Let's say I have two variables $B_1$ and $B_2$ such as $B_1 = (b_{1,1}, b_{1,2}, ..., b_{1,n})$ where $\forall i, \space 0 \leq b_{1,i} \leq x$ and $B_2 = x - B_1 = (x - b_{1,1}, x - b_{1,2}, ..., x - b_{1,n})$.
I experimented that the two Pearson correlation coefficients $\rho_1 = corr(A, B_1)$ and $\rho_2 = corr(A, B_2)$ where $A$ is an arbitrary variable where symmetric (i.e. $\rho_1 = -\rho_2$).
I tried to prove this assumption with the definition of correlation but without any success...
Could someone help me how to prove this result from the definition which is: $corr(A,B) = \frac{cov(A, B)}{\sigma_A\sigma_B}$?
Examine the definitions of correlation, and covariance: $$\begin{align}\rho_i~=~&\mathsf {Corr}(A,B_i)\\[1ex] ~=~& \dfrac{\mathsf {Cov}(A,B_i)}{\sigma_A\sigma_{B_i}} \\[2ex] ~=~& \dfrac{\mathsf E(A~B_i)-\mathsf E(A)~\mathsf E(B_i)}{\sigma_A~\sqrt{~\mathsf E\Big(\big(B_i-\mathsf E(B_i)\big)^2\Big)~}} \end{align}$$
So on inspection we should find that if $B_2 = x-B_1$, then $\rho_2=-\rho_1$ for any $A$. $$\begin{align}\rho_2 ~=~& \dfrac{\mathsf E(A~B_2)-\mathsf E(A)~\mathsf E(B_2)}{\sigma_A~\sqrt{~\mathsf E\Big(\big(B_2-\mathsf E(B_2)\big)^2\Big)~}} \\[1ex] ~=~& \dfrac{\mathsf E(A~(x-B_1))-\mathsf E(A)~\mathsf E(x-B_1)}{\sigma_A~\sqrt{~\mathsf E\Big(\big((x-B_1)-\mathsf E(x-B_1)\big)^2\Big)~}} \\[1ex] \ddots &\end{align}$$