Is there a set of point $S \subset \mathbb R^2$ such that $|\{C: C \text{ is unit circle boundary }, |C \cap S| = 10\}| > |S|$

Question

There are some blue points and red points on the plane such that in the boundary of every unit circle centered at one blue point there are exactly 10 red point. Can the number of blue points strictly greater than the number of red points ?

This is a problem I found a very long time ago with any solution the only hint I found is this problem will use probabilistic method somewhere

I figured out that if we change exactly 10 to at least 10, the answer is yes, it is because we can chessboard coloring a large grid and scale it down by a factor of $1/ \sqrt{N}$ where $N$ satisfy that the solution $x^2+y^2 = N$ has at least 10 positive integer solution.

Translating the problem into a set-intersecting problem will not work here, there exist a large number $n$ and a family of set $\mathcal F$ with $|\mathcal F| = O(n^2)$ such that $|C| = 10$ and $C \subset \{1,2,\ldots,n\}$ for every $C \in \mathcal F$ and $|C_i \cap C_j| < 2 $ for all $C_i, C_j \in \mathcal F, C_i \not = C_j$

Answer 1

Yes, it is possible, using a modified version of the grid construction from the question (thanks to jackdean for the more elegant version of my argument).

Firstly, let $N=5^{24}$, so that $x^2+y^2=N^2$ has $100$ integer solutions with $x+y\equiv1$ mod $2$.

Now consider a grid $X=\{1,\dots,M\}\times\{1,\dots,M\}$, where $M$ is a big even natural number. We will color each even point $(x,y)\in X$ (by which I mean points with $x+y$ even) red with probability $\frac{1}{10}$. So the expected number of red points is $M^2/20$.

Now, each odd point $p$ in $X$ (except the ones near the edges of the grid, but there are $<M^2/10^{10}$ of those) has $100$ even points of $X$ at distance $\sqrt{N}$. So the probability that there are exactly $10$ red points at distance $\sqrt{N}$ of $p$ is $\binom{100}{10}0.9^{90}0.1^{10}>0.131$.

Now color the odd points in $X$ which have exactly $10$ red points at distance $\sqrt{N}$ blue. Then the expected number of blue points is at least $\frac{M^2}{2}(1-\frac{1}{10^9})\cdot0.131>0.065M^2$.

As the expected number of blue points is bigger than the expected number of red points, we certainly can have more blue points than red points, so scaling the grid by $\frac{1}{\sqrt{N}}$, we are done.

Answer 2

An alternative solution is the following:

Take 10 unit vectors in the generic position. Consider all $1024$ possible sums from the empty one (the zero vector) to the full sum of all $10$ vectors. Color sums with odd number of vectors blue and sums with even number of vectors red. You'll get almost what you want except the number of blue points is now exactly the same as the number of red points. To fix it, add another configuration of the same type in the generic position with respect to the first one. The red origin will be common, so now the blue count will exceed the red one by $1$.