8. (51st Czech and Slovak Mathematical Olympiad (Open Book) Question) Find all tetrahedra that have exactly four edges of length $a$, and when unfolded into a plane figure, form a kite that is not a rhombus.
8. Let each edge length be as shown in Figure (a), and discuss in two cases.
(1) If in the unfolded diagram there is one triangle adjacent to the other three triangles, as shown in Figure (b), since $\angle D_{1} 、 \angle D_{2} 、 \angle D_{3}$ cannot be straight angles, assume without loss of generality that $\angle D_{2} A D_{3}=\angle D_{3} B D_{1}=180^{\circ}$, hence it can only be $u=v$, and $2 u \neq w$, as shown in Figure (c), thus $y=x$, the resulting tetrahedron is as shown in Figure (d), and $z=\frac{1}{2} D_{1} D_{2}x$, then it must be $2 u=x+v, x=v$, so, $x=u=v$, as shown in Figure (f), and $w=2 z$. The corresponding tetrahedron is as shown in Figure (g).

For Figure (d), the six edges are $x 、 x 、 z 、 u 、 u 、 w$, then it must be $x=a$ (otherwise $z=u=u=a$, and $z2 z$, i.e., $x>z$. Since $\triangle A B D_{1}$ is an isosceles triangle, $\angle A B C$ is an obtuse angle, $y>\max \{x, z\}$, hence $y>x>z$. To make the conclusion hold, it can only be $2 z=a$. In $\triangle C D_{1} D_{2}$, the median on side $D_{1} D_{2}$ is $y=\frac{\sqrt{6}}{2} a$.
Therefore, there is a unique tetrahedron $A B C D$, where $A D=B D=C D=B C=a, A B=\frac{a}{2}, A C=\frac{\sqrt{6}}{2} a$.