Show that two points from four are at a distance $\leq \sqrt{3}$ in an equilateral triangle.

Question

In a given equilateral triangle of sides length $3$, we locate 4 points. Prove that there are two of them are located at a distance less or equal to $\sqrt{3}$.

I arranged the four points in this configuration:

As the picture shows, I proved that if tree points are located in the triangle vertexes and the fourth point lies in the intersection of the medians, the distance $P_1P4=P_2P_4=P_3P_4=\sqrt{3}$.

I tried to argue with the hinge theorem and it sort of make sense, but this is not a proof. Can anyone help me how can I approach to this problems proof ? and How can I show that this case is the general one ? or isn't it ?

Answer 1

Your lines partition the triangle into three "kites". By pigeon-hole, one kite must contain at least two points. The kites are contained within disks of diameter $\sqrt 3$ (thanks to Thales), hence two points in the same kite cannot be further apart than $\sqrt 3$.

Answer 2

Hopefully this will get you started.

We can start by assuming that the first three points are more than $\sqrt{3}$ apart. If this were not the case, we would be done. Then we can place balls around these three points, with radius $\sqrt{3}$. Look at how these balls cover the triangle when you move the points around, but never letting the center of one of the balls be in the other two. (This would give you that two points are closer than $\sqrt{3}$ apart.) Then think about where that fourth point could be.

Answer 3

This is just the pigeonhole principle: divide the triangle into 3 triangles like you've done already. One of these triangles must contain 2 points, which are at most $\sqrt 3 $ away from each other.