1. Line segment $A B$ divides a square into two polygons (points $A$ and $B$ are on the sides of the square), each polygon has an inscribed circle, one of which has a radius of 6, while the other has a radius greater than 6. What is the difference between the side length of the square and twice the length of line segment $A B$?
If line segment $A B$ divides the square into two triangles, then $A B$ can only be the diagonal of the square. But in this case, the radii of the two incircles are equal, which contradicts the condition.

If one of the polygons is a quadrilateral, then the sum of the lengths of $A B$ and its opposite side is greater than the sum of the lengths of the other two sides, which contradicts the fact that the quadrilateral has an incircle.

Therefore, it can only be that $A B$ divides the square into a triangle and a pentagon, and the incircle of the pentagon is also the incircle of the original square (as shown in Figure 9).

As shown in Figure 10, let the incircle of $\triangle A B C$ touch $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$, respectively, and the incircle of the original square, which is also the excircle of $\triangle A B C$, touch $C A$, $B C$, and $A B$ at points $P$, $Q$, and $R$, respectively. Then the side length of the original square is $C P + C Q$.
$$
\begin{aligned}
\text { Hence } & C P + C Q - 2 A B \\
= & (A P + A E + 6) + (B Q + B D + 6) - \\
& (A F + B F + A R + B R) \\
= & 12 + (A P - A R) + (B Q - B R) + \\
& (A E - A F) + (B D - B F) \\
= & 12 .
\end{aligned}
$$