In right $\triangle ABC$ with legs $5$ and $12$, arcs of circles are drawn, one with center $A$ and radius $12$, the other with center $B$ and radius $5$. They intersect the [hypotenuse](https://artofproblemsolving.com/wiki/index.php/Hypotenuse) in $M$ and $N$. Then, $MN$ has length

$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }\frac{13}{5} \qquad \mathrm{(C) \  } 3 \qquad \mathrm{(D) \  } 4 \qquad \mathrm{(E) \  }\frac{24}{5}$
First of all, from the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$. Also, since $AM$ and $AC$ are radii of the same circle, $AM=AC=12$. Therefore, $MB=AB-AM=13-12=1$. Also, since $BN$ and $BC$ are radii of the same circle, $BN=BC=5$. We therefore have $MN=BN-BM=5-1=4, \boxed{\text{D}}$.