2018
Problem - 4764
A circle of radius $2$, center on the origin, is drawn on a grid of points with integer coordinates. Let $n$ be the grid points that lie within or on the circle. What is the smallest amount of radius needs to increase by for there to be $(2n-5)$ grid points within or on the circle?
First, $n$ is the number of integer solutions for the inequality $x^2 + y^2 = 2^2$. Therefore, $n=13$. The next step is to find the minimal radius such that there are $2\times 13-5=21$ points inside the circle, or exactly $8$ more points. We note that the next set of points closet to the origin are $(\pm 1, \pm 2)$ and $(\pm 2, \pm 1)$ which contain exactly $8$ points and their distances to the origin are $5$. Hence, the desired radius is $\sqrt{5}$ which means the answer is $\boxed{\sqrt{5}-2}$.