Let $A(2,2)$ and $B(7,6)$ be points in the plane. Define $R$ as the region in the first quadrant consisting of those points $C$ such that $\triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $R$?
The angles at $A$ and $B$ are acute exactly when $18<5x+4y<59$, forming a trapezoid in the first quadrant with area $\frac{59^2-18^2}{40}=\frac{3157}{40}$.
For the angle at $C$ to be acute, $C$ must lie outside the circle with diameter $AB$. This circle lies entirely inside the trapezoid and has area $\pi(\frac{\sqrt{41}}2)^2=\frac{41\pi}{4}$.
Thus $[R]=\frac{3157}{40}-\frac{41\pi}{4}\approx46.72$, so the closest integer is $\boxed{47}$.
