If 9 points are chosen from within the rectangle, explain why two of the points must be at most 18cm away from each other.

Question

I've been revising for an exam and the following question has stumped me. I thought it might involve pythagorean theorem but after trying it out it doesn't seem to...

"A rectangle has width 6 cm and height 12 cm. If 9 points are chosen from within the rectangle, explain why two of the points must be at most 18cm away from each other. [Hint: divide the rectangle into squares of equal area.]"

Any help would be appreciated.

Answer 1

Well I can find much more stronger bound $\sqrt{18}$. First one half $6\times 6$ must have at least 5 points. Then we divide this part on 4 equal parts $3\times 3$. Now in one of those there are 2 points. So they are apart at most $\sqrt{18}<18$.

Answer 2

Divide the rectangle into squares $3cm\times 3 cm$, there are $2 \times 4 = 8$ of them. Two of the points will land in the same square, so the distance between them is at most the diagonal of the square $= 3\sqrt{2} cm = 4.24\ldots cm$.

Obs: Maybe the problem stated $\sqrt{18} cm$ ?