prime }$ is larger than that of the circle through points $M$, $N$ and $P$. Therefore $\angle MPN>\angle M{P}^{\prime }N$. Thus $P(1,0)$ is the point we want to find, and the $x$-axis of point $P$ is 1.