2002 AMC 12B Problem 23

Let $M$ be the midpoint of $BC,$ set $AM=2a,$ and let $\theta=\angle AMB,$ so $\angle AMC=180^\circ-\theta.$ With $BM=CM=a,$ so that $BC=2a,$ the Law of Cosines in $\triangle ABM$ and $\triangle AMC$ gives $$a^2+4a^2-4a^2\cos\theta=1,$$ $$a^2+4a^2+4a^2\cos\theta=4.$$ Adding, $10a^2=5,$ so $a=\dfrac{\sqrt2}{2}$ and $BC=2a=\sqrt2.$ Thus, the correct answer is C.