Difference between revisions of "2019 USAJMO Problems/Problem 6"
m (→Solution) |
(→Solution) |
||
Line 16: | Line 16: | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=hIX89vyuGD8&list=PLa8j0YHOYQQK1xH_fn0WSXR-AW9uWdDEf&index=1&t=1s - AMBRIGGS | ||
==See also== | ==See also== | ||
{{USAJMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}} | {{USAJMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}} |
Revision as of 18:25, 30 July 2022
Two rational numbers and are written on a blackboard, where and are relatively prime positive integers. At any point, Evan may pick two of the numbers and written on the board and write either their arithmetic mean or their harmonic mean on the board as well. Find all pairs such that Evan can write on the board in finitely many steps.
Proposed by Yannick Yao
Solution
We claim that all odd work if is a positive power of 2.
Proof: We first prove that works. By weighted averages we have that can be written, so the solution set does indeed work. We will now prove these are the only solutions.
Assume that , so then for some odd prime . Then , so . We see that the arithmetic mean is and the harmonic mean is , so if 1 can be written then and which is obviously impossible, and we are done.
-Stormersyle
This proof is wrong because and are not integers, so you cannot say that they are and work .
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Video Solution
https://www.youtube.com/watch?v=hIX89vyuGD8&list=PLa8j0YHOYQQK1xH_fn0WSXR-AW9uWdDEf&index=1&t=1s - AMBRIGGS
See also
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 5 |
Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |