- Let $S$ be the set formed by these $n$ centers. We first prove: there exists a line $L$ such that the projections of the points in $S$ onto $L$ fall within an interval of length $\sqrt{4n}$.

證明：設 $A, B$ 為 $S$ 中相距最遠的兩個點，令其距離為 $d$。

Proof: Let $A, B$ be the two points in $S$ that are farthest apart, and let their distance be $d$.

* Take any point $C$ in $S$ different from $A, B$; since $|\triangle ABC| \le n$, the distance from $C$ to line $AB$ is at most $2n/d$.

* Hence, if $L$ is perpendicular to line $AB$ at $D$, then the projection of any point in $S$ onto $L$ will fall within an interval centered at $D$ with length $4n/d$.

* Also, since the maximum distance between two points in $S$ is $d$, the projections of the points in $S$ onto $L$ must fall within an interval of length $d$, so this interval has length at most $\min\{d, 4n/d\} \le \sqrt{4n}$. Q.E.D.!

- Note that the projections of these $n$ circles onto $L$ are all intervals of length 2, and these intervals are all contained within an interval of length at most $\sqrt{4n} + 2$. Let this interval be $I$. Let $C_i$ be the projection of the $i$-th circle onto $L$.

- Now note that the total length of all the $C_i$ is $2n$, and they all fall within $E$. This means there must exist at least one point $x$ in $E$ that belongs to at least $\frac{2n}{\sqrt{4n+2}}$ of the $C_i$ simultaneously (otherwise, by the averaging principle, a contradiction would result.)

- Therefore, taking the line $L$ perpendicular to $AB$ passing through the point $x$ suffices. Q.E.D.

Substituting $n = 100$, we get at least $\frac{200}{\sqrt{400+2}} = \frac{200}{22} > 9$, so it intersects at least 10 circles.