CEMC Pascal, 2025

Label the remaining faces using the variables $v$, $w$, $y$, and $z$, as shown in the diagram below.

The faces that share an edge with the face labelled by $z$ have values of $-20$, $-56$, and $0$.

By the condition on the integers labelling the faces, we get $z=-20-56+0=-76$.

Now consider the face labelled by $-56$. The faces with which it shares an
edge have labels $y$, $z$, and $v$, and so $-56=y+z+v$.

Since $z=-76$, we get the equation
$-56=y-76+v$ or $v+y=20$.

Next, consider the face labelled with $0$. The faces with which it shares an edge
have labels $x$, $y$, and $z$. Therefore, $0=x+y+z$, and since $z=-76$, we get $x+y=76$.

Finally, consider the face labelled with $-20$. The faces with which it shares an
edge have labels $v$, $x$, and $z$. Therefore, $-20=v+x+z$, and since $z=-76$, we get $v+x=56$.

We have now derived the following three equations: $$\begin{align*} v +y &= 20 \\ x+y &= 76 \\ v+x &= 56\end{align*}$$ Adding these three equations gives
$(v+y)+(x+y)+(v+x)=20+76+56$ or $2(v+x+y)=152$.

Dividing by two, we get $v+x+y = 76$.

Using that $v+y=20$, we can subtract
from $v+x+y=76$ to get $(v+x+y)-(v+y) = 76-20$ or $x=56$.

We will stop here since the question only asked for the value of $x$, but it is possible show that $v=0$, $w=76$, $y=20$, and $z=-76$.