0

An NFL playing field is 360 ft by 160 ft. Prove that during an NFL game:

(a) Two of the 22 players are within 75 ft of each other.

(b) Four of the 22 players are within 50 yards of each other.

My question is can I divide the field into squares of 50 and 75 ft so I can use the pigeonhole principle as a proof? Is the proof that simple or is it more complex?

Daniel Awbrey
  • 1
  • Using squares, be sure to look at the diagonal, not the sidelength. For the first question, you can't use squares of side length $75$ because the diagonal is $75\times \sqrt 2$. – lulu Apr 10 '17 at 23:08
  • Also: it's not immediately clear to me how you intend to arrange those $22$ squares. You have to cover the field... – lulu Apr 10 '17 at 23:11
  • I was actually going to divide the each side by 75ft and 50yd respectively using a ceiling function so that each square is equal to or in most cases less than my desired distance. However, you do raise a good point about making sure the length of the hypotenuse in each of the boxes is less than or equal to 50yd or 75ft. – Daniel Awbrey Apr 10 '17 at 23:16
  • As I say, it's no use taking squares of side length $75$ because the diagonal is too long. (you could have two points in the same square which are further than $75$ apart). Now, to make the diagonal $75$ you need the side length around $53$ and $\frac {160}{53} \approx 3$ and $\frac {360}{53}\approx 7$ so it's very, very close. Not clear to me that it works. – lulu Apr 10 '17 at 23:20
  • You might do better with rectangles of diagonal $75$...or, if necessary, regular hexagons. – lulu Apr 10 '17 at 23:22
  • Just wrote it out...rectangles work. Squares of side length $\frac {75}{\sqrt 2}\approx 53.033$ don't quite work. But you can stretch them a bit (keeping the diagonal constant) and get a solution. – lulu Apr 10 '17 at 23:27
  • Yeah, I ended up going with a diagonal of 75 ft and 50 yds then I derived the other sides using trig and took the ceiling function to get a whole number of squares. Thank you for your help, I did consider that two players standing on opposite sides of the hypotenuse would be great than 75 ft given my initial proposal. – Daniel Awbrey Apr 11 '17 at 01:49
  • As I say, I can't see how to do it with squares. Rectangles work: if you take rectangles with width $\frac {160}3$ and length $\frac {360}7$ you can exactly cover the field with $3\times 7 = 21$ of them and the diagonal is $<75$. Good thing...the hexagon tiling is more efficient but a lot harder to calculate with. – lulu Apr 11 '17 at 11:07

0 Answers0