[Algebra]
Problem: Let $T=256$. Let $\mathcal{R}$ be the region in the plane defined by the inequalities $x^{2}+y^{2} \geq T$ and $|x|+|y| \leq \sqrt{2 T}$. Compute the area of region $\mathcal{R}$.

Solution: The first inequality states that the point $(x, y)$ is outside the circle centered at the origin with radius $\sqrt{T}$, while the second inequality states that $(x, y)$ is inside the tilted square centered at the origin with diagonal $2 \sqrt{2 T}$. The area of the square is $4 \cdot \frac{1}{2}(\sqrt{2 T})^{2}=4 T$, while the area of the circle is simply $\pi T$, so the area of $\mathcal{R}$ is $4 T-\pi T=\mathbf{1 0 2 4}-\mathbf{2 5 6 \pi}$.

Final answer: $1024-256 \pi$