AustriaMO2011

Choosing an arithmetic set $\{u,v,w\}$, we can assume that $u<v<w$ holds, and we can therefore write $u=a-d$, $v=a$ and $w=a+d$ with $d>0$. We now wish this set to also be harmonic. If some number $q$ is the harmonic mean of numbers $p$ and $r$, we have $\frac{1}{p}+\frac{1}{r}=\frac{2}{q}$, which is equivalent to $qr-2rp+pq=0$. If $v$ is the harmonic mean of $u$ and $w$, this equation yields $a(a+d)-2(a+d)a+(a-d)a=2d^2=0$, which means that all three elements of the set are equal, which is a contradiction. There can therefore be no such subsets. If $w$ is the harmonic mean of $u$ and $v$, we obtain $$(a+d)a-2a(a-d)+(a-d)(a+d)=3ad-d^2=d(3a-d)=0,$$ which yields $d=3a$, since $d=0$ is not possible. We see that any set of the form $\{-2a,a,4a\}$ has the required properties. Finally, if $u$ is the harmonic mean of $v$ and $w$, we obtain $(a-d)(a+d)-2(a+d)a+a(a-d)=-3ad-d^2=-d(3a+d)=0$, which yields $d=-3a$ and therefore the same sets as the previous case. Since $a$ can assume any integer value not equal to $0$ such that $-2011<4a<2011$, we have $-502\le a\le 502$, and we see that there are $1004$ subsets with the required properties.