2017 AMC 10A Problem 17

The integer solutions of $x^2 + y^2 = 25$ come from $25 = 0^2 + 5^2 = 3^2 + 4^2$, giving the $12$ points $(0, \pm5)$, $(\pm5, 0)$, $(\pm3, \pm4)$, $(\pm4, \pm3)$.

The squared distance between two lattice points is an integer, so a distance is irrational exactly when that integer is not a perfect square. To maximize $PQ / RS$, make $PQ$ as large as possible and $RS$ as small as possible.

Largest: the diameter has length $10$, which is rational and not allowed. The next-longest chords join nearly opposite points such as $(3, 4)$ and $(-4, -3)$: squared distance $7^2 + 7^2 = 98$, so $PQ = \sqrt{98} = 7\sqrt2$. (Any other pair gives a smaller squared distance, e.g. $(3,4)$ to $(0,-5)$ gives $90$.)

Smallest: neighboring points such as $(-3, 4)$ and $(-4, 3)$ have squared distance $1 + 1 = 2$, so $RS = \sqrt2$. No two distinct lattice points are closer than $1$, and distance $1$ does not occur on this circle.

These four points are distinct, and
$$\frac{PQ}{RS} = \frac{7\sqrt2}{\sqrt2} = 7.$$

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

A finite, listable set of points turns an optimization into an enumeration: maximize the numerator and minimize the denominator separately, subject to the irrationality filter. Squared distances between lattice points are integers, so "irrational" simply means "not a perfect square," and the diameter is the one natural candidate that fails.

• Using the diameter 10 as PQ, forgetting that PQ must be irrational.
• Missing the points $(\pm4, \pm3)$ and working only with $(\pm3, \pm4)$ and the axis points, so the shortest chord $\sqrt2$ is never found.