11.162. In a triangular pyramid, all four faces are equal isosceles triangles with base $a$ and lateral side $b$. Calculate the volume of the pyramid. Does the problem have a solution for any $a$ and $b$?
## Solution.

In the pyramid $SABC$ (Fig. 11.57), $AB = AC = BS = CS = b, BC = AS = a$: $E$ is the midpoint of $BC$. Then $AE \perp BC, SE \perp BC$. Therefore, the line $BC$ is perpendicular to the plane $ASE$. Then the plane $ABC$ is also perpendicular to the plane $ASE$. In the plane $ASE$, drop the perpendicular $SO$ to $AE$. Then the segment $SO$ is perpendicular to the plane $ABC$ and is the height of the pyramid.

From $\triangle AEB\left(\angle AEB=90^{\circ}\right): AE=\sqrt{AB^{2}-BE^{2}}=\sqrt{b^{2}-\frac{a^{2}}{4}}=\frac{1}{2} \sqrt{4b^{2}-a^{2}}$. $M$ is the midpoint of $AS$. Then $EM$ is the height of the triangle $\triangle AES$ and $EM=\sqrt{AE^{2}-AM^{2}}=\sqrt{\left(b^{2}-\frac{a^{2}}{4}\right)-\frac{a^{2}}{4}}=\frac{\sqrt{4b^{2}-2a^{2}}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0867.jpg?height=676&width=565&top_left_y=110&top_left_x=81)

Fig. 11.57

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0867.jpg?height=775&width=539&top_left_y=67&top_left_x=736)

Fig. 11.58

$S_{\triangle AES}=\frac{1}{2} AS \cdot EM=\frac{1}{2} AE \cdot SO$.

Then $SO=\frac{AS \cdot EM}{AE}=\frac{a \frac{1}{2} \sqrt{4b^{2}-2a^{2}}}{\frac{1}{2} \sqrt{4b^{2}-a^{2}}}=\frac{a \sqrt{4b^{2}-2a^{2}}}{\sqrt{4b^{2}-a^{2}}}$. The volume of the pyramid $V=\frac{1}{3} S_{\triangle ABC} \cdot SO=\frac{1}{3} \cdot \frac{1}{4} a \sqrt{4b^{2}-a^{2}} \frac{a \sqrt{4b^{2}-2a^{2}}}{\sqrt{4b^{2}-a^{2}}}=\frac{a^{2} \sqrt{4b^{2}-2a^{2}}}{12}$.

The problem has a solution when $\left\{\begin{array}{l}4b^{2}-2a^{2}>0 \\ 4b^{2}-a^{2}>0\end{array}\right.$, i.e., when $0<a<b \sqrt{2}$.

Answer: $\frac{a^{2} \sqrt{4b^{2}-2a^{2}}}{12} ; 0<a<b \sqrt{2}$.