Difference between revisions of "2018 USAJMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
| + | Note that in this solution, the term "inversions" is used synonymously with "swapped pairs." | ||
| − | We define a new process <math>P'</math> where, when re-inserting card <math>i</math>, we additionally change its label from <math>i</math> to <math>n+i</math>. For example, an example of <math>P'</math> also starting with <math>3142</math> is: <cmath> 3142 \longrightarrow 3452 \longrightarrow 6345 \longrightarrow 6475 \longrightarrow 6785. </cmath>Note that now, each step of <math>P'</math> preserves the number of inversions. Moreover, the final configuration of <math>P'</math> is the same as the final configuration of <math>P</math> with all cards incremented by <math>n</math>, and of course | + | We define a new process <math>P'</math> where, when re-inserting card <math>i</math>, we additionally change its label from <math>i</math> to <math>n+i</math>. For example, an example of <math>P'</math> also starting with <math>3142</math> is: |
| + | <cmath>3142 \longrightarrow 3452 \longrightarrow 6345 \longrightarrow 6475 \longrightarrow 6785.</cmath> | ||
| + | Note that now, each step of <math>P'</math> preserves the number of inversions. Moreover, the final configuration of <math>P'</math> is the same as the final configuration of <math>P</math> with all cards incremented by <math>n</math>, and thus, of course, has the same number of inversions. | ||
| − | ~ v_enhance | + | ~ v_enhance (clarified by integralarefun) |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:30, 7 April 2023
Problem 6
Karl starts with 
cards labeled 
lined up in a random order on his desk. He calls a pair 
of these cards swapped if 
and the card labeled 
is to the left of the card labeled 
. For instance, in the sequence of cards 
, there are three swapped pairs of cards, 
, 
, and 
.
He picks up the card labeled 1 and inserts it back into the sequence in the opposite position: if the card labeled 1 had 
card to its left, then it now has cards to its right. He then picks up the card labeled 
and reinserts it in the same manner, and so on until he has picked up and put back each of the cards 
exactly once in that order. (For example, the process starting at would be 
.)
Show that no matter what lineup of cards Karl started with, his final lineup has the same number of swapped pairs as the starting lineup.
Solution
Note that in this solution, the term "inversions" is used synonymously with "swapped pairs."
We define a new process 
where, when re-inserting card , we additionally change its label from to 
. For example, an example of also starting with 
is:
![\[3142 \longrightarrow 3452 \longrightarrow 6345 \longrightarrow 6475 \longrightarrow 6785.\]](http://latex.artofproblemsolving.com/0/f/8/0f8c4e029950c6775fa0cfbb768155bb956ef0ce.png)
Note that now, each step of preserves the number of inversions. Moreover, the final configuration of is the same as the final configuration of 
with all cards incremented by , and thus, of course, has the same number of inversions.
~ v_enhance (clarified by integralarefun)
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
See also
| 2018 USAJMO (Problems • Resources) | ||
| Preceded by Problem 5 |
Last Problem | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAJMO Problems and Solutions | ||
