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Athletes arrange themselves into $x$ rows and $y$ columns.
We pick the tallest of every column and name the shortest of them A.
We pick the shortest of each row and name the tallest of them B.

Prove $\operatorname{height}(A)\ge \operatorname{height}(B)$.

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I think the answer is

$A \geq B$, with equality being guaranteed if $x=1$ or $y=1$

Reasoning

It's clear that if $x=1$ then both $A$ and $B$ will be the shortest in the row and if $y=1$ then both $A$ and $B$ will be the tallest in the column.

Now suppose $x,y >1$ and pick $C$ so that $C$ is in the same column as $A$ and the same row as $B$. Then, $A \geq C$ since $A$ is the tallest in the column and $C \geq B$ since $B$ is the shortest in the row. Hence $A \geq C \geq B$.

To illustrate why equality doesn't hold in general, consider the following $2 \times 2$ case:

23 35
31 24

Clearly $A=31$ and $B=24$ so that $A > B$ and this is easily generalised to larger rectangles.

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  • $\begingroup$ what's C in your 2x2 example? $\endgroup$ – JMP Sep 26 '18 at 11:56
  • $\begingroup$ @JonMarkPerry In this example $C$ is $31$ (i.e, $C=A$ here). $\endgroup$ – hexomino Sep 26 '18 at 11:59
  • $\begingroup$ Yeah I see. I was getting confused with row containing A, column containing B $\endgroup$ – JMP Sep 26 '18 at 19:28
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The answer

Cannot be determined

Consider

1  2  3  8 
8  9 10 11 
5  6  7  9 
6  8  9 10
A: 8, B: 8 => A = B

Consider

1  2  3  9 
8  9 10 12 
5  6  7 11 
6  8  9 10
A: 9, B: 8 => A > B

Consider

1  2  3  7 
8  9 10 11 
5  6  7  9 
6  8  9 10
A: 7, B: 8 => A < B

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  • $\begingroup$ I just realised I might have read the question wrong... $\endgroup$ – nikki Sep 26 '18 at 5:42
  • $\begingroup$ .A=min(max(c1),max(c2),max(c3),max(c4))=min(8,9,10,11)=8. B=max(min(r1),min(r2),min(r3),min(r4))=max(1,8,5,6)=8=A $\endgroup$ – JMP Sep 26 '18 at 19:26
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I think that:

A and B are the same person EDIT This doesn't hold up lol

Because

1  2  3  4 
2  3  4  5 
3  4  5  6 
4  5  6  7
The tallest of each column would be the bottom row. The shortest of each row would be the left column. The tallest of the left column and the shortest of the bottom row would both be the bottom left 4.

Another Example, but less regular:

2  5  3  7 
6  3  4  5 
3  9  5  2 
8  1  6  3
The tallest of each column are 8, 9, 6, 7, the shortest of which is 6. The shortest of each row is 2, 3, 2, 1, the tallest of which is 3 Aaaaaand this is where my answer doesn't work. If you've made it this far, nevermind lol.

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  • $\begingroup$ you should put random numbers to see more cases $\endgroup$ – stelioball Sep 26 '18 at 10:04
  • $\begingroup$ @stelioball i did and my theory fell through :'( $\endgroup$ – AHKieran Sep 26 '18 at 10:06
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One answer could be

They are both the tallest of one group and the shortest of another group. Both are sure to be neither the tallest nor the shortest of all the athletes. (I am aware that this answer may have too little to do with :S)

Another (even) less serious answer is

They are both named as a letter of the alphabet.

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