Question 1210147
Let $\triangle ABC$ be a triangle with $AB = AC = 4$ and $\angle BAC = 45^\circ$. Let $M$ be the midpoint of $BC$. We want to find $AM^2$.

We can use the Law of Cosines to find $BC$.
$$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)$$
$$BC^2 = 4^2 + 4^2 - 2(4)(4)\cos(45^\circ)$$
$$BC^2 = 16 + 16 - 32\left(\frac{\sqrt{2}}{2}\right)$$
$$BC^2 = 32 - 16\sqrt{2}$$
$$BC = \sqrt{32 - 16\sqrt{2}} = 4\sqrt{2 - \sqrt{2}}$$

Since $M$ is the midpoint of $BC$, we can use Apollonius' Theorem:
$$AB^2 + AC^2 = 2(AM^2 + BM^2)$$
$$4^2 + 4^2 = 2(AM^2 + BM^2)$$
$$32 = 2(AM^2 + BM^2)$$
$$16 = AM^2 + BM^2$$

We know $BM = \frac{1}{2}BC$, so $BM^2 = \frac{1}{4}BC^2 = \frac{1}{4}(32 - 16\sqrt{2}) = 8 - 4\sqrt{2}$.

Substituting $BM^2$ into the equation from Apollonius' Theorem:
$$16 = AM^2 + (8 - 4\sqrt{2})$$
$$AM^2 = 16 - (8 - 4\sqrt{2})$$
$$AM^2 = 8 + 4\sqrt{2}$$

Thus, $AM^2 = 8 + 4\sqrt{2}$.

Final Answer: The final answer is $\boxed{8 + 4 \sqrt{2}}$