, 2020

Solution:

Consider a coordinate system on any line $\ell$ where 0 is placed at the foot from $(0,0)$ to $\ell$. Then, by the Pythagorean theorem, a point $(x, y)$ on $\ell$ is assigned a coordinate $u$ for which $x^{2}+y^{2}=u^{2}+a$ for some fixed $a$ (dependent only on $\ell$). Consider this assignment of coordinates for our segment.

First, suppose that along the line segment $u$ never changes sign; without loss of generality, assume it is positive. Then, if $u_{0}$ is the minimum value of $u$, the length of the interval covered by $u^{2}$ is $\left(u_{0}+10\right)^{2}-u_{0}^{2}=100+20 u_{0} \geq 100$, meaning that at least 100 points lie on the given circles.

Now suppose that $u$ is positive on a length of $k$ and negative on a length of $10-k$. Then, it must intersect the circles at least $\left\lfloor k^{2}\right\rfloor+\left\lfloor(10-k)^{2}\right\rfloor$ points, which can be achieved for any $k$ by setting $a=2020+\varepsilon$ for very small $\varepsilon$.

To minimize this quantity note that $k^{2}+(10-k)^{2} \geq 50$, so $\left\lfloor k^{2}\right\rfloor+\left\lfloor(10-k)^{2}\right\rfloor>k^{2}+(10-k)^{2}-2 \geq 48$, proving the bound. For a construction, set $k=4.99999$.