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Problem 983

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Consonant Circle Crossing

We say two circles on the plane harmonise if the circles intersect at two grid points (A point with integer coordinates), in which case the two intersection points are called the harmony points.

A set of circles on the plane is called consonant if it satisfies all the following requirements:

1. There are at least two circles in the set.
2. The center point of every circle is a grid point.
3. All circles have the same radius.
4. No circle is tangent to any other circle.
5. The circles are connected in the sense that a chain of circles can be formed between every pair of circles such that each circle harmonises with the next circle.

It can be proven that the number of unique harmony points of a consonant set of circles cannot be smaller than the number of circles. If the number of unique harmony points equals the number of circles, we say the consonant set is perfect.

For example, here are two perfect consonant sets of circles:

Let $R(n)$ be the minimal radius $r$ such that a perfect consonant set of $n$ or more circles with radius $r$ exists.

You are given $R(2) = 1$ and $R(4) = \sqrt{5}$.

Find $R(500)^2$.

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完美协和圆集

若平面上的两个圆相交于两个格点（坐标均为整数的点），则称这两个圆是和谐的，并称这两个交点为和谐点。

若平面上的一组圆所构成的集合满足以下所有条件，则称这个圆集是协和的：

1. 圆集中至少有两个圆；
2. 每个圆的圆心都是格点；
3. 所有圆的半径都相同；
4. 任意两个圆均不相切；
5. 这些圆可以构成一条连通的圆链，使得链上任意相邻的两个圆都是和谐的。

可以证明，在协和圆集中不同和谐点的数量不会少于其中圆的数量；如果不同和谐点的数量恰好等于圆的数量，则进一步称这个协和圆集是完美的。

例如，下图展示了两个完美协和圆集：

记$R(n)$为使至少$n$个圆构成完美协和圆集所需的最小半径$r$。

已知$R(2) = 1$，$R(4) = \sqrt{5}$。

求$R(500)^2$。

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