Difference between revisions of "2019 USAJMO Problems/Problem 6"
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− | Two rational numbers \ | + | Two rational numbers <math>\frac{m}{n}</math> and <math>\frac{n}{m} </math> are written on a blackboard, where <math>m</math> and <math>n</math> are relatively prime positive integers. At any point, Evan may pick two of the numbers <math>x</math> and <math>y</math> written on the board and write either their arithmetic mean <math>\frac{x+y}{2}</math> or their harmonic mean <math>\frac{2xy}{x+y}</math> on the board as well. Find all pairs <math>(m,n)</math> such that Evan can write <math>1</math> on the board in finitely many steps. |
Proposed by Yannick Yao | Proposed by Yannick Yao | ||
+ | |||
+ | ==Solution== | ||
+ | We claim that all odd <math>m, n</math> work if <math>m+n</math> is a positive power of 2. | ||
+ | |||
+ | Proof: | ||
+ | We first prove that <math>m+n=2^k</math> works. By weighted averages we have that <math>\frac{n(\frac{m}{n})+(2^k-n)\frac{n}{m}}{2^k}=\frac{m+n}{2^k}=1</math> can be written, so the solution set does indeed work. We will now prove these are the only solutions. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | -Stormersyle | ||
+ | |||
+ | {{MAA Notice}} | ||
+ | |||
+ | ==See also== | ||
+ | {{USAJMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}} |
Latest revision as of 23:57, 19 April 2019
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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 5 |
Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |