7. What is the minimum volume of a pyramid whose base is an equilateral triangle with a side length of 6, and all planar angles at the vertex are equal to each other and do not exceed $2 \arcsin \frac{1}{3}$?
Answer: $5 \sqrt{23}$. Solution. Let's introduce the following notations: $a$ - the length of the base edge, $S A=x$, $S B=y$ and $S C=z$ - the lengths of the lateral edges, $\alpha$ - the magnitude of the dihedral angle at vertex $S$, $S O$ - the height of the pyramid, $h=A A_{1}$ - the height of the base.

From the cosine theorem for the lateral faces, we have $\left\{\begin{array}{l}a^{2}=x^{2}+y^{2}-2 x y \cos \alpha, \\ a^{2}=y^{2}+z^{2}-2 z y \cos \alpha, \\ a^{2}=x^{2}+z^{2}-2 x z \cos \alpha .\end{array}\right.$

Subtracting the second equation of the system from the first, we get $(x-z)(x+z)=2(x-z) y \cos \alpha$. Consequently, there are two cases: 1) $x=z$; 2) $x+z=2 y \cos \alpha$ (from which it follows that $\cos \alpha>\frac{1}{2} \Leftrightarrow 0<\alpha<\frac{\pi}{3}$). If all values $x, y$, and $z$ are different, we necessarily obtain $\left\{\begin{array}{l}x+y=2 z \cos \alpha, \\ y+z=2 x \cos \alpha, \\ x+z=2 y \cos \alpha,\end{array}\right.$ from which it follows that $x-z=0$, i.e., at least two of the values $x, y$, and $z$ are equal.

Consider two cases: 1) $x=y=z$; 2) $y=z \neq x$.

In the first case, the pyramid $S A B C$ is regular, and its volume is $V_{1}=\frac{a^{3}}{12} \cdot \sqrt{\frac{1}{4 \operatorname{tg}^{2} \frac{\alpha}{2}}-\frac{1}{12}}$ $\left(S_{\text {base }}=\frac{a^{2} \sqrt{3}}{4}, S A_{1}=\frac{a}{2 \operatorname{tg} \frac{\alpha}{2}}, A_{1} O=\frac{h}{3}=\frac{a \sqrt{3}}{6}\right.$, hence $\left.S O=a \sqrt{\frac{1}{4 \operatorname{tg}^{2} \frac{\alpha}{2}}-\frac{1}{12}}\right)$. We get that the volume is a decreasing function of $\alpha$ and for a regular pyramid, it will be minimal when $\alpha=2 \arcsin \frac{1}{3}$. For $a=6$, we get $h=3 \sqrt{3}, S O=\sqrt{69}, S_{\text {base }}=9 \sqrt{3}$, and $V_{S A B C}=9 \sqrt{23}$.

In the second case, if $y=z \neq x$, we get the pyramid $S B C A^{\prime}$, where points $A$ and $A^{\prime}$ lie on the same circle with center at point $A_{1}$ and radius equal to the length of segment $A A_{1}$, i.e., $A_{1} A^{\prime}=A_{1} A=h$. Triangle $A^{\prime} B C$ is equilateral and $V_{S A^{\prime} B C}<V_{S A B C}$. To find the volume of the pyramid $A^{\prime} A B C$, it is sufficient to find its height $A^{\prime} O_{1}$. Construct the section of the pyramid by the plane $\left(S A_{1} A\right) . A^{\prime} O_{1}=h \sin \left(\pi-2 \angle A_{1} A A^{\prime}\right)=h \sin 2 \angle A_{1} A A^{\prime}=\frac{2 h \cdot \operatorname{tg} \angle A_{1} A A^{\prime}}{1+\operatorname{tg}^{2} \angle A_{1} A A^{\prime}}=\frac{12 h^{2} \cdot S O}{4 h^{2}+9 S O^{2}}=\frac{4 \sqrt{69}}{9}$, since $\operatorname{tg} \angle A_{1} A A^{\prime}=\frac{3 S O}{2 h}$.

Finally, we get $V_{A^{\prime} A B C}=4 \sqrt{23}$ and $V_{A^{\prime} B C}=V_{\text {SABC }}-V_{A^{\prime} A B C}=5 \sqrt{23}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_0f7f3ab80da5a0ce1d2dg-4.jpg?height=612&width=1148&top_left_y=2144&top_left_x=564)

Answer for variant 152: $\frac{\sqrt{11}}{2}$.

Answer for variant 153: $21 \sqrt{39}$.

Answer for variant 154: $32 \sqrt{11}$.