If corr(A,B) = x and corr(B,C) = y, what is corr(A,C)? If I know the correlation between two pairs of random variables (A,B) and (B,C), can I determine the correlation of the pair (A,C)? If not, can I at least constrain it to some range?
I'm interesting in generating a covariance matrix where certain pairwise correlation values are determined, and the rest are as small (i.e. close to 0) as possible.
 A: Unfortunately no, we cannot determine $\rho_{AC}$ given just $\rho_{AB}$ and $\rho_{BC}$. You can derive the theoretical bounds
\begin{align*}
\rho_{AC} \ge \max\{2(\rho_{AB} + \rho_{BC}) - 3, 2\rho_{AB}\rho_{BC} - 1\} 
\end{align*}
Proof. Some notation. I let $\sigma_{AB} = \text{Cov}(A,B)$ and $\sigma_A^2 = \text{Var}(A)$. 
Let's first prove $\rho_{AC} \ge 2(\rho_{AB} + \rho_{BC}) - 3$. Recall the identity
\begin{align*}
2 E[X^2] + 2E[Y^2] = E[(X+Y)^2] + E[(X-Y)^2]
\end{align*}
hence $2E[Y^2] \le E[(X+Y)^2] + E[(X-Y)^2]$. Set 
\begin{align*}
X = \widetilde{B} - (\widetilde{A} + \widetilde{C})/2 \quad \text{and} \quad Y =
 (\widetilde{A} - \widetilde{C})/2
\end{align*}
where $\widetilde{C} = (C - E[C])/\sigma_C$, the normalized random variable, and similarly for $\widetilde{A}, \widetilde{B}$. Upon substitution and simplification, we get
\begin{align*}
\frac{1}{2}(2 - 2\rho_{AC}) \le (2 - 2\rho_{AB}) + (2 - 2\rho_{BC}) \iff \rho_{AC} \ge 2(\rho_{AB} + \rho_{BC}) - 3
\end{align*}
To prove $\rho_{AC} \ge 2\rho_{AB}\rho_{BC} - 1$, consider the random variable
\begin{align*}
W = 2 \frac{\sigma_{AB}}{\sigma_B^2}B - A
\end{align*}
We can verify $\sigma_W^2 = \sigma_A^2$, and hence $\sigma_{WC} \le \sigma_{W}\sigma_{C} = \sigma_ A \sigma_C$ by the Cauchy-Schwarz inequality. On the other hand, you may compute
\begin{align*}
\sigma_{WC} = 2 \frac{\sigma_{AB}}{\sigma_B^2}\sigma_{BC} - \sigma_{AC}
\end{align*}
Reorganizing all this, we prove $\rho_{AC} \ge 2\rho_{AB}\rho_{BC} - 1$.
