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Difference between revisions of "2019 USAJMO Problems/Problem 6"
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Assume that <math>m+n\ne 2^k</math>, so then <math>m+n\equiv 0\pmod{p}</math> for some odd prime <math>p</math>. Then <math>m\equiv -n\pmod{p}</math>, so <math>\frac{m}{n}\equiv \frac{n}{m}\equiv -1\pmod{p}</math>. We see that the arithmetic mean is <math>\frac{-1+(-1)}{2}\equiv -1\pmod{p}</math> and the harmonic mean is <math>\frac{2(-1)(-1)}{-1+(-1)}\equiv -1\pmod{p}</math>, so if 1 can be written then <math>1\equiv -1\pmod{p}</math> and <math>2\equiv 0\pmod{p}</math> which is obviously impossible, and we are done.
 
Assume that <math>m+n\ne 2^k</math>, so then <math>m+n\equiv 0\pmod{p}</math> for some odd prime <math>p</math>. Then <math>m\equiv -n\pmod{p}</math>, so <math>\frac{m}{n}\equiv \frac{n}{m}\equiv -1\pmod{p}</math>. We see that the arithmetic mean is <math>\frac{-1+(-1)}{2}\equiv -1\pmod{p}</math> and the harmonic mean is <math>\frac{2(-1)(-1)}{-1+(-1)}\equiv -1\pmod{p}</math>, so if 1 can be written then <math>1\equiv -1\pmod{p}</math> and <math>2\equiv 0\pmod{p}</math> which is obviously impossible, and we are done.
  
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-Stormersyle  
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-Stormersyle
  
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{{MAA Notice}}
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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==
 
==See also==
 
{{USAJMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}}
 
{{USAJMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}}
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{{MAA Notice}}

Latest revision as of 11:51, 8 November 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

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

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