A team of three high jumpers all have a personal record that is at least 6 feet and less than 7 feet. Is it necessarily true that two of the team members must have personal records that are within four inches of each other? What if there are four jumpers?
For this problem, I tried solving like let x, y, z are three jumpers, since their jump is at least 6 feet and but less then 7, hence their average is
(x+y+z)/3 < 7, with x,y,z taking the value greater than 6 and less then 7.
if x=6 feet, y+z < 15 now for possible values y and z can take this seems to be true.
I don't know I am following right path, as I have started working on discrete math after long time.
For 3 jumpers, the statement is not true. A counterexample would be $6$ feet, $6$ feet $5$ inches, and $6$ feet $10$ inches. They're all between $6$ and $7$ but not within $4$ inches of each other.
For 4 jumpers, you can split the interval into $[6, 6\frac{1}{3})$, $[6\frac{1}{3}, 6\frac{2}{3})$ and $[6\frac{2}{3}, 7)$. The length of each interval is $4$ inches. By pigeonhole principle, since there are $3$ intervals and $4$ people, there must be an interval with at least two people, and hence the personal records of these two people must be at least $4$ inches of each other.