2017 AMC 10A Problem 17

To make the fraction $\frac{PQ}{RS}$ as big as possible, make the top as long as possible and the bottom as short as possible.

1. List the points. We need whole numbers with $x^2 + y^2 = 25$. The only ways to write $25$ as a sum of two squares are $0 + 25$ and $9 + 16$. That gives $12$ points:
$$(0, \pm 5),\ (\pm 5, 0),\ (\pm 3, \pm 4),\ (\pm 4, \pm 3).$$
2. Which lengths are irrational? For two such points, the squared distance ($\text{across}^2 + \text{up}^2$) is a whole number. The length is irrational exactly when that whole number is not a perfect square.
3. Longest allowed $PQ$. The longest chord is a diameter, length $10$, but $10$ is not irrational, so it is not allowed. The next best is to join two points that are almost opposite, such as $(3, 4)$ and $(-4, -3)$. They differ by $7$ across and $7$ up:
$$PQ^2 = 7^2 + 7^2 = 98, \qquad PQ = \sqrt{98} = 7\sqrt2.$$
$98$ is not a perfect square, so this is allowed. No other pair (except diameters) gets farther apart; for example $(5, 0)$ to $(-4, 3)$ gives only $81 + 9 = 90$.
4. Shortest allowed $RS$. Neighbors on the circle such as $(-3, 4)$ and $(-4, 3)$ differ by $1$ across and $1$ up:
$$RS^2 = 1 + 1 = 2, \qquad RS = \sqrt2.$$
Two different points on this circle are never just $1$ apart, so $\sqrt2$ is the shortest possible.
5. The four points are all different, as required, so the best ratio is
$$\frac{PQ}{RS} = \frac{7\sqrt2}{\sqrt2} = 7.$$

The answer is $\boxed{\textbf{(D)}\ 7}$.

When only a few points are possible, write them all down and search directly. Maximizing a fraction means pushing the top up and the bottom down separately. The "irrational" condition is easy to check with squared distances: a length is irrational exactly when its square is a whole number that is not a perfect square.