Is this equivalent to RH - Riemann hypothesis? $$\pi = 3\prod_{\zeta(1/2+it) = 0}\frac{9+4t^2}{1+4t^2}\iff\text{RH is true}.$$
 A: Yes, this is equivalent to RH (but not in any significant way).  Recall the completed Riemann $\xi$-function 
$$ 
\xi(s) = s(s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s), 
$$ 
which, by Hadamard's factorization formula can be written as
$$ 
e^{A+Bs} \prod_{\rho}\Big(1-\frac{s}{\rho}\Big) e^{s/\rho},  
$$ 
where the product is over all non-trivial zeros $\rho$ of $\zeta(s)$.
Now one can check that $A=0$ (plug in $s=0$) and that $B= -\sum_{\rho}\text{Re }(1/\rho)$.  It follows that 
$$ 
|\xi(s)| = \prod_{\rho: \text{Im}(\rho) >0} \Big| \frac{s-\rho}{\rho}\Big|^2,
$$ 
by grouping complex conjugate zeros (and the product now converges). Now evaluate this at $s=2$: thus 
$$ 
\xi(2) =2 \times 1 \times \pi^{-1} \times \Gamma(1) \times \zeta(2) = \frac{\pi}{3} 
$$ 
equals 
$$ 
\prod_{\rho: \text{Im}(\rho) >0} \Big| \frac{2-\rho}{\rho}\Big|^2. 
$$ 
Split the product over zeros into two factors: the first one from zeros on the critical line, and the second one over zeros not on the critical line (if any).  The first factor is simply 
$$
\prod_{\substack{\rho = 1/2 +i\gamma \\ \gamma >0}} \frac{(3/2)^2+\gamma^2}{(1/2)^2 +\gamma^2} = \prod_{\substack{\rho = 1/2 +i\gamma \\ \gamma >0}} \frac{9+4\gamma^2}{1+4\gamma^2}.
$$ 
If RH is true then the second factor is $1$.  If RH is false, then note that the contribution of the zeros $\beta+i\gamma$ and $1-\beta+i\gamma$ together to the second product is 
$$ 
\frac{(2-\beta)^2+\gamma^2}{\beta^2+\gamma^2} \frac{(1+\beta)^2+\gamma^2}{(1-\beta)^2 +\gamma^2} > 1;
$$ 
(both factors are $\ge 1$ since $0\le \beta \le 1$, and at least one of them must be strictly larger than $1$). 
There's a little bit more fun to be had with this problem.  It can be used to show easily that if $\gamma_0$ is the first ordinate of a zero (not necessarily on the critical line) of $\zeta(s)$ then 
$$
\frac{\pi}{3} \ge \frac{9+4\gamma_0^2}{1+4\gamma_0^2}, 
$$ 
and since $\pi$ is so close to $3$, one can extract from this the fairly good bound that $\gamma_0 \ge 6.49\ldots$.  This general idea is of course well known, but I thought this particular choice was pretty!  
