Pigeonhole question and generalization

Question

Let H be a regular hexagon with side length 1 unit.

(a) Show that if more than 6 points are specied inside H then the points of at least one pair of them are at most 1 unit apart.

(b) State and prove a generalization of the result in (a) to the situation where there are in excess of $2^{2n+1}3$ points inside H.

I did part (a), I solved it using the Pigeonhole Principle. On the other side, I don't understand what part (b) is about. Any suggestions is more that appreciate it!!

Answer 1

Hint: An equilateral triangle with sides of unit length can be split into $2^{2n}$ triangles of side length $\frac{1}{2^n}$.

As you noticed before, a regular hexagon with side lengths 1 can be split into 6 equilateral triangles of unit length. Thus, a regular hexagon with side lengths 1 can be split into $2^{2n+1}3$ equilateral triangles of side length $\frac{1}{2^n}$.

Now, how can you use the pigeonhole principle to prove a more general result for your part (b)?

Answer 2

In case $n=0$ you have more than $6$ points in $H$. You solved that by dissecting $H$ into $6$ equilateral triangles of side $1$; by the pigeonhole principal, two of the points must be in the same triangle.

In case $n=1$ you have more than $24$ points in $H$. Now you want to dissect $H$ into $24$ congruent equilateral triangles. You do that by subdividing each of the $6$ triangles into four smaller triangles of side $\frac12$.