11. Find all tetrahedra that have exactly four edges of length \( a \), and when unfolded into a plane, form a kite that is not a rhombus.
(51st Czech and Slovak Mathematical Olympiad (Open Book))
Solution: Let each edge length be as shown in Figure 9, and we discuss in two cases.
(1) If in the unfolded diagram there is one triangle adjacent to the other three triangles, as shown in Figure 10. Since $\angle D_{1} 、 \angle D_{2} 、 \angle D_{3}$ cannot be straight angles, assume without loss of generality that $\angle D_{2} A D_{3}=\angle D_{3} B D_{1}=180^{\circ}$. Therefore, it can only be that $u=v$, and $2 u \neq w$, as shown in Figure 11. Thus, $y=x$, and the resulting tetrahedron is as shown in Figure 12, and $z=\frac{1}{2} D_{1} D_{2} x$, so there must be $2 u=x+v, x=v$, hence, $x=u=v$, as shown in Figure 14. And $w=2 z$. The corresponding tetrahedron is as shown in Figure 15.

For Figure 12, the six edges are $x 、 x 、 z 、 u 、 u 、 w$, so there must be $x=a$ (otherwise $z=u=w=a$, which contradicts $z<2 z$, i.e., $x>z$. Since $\triangle A B D_{1}$ is an isosceles triangle, $\angle A B C$ is an obtuse angle, so $y>\max \{x, z\}$. Therefore, $y>x>z$. For the conclusion to hold, it can only be that $2 z=a$. 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 $A D=$ $B D=C D=B C=a, A B=\frac{a}{2}, A C=\frac{\sqrt{6}}{2} a$.