We're tasked to prove using the pigeonhole principle that, given an isosceles triangle with the two equal side lengths being $l = 2$, you can always choose $5$ random points of which two will have a distance $d < 1$ from each other.
Now I've attempted this problem, but I haven't come to a satisfying answer. I've tried reverse-engineering it. Obviously, we've got $5$ points, and we probably want to get $2$ for the amount of points which have a distance less than $1$. So I assume it will come down to having $3$ pigeon holes, as $\left\lceil \frac{5}{3} \right\rceil = 2$.
It seems like it might be helpful to divide the area of the triangle into three smaller shapes, in each of which one can be sure that two points will always have a distance less than $1$ from each other. Unfortunately, I lack an answer to this question.
What's the proof for this problem?
Counterexample:
Consider the triangle with vertexes $(0,0),(0,2),(2,0)$
then select the points:
No two points are within 1 unit of each other.
More generally, if our vertex angle is greater than 60 degrees.
We can select 3 points, one near each vertexes, we can find 2 points in the grey region, such that no point is within 1 of each other.