Problem 264

Solution:

The answer is $\mathbf{(B)}$.

Since $M$ is the midpoint we obtain that $BM = MC = 2$. Let $H$ be the midpoint of $MC$, and in particular $MH = HC = 1$; since triangle $AMC$ is isosceles on base $MC$, $H$ is also the foot of the altitude and therefore angle $\angle AHB$ is right.

Thus triangle $ABH$ is right-angled, with hypotenuse $AB$ of length $5$ and leg $BH$ of length $3$; it follows that $AH = 4$. Finally $AC$ is the hypotenuse of triangle $AHC$, so
$$AC = \sqrt{4^2 + 1^2} = \sqrt{17}$$