I looked at this page: 51 points lie inside an square of side 1.Prove that it's possible to draw a circle of radius $\frac17$ covering at least 3 of theses points
Instead of using the Pigeonhole Principle, I will attempt to find a proof using expected value. If I show that the expected number of points a random circle will cover is more than 2, then I know that it is always possible for the circle to cover at least 3.
Given any given point, if the center of the circle is within $\frac{1}{7}$ of the point, then it will cover it. So given any point, the area of the set of all possible centers that cover the point is $\frac{\pi}{49}$.
To find the total area of the set all possible centers, we need to find the set of all centers such that the circle intersects with the square. (Otherwise it would never cover a point.) This means that the center must be within $\frac{1}{7}$ of the square. Below is all possible regions for the center. 
The big square in the middle is the unit square with area 1, the 4 rectangles has its base equal to 1 and its height equal to $\frac{1}{7}$, and the four quarter-circles make 1 circle with radius $1/7$, so the total possible area is $1 + \frac{4}{7} + \frac{\pi}{49} = \frac{77}{49} + \frac{\pi}{49} = \frac{\pi+77}{49}$.
Therefore, the probability that a randomly chosen circle will cover a point is $\frac{\pi/49}{(\pi+77)/49} = \frac{\pi}{\pi+77}$
Because there are 51 points, the expected number of points a random circle would cover is $51 \cdot \frac{\pi}{\pi+77} \approx 1.9992$. So close, but not enough.