In parallelogram $ABCD$, $AB = 10$, and $AB = 2BC$. Let $M$ be the midpoint of $CD$, and suppose that $BM = 2AM$. Compute $AM$.
1. Given that $ABCD$ is a parallelogram with $AB = 10$ and $AB = 2BC$, we can deduce that $BC = \frac{AB}{2} = \frac{10}{2} = 5$.

2. Let $M$ be the midpoint of $CD$. Since $M$ is the midpoint, $CM = MD$.

3. We are given that $BM = 2AM$. Let $AM = x$. Then $BM = 2x$.

4. Since $M$ is the midpoint of $CD$, $M$ divides $CD$ into two equal segments. Therefore, $CD = 2CM = 2MD$.

5. In parallelogram $ABCD$, opposite sides are equal, so $CD = AB = 10$.

6. Since $M$ is the midpoint of $CD$, $CM = MD = \frac{CD}{2} = \frac{10}{2} = 5$.

7. Now, consider triangle $AMB$. We know that $AM = x$ and $BM = 2x$. We need to find the length of $AM$.

8. Since $AB = 10$, we can use the Pythagorean theorem in triangle $AMB$:
   \[
   AB^2 = AM^2 + BM^2
   \]
   Substituting the known values:
   \[
   10^2 = x^2 + (2x)^2
   \]
   Simplifying:
   \[
   100 = x^2 + 4x^2
   \]
   \[
   100 = 5x^2
   \]
   \[
   x^2 = \frac{100}{5}
   \]
   \[
   x^2 = 20
   \]
   \[
   x = \sqrt{20} = 2\sqrt{5}
   \]

The final answer is $\boxed{2\sqrt{5}}$