# Pigeonhole Principle: Proof that in an isosceles triangle with both side lengths 2 amongst 5 random points within there's two with distance $d < 1$

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?

• Hint: draw an isosceles triangle of the above type. Then draw circular arcs of radius one centered at each vertex. What do you see? (You might have to draw two or three different triangles - fat ones and skinny ones). – stochasticboy321 Dec 6 '18 at 23:56
• Well, considering the angle at the vertex where the two equal sides touch, if that angle is $90^{\circ}$, the circles barely touch. If it's greater than that, the circles don't touch at all, and if it's less than that, the circles do intersect. Now obviously when you put one point at a vertex, you can position another one along the part of the arc that's inside the big triangle. Maybe you can even try the same with another arc and hope the points aren't too close to one another. Still, I don't know how this may help prove that you can't place fifth point without it having a distance $d < 1$ – StckXchnge-nub12 Dec 7 '18 at 0:10

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.

• I suppose you can assume each and every of these triangles to be similar, i.e. all of them to be isosceles triangles with the equal sides having the length $1$? – StckXchnge-nub12 Dec 7 '18 at 0:17
• If you connect the mid-lines of any triangle, it will create 4 congruent triangles. And, all similar to the orignal triangle. – Doug M Dec 7 '18 at 0:20
• Well, alright. Assuming a right-angle at the vertex of the equal sides, that'd mean each of the remaining sides of the smaller triangles has a length of $\sqrt{1^2 + 1^2} = \sqrt{2} > 1$. What prevents us from conveniently placing two points within the smaller triangle so that the distance is greater than one, since obviously we could place both of them close to the vertexes that have a distance of $\sqrt{2}$? I'm sure the position of the points within the other small triangles would somehow prevent this or cause by themselves a distance smaller than 1 between those, but how to show it? – StckXchnge-nub12 Dec 7 '18 at 0:28
• You are right, the proposition is not true. – Doug M Dec 7 '18 at 0:55