Difference between revisions of "2019 USAJMO Problems/Problem 6"
Kevinmathz (talk | contribs) |
Stormersyle (talk | contribs) |
||
Line 2: | Line 2: | ||
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 |
Revision as of 19:50, 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