Kuzneuov A.

In space, there are 2016 spheres, no two of which coincide. Some of the spheres are red, and the rest are green. Each point of contact between a red and a green sphere was painted blue. Find the maximum possible number of blue points.
Evaluation. Let there be $r$ red spheres and $2016-r$ green spheres. Since any two spheres touch at no more than one point, the number of blue points does not exceed $r(2016-r)=1008^{2}-(1008-r)^{2} \leq 1008^{2}$.

We will provide an example with this number of blue points. Let $l$ be some line, $\alpha$ a plane perpendicular to $l$ and intersecting it at point $O$, and $\omega$ a circle with center $O$ and radius 1, lying in $\alpha$. Construct 1008 red spheres of the same radius $r<1$ with different centers $R_{1}, R_{2}, \ldots, R_{1008}$, lying on $\omega$.

Let $G_{1}, G_{2}, \ldots, G_{1008}$ be different points on $l$, at distances $d_{1}, d_{2}, \ldots, d_{1008}$ from $O$. Then the distance between $G_{i}$ and any point $R_{j}$ is $\sqrt{1+d_{i}^{2}}$. Therefore, if we construct a green sphere with center $G_{i}$ and radius $\sqrt{1+d_{i}^{2}}-r$, it will touch all the red spheres. Moreover, all points of contact will be distinct, as they lie on segments of the form $R_{j} G_{i}$, which do not have common points except the endpoints. Thus, in our construction, there will indeed be $1008^{2}$ blue points.

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-46.jpg?height=492&width=626&top_left_y=1898&top_left_x=722)

## Answer

$1008^{2}=1016064$ points.

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