China Mathematical Competition

Suppose that $A(a,9-a)$. Then the distance from the center of $M$ to the line $AC$ is $$\begin{gathered} d=|AM|\times \sin \angle BAC \\ =\sqrt{(a-2{)}^2+(9-a-2{)}^2}\times \sin 45^{\circ } \\ =\sqrt{2a^2-18a+53}\times \frac{\sqrt{2}}{2}. \end{gathered}$$ On the other hand, since the line $AC$ intercepts $M$, it follows that $d\le$ the radius of $M=\sqrt{\frac{17}{2}}$, i.e. $$\sqrt{2a^2-18a+53}\times \frac{\sqrt{2}}{2}\le \sqrt{\frac{17}{2}}$$ The solution is $3\le a\le 6$.