11.6. Given a tetrahedron \(ABCD\), all of whose faces are similar right triangles with acute angles at vertices \(A\) and \(B\). The edge \(AB\) is equal to 1. Find the length of the shortest edge of the tetrahedron.
Answer: $\left(\frac{\sqrt{5}-1}{2}\right)^{\frac{3}{2}}$.

Solution. Let $\angle C B D=\alpha, \angle C D B=\beta, \alpha+\beta=90^{\circ}$ (see Fig. 11.6a). Right triangles $D B C$ and $A B C$ share the side $B C$. If $\angle A B C=\alpha$, then triangles $A B C$ and $D B C$ are congruent, which implies $C D=A C$, which is impossible because $A C$ is the hypotenuse of the right triangle $A C D$.

Therefore, $\angle A B C=\beta, \angle B A C=\alpha$. Since $B A C$ is the angle between the line $A B$ and the plane $A C D$, then $\alphaC D$, hence $C DA D$.

Therefore, $\angle B A D=\beta, \quad \angle A B D=\alpha, \quad$ then $A D=\cos \beta$; $C D=B D \cos \beta=A B \cos \alpha \cos \beta=\cos \alpha \cos \beta$. Thus, $C D<A D$.

From $\triangle A C D: C D<A D$ and $\alpha<\beta$. Therefore, $\angle C A D=\alpha$. From $\triangle A B C$ and $\triangle A B D$: $A D=A C \cos \alpha=A B \cos ^{2} \alpha$. From $\triangle A B D \quad A D=A B \cos \beta$. We obtain the equation: $\cos \beta=\cos ^{2} \alpha$.

Since $\alpha+\beta=90^{\circ}$, then $\cos \beta=\sin \alpha$. The obtained equation becomes: $\sin \alpha=1-\sin ^{2} \alpha$. Introducing the notation $\sin \alpha=t$, where $0<t<1$, we get the quadratic equation $t^{2}+t-1=0$, whose roots are $t_{1}=\frac{-1-\sqrt{5}}{2}<0$ and $t_{2}=\frac{-1+\sqrt{5}}{2}, 0<t_{2}<1$. Therefore, $\sin \alpha=\frac{\sqrt{5}-1}{2}$.

Since $\cos \beta=\sin \alpha$, then $C D=\cos \alpha \cos \beta=\cos \alpha \sin \alpha$. Moreover, $\sin \alpha=\cos ^{2} \alpha$, and the angle $\alpha$ is acute, so $\sqrt{\sin \alpha}=\cos \alpha$. Thus, $C D=\left(\frac{\sqrt{5}-1}{2}\right)^{\frac{3}{2}}$.

For a series of calculations (for example, to determine the arrangement of planar angles at vertex A), the formula of three cosines can be used. Let the line $B C$ be perpendicular to the plane $A C D, \angle B A C=\alpha$, $\angle C A D=\beta, \quad \angle B A D=\varphi \quad$ (see Fig. 11.6b). Then $\cos \varphi=\cos \alpha \cos \beta$.

The tetrahedron described in the problem exists. This follows, for example, from the fact that $\sin \alpha=\frac{\sqrt{5}-1}{2}<\frac{\sqrt{2}}{2}=\sin 45^{\circ}$, which means that indeed $\alpha<\beta$. This does not need to be justified by students.

Grading criteria.

«+» A complete and justified solution is provided

![](https://cdn.mathpix.com/cropped/2024_05_06_51c28cf2652dc1e7d646g-5.jpg?height=671&width=546&top_left_y=838&top_left_x=1360)
«士» A generally correct reasoning is provided, containing minor gaps or inaccuracies «» The arrangement of planar angles $\alpha$ and $\beta$ is justified, it is justified that $C D$ is the smallest edge, the quadratic equation is correctly derived, but a computational error is made further

«Ғ» Only calculations are provided and the correct answer is obtained, but no justification is given

«-» Only the answer is provided

«-» The problem is not solved or is solved incorrectly