Track / Stage 6 / 153 of 400 #1153 of 2000

Problem 1153

National olympiad, first round

Geometry Difficulty 6.2 Prove it

13.17 On a grid paper with small squares of size $1 \times 1$ , a circle of radius $R$ is drawn with one of the nodes (the intersection points of the grid lines) as the center. Prove that if there are exactly 1988 nodes on the circumference, then either $R$ or $\sqrt{2} R$ is an integer.
(14th All-Russian Mathematical Olympiad, 1988)

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

[Proof] With the center of the circle as the origin, and the coordinate axes parallel to the grid lines, establish a coordinate system on the grid paper. Then all nodes are integer points.

If the point $(x, y)$ lies on the circumference of the circle, by symmetry, all points of the form $(\pm x, \pm y)$ and $(\pm y, \pm x)$ also lie on the circumference of the circle. If $x \neq y$ and $xy \neq 0$ , then there are 8 such points; if $x = y$ or $xy = 0$ , then there are 4 such points.
Since
$1988 \equiv 4 \quad(\bmod 8) .$

Therefore, among the given 1988 points, there must be points satisfying $x = y$ or points satisfying $xy = 0$ .
If there exists a point satisfying $x = y$ , then by
$x^{2} + y^{2} = 2x^{2} = R^{2}$

it follows that $\frac{R}{\sqrt{2}}$ is an integer, hence $\sqrt{2} R$ is an integer.
If there exists a point satisfying $xy = 0$ , then by
$R = |x| \quad \text{or} R = |y|$

it follows that $R$ is an integer.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.