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Difference between revisions of "2019 USAJMO Problems/Problem 6"

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==Video Solution==
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https://www.youtube.com/watch?v=hIX89vyuGD8&list=PLa8j0YHOYQQK1xH_fn0WSXR-AW9uWdDEf&index=1&t=1s - AMBRIGGS
  
 
==See also==
 
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Revision as of 17:25, 30 July 2022

Two rational numbers $\frac{m}{n}$ and $\frac{n}{m}$ are written on a blackboard, where $m$ and $n$ are relatively prime positive integers. At any point, Evan may pick two of the numbers $x$ and $y$ written on the board and write either their arithmetic mean $\frac{x+y}{2}$ or their harmonic mean $\frac{2xy}{x+y}$ on the board as well. Find all pairs $(m,n)$ such that Evan can write $1$ on the board in finitely many steps.

Proposed by Yannick Yao

Solution

We claim that all odd $m, n$ work if $m+n$ is a positive power of 2.

Proof: We first prove that $m+n=2^k$ works. By weighted averages we have that $\frac{n(\frac{m}{n})+(2^k-n)\frac{n}{m}}{2^k}=\frac{m+n}{2^k}=1$ can be written, so the solution set does indeed work. We will now prove these are the only solutions.

Assume that $m+n\ne 2^k$, so then $m+n\equiv 0\pmod{p}$ for some odd prime $p$. Then $m\equiv -n\pmod{p}$, so $\frac{m}{n}\equiv \frac{n}{m}\equiv -1\pmod{p}$. We see that the arithmetic mean is $\frac{-1+(-1)}{2}\equiv -1\pmod{p}$ and the harmonic mean is $\frac{2(-1)(-1)}{-1+(-1)}\equiv -1\pmod{p}$, so if 1 can be written then $1\equiv -1\pmod{p}$ and $2\equiv 0\pmod{p}$ which is obviously impossible, and we are done.

-Stormersyle

This proof is wrong because $\frac{m}{n}$ and $\frac{n}{m}$ are not integers, so you cannot say that they are $-1\pmod{p}$ and work $\pmod{p}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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
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All USAJMO Problems and Solutions
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