Consider the graph representation: We are given a graph $G=(V,E)$ with $V$ being the set of vertices (i.e. mathematicians) $\{v_1, v_2, \dots, v_{12}\}$ and $E$ being the set of edges (i.e. mathematicians that are friends). The problem's condition give us that $\deg(v_i), \forall 1\leq i \leq 12$, is odd. Moreover, color the vertices red if they are in $\sqrt{2}$-clan and blue if they are in $\pi$-clan. Now let $x_i$ be the number of vertices that are connected with $v_i$ but have a different color. Define $S=\sum_{k=1}^{12}x_i$. We see the following: whenever $v_i$ switches colors, then $x_i$ decreases by at least one (since he switches to the color that more friends have). The friends $v_{i_1}, v_{i_2}, \dots, v_{i_k}$ that have the color that $v_i$ switched to, now have $x_{i_1}, x_{i_2}, \dots, x_{i_k}$ decreased by one. The friends $v_{i_1}, v_{i_2}, \dots, v_{i_j}$ that have the color that $v_i$ had before the switch, now have $x_{i_1}, x_{i_2}, \dots, x_{i_j}$ increased by one. However, since $j < k$, the increase is smaller than the decrease, $S$ must decrease for every color switch that happens (when the color does not switch, $S$ obviously remains unchanged). As long as we can change a color (if we could not, than our process already terminated), $S$ decreases monotonically and hence reaches $0$ at some point, which means, that all vertices have the same color. This concludes the proof. $\square$