Harvard-MIT November Tournament

Solution:

We can rewrite $x^{2}+y^{2}-3y-1=0$ as
$$x^{2} + \left(y - \frac{3}{2}\right)^{2} = \frac{13}{4}.$$
We then see that the set of solutions to $x^{2}+y^{2}-3y-1=0$ is the circle of radius $\frac{\sqrt{13}}{2}$ and center $\left(0, \frac{3}{2}\right)$.

This can be written as
$$x = \frac{\sqrt{13}}{2} \cos(\theta), \quad y = \frac{\sqrt{13}}{2} \sin(\theta) + \frac{3}{2}.$$
Thus,
$$x + y = \frac{3}{2} + \frac{\sqrt{13}}{2}(\cos(\theta) + \sin(\theta)) = \frac{3}{2} + \frac{\sqrt{13}}{2} \sqrt{2} \sin\left(\theta + 45^{\circ}\right).$$
This is maximized for $\theta = 45^{\circ}$ and gives
$$\frac{\sqrt{26} + 3}{2}.$$
(We could also solve this geometrically by noting that if $x+y$ attains a maximum value of $s$ then the line $x+y=s$ is tangent to the circle.)