# Prove there is a line intersecting at least three of these circles.

One hundred circles of radius one are positioned in the plane so that the area of any triangle formed by the centres of three of these circles is at most 2017. Prove there is a line intersecting at least three of these circles.

I found a solution below, but I have a question at the end that would help me understand the solution better.

The distance between two projections of two centers on any line is at most the actual distance between the centers, and this can be shown using the triangle inequality.

Why is it that in the solution, each point of this interval belongs to on average $$\dfrac{2n}{\sqrt{8068} + 2}$$ of the subintervals of length 2? Also, isn't the set of points in this interval uncountable? So shouldn't an integral be used for the average?

We already know that we have $$100$$ intervals of length $$2$$ which are contained in an interval of length $$92$$.
Consider all the intervals on the $$x$$-axis with $$0\le x_1\le x_2 \le ... \le x_{100}$$, where $$x_i$$ ($$1\le i \le 100$$) is the first point of the $$i$$ th interval ($$[x_i, x_i+2]$$).
Now assume there is no point belonging to three intervals. In this manner if $$x_1+2\le x_2$$, then $$2 \le x_2\le x_3$$, and if $$x_1+2\ge x_2$$, then we must have $$x_1+2\le x_3$$. Hence, in either case, we get $$2\le x_3$$. Again, considering $$2\le x_3\le x_4 \le x_5$$, we reach $$4\le x_5$$. By repeating this inductive comparison, we conclude that $$6\le x_7$$, $$8\le x_9$$, ..., and $$98\le x_{99}$$, which is a contradiction. Therefore there is some point which is contained in three of those intervals.