10. Glue the bases of two given congruent regular tetrahedra together, precisely to form a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$
As shown in the figure, $E-ABC$ and $F-ABC$ are two congruent regular triangular pyramids. A line $BG \perp AF$ is drawn through $B$ at point $G$, and $CG$ is connected. Since $\triangle ABG \cong \triangle ACG \Rightarrow \angle BGC$ is the plane angle of the dihedral angle $B-AF-C$. $H$ is the midpoint of $AB \Rightarrow \angle EHF$ is the plane angle of the dihedral angle $E-AB-F$. Let $AB=BC=x, AF=BF=y$, $EF=z, AH=h_{1}, BG=h_{2} \Rightarrow\left\{\begin{array}{l}x h_{1}=y h_{2}, \\ \frac{h_{1}}{z}=\frac{h_{2}}{x}\end{array} \Rightarrow y=z=2\right.$ $\Rightarrow\left\{\begin{array}{l}h_{1}^{2}=1+\left(\frac{\sqrt{3} x}{6}\right)^{2}, \\ h_{1}^{2}+\left(\frac{x}{2}\right)^{2}=4\end{array} \Rightarrow 1+\frac{x^{2}}{12}+\frac{x^{2}}{4}=4 \Rightarrow \frac{x^{2}}{3}=3 \Rightarrow x=3\right.$. Therefore, the distance between the farthest two vertices is 3.