Find the volume of a regular triangular pyramid with a lateral edge $b$ and the radius $R$ of the circumscribed sphere.

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Let $O$ be the center of the sphere of radius $R$, circumscribed around a regular triangular pyramid $ABCD$ with lateral edges $DA=DB=DC=b$. The point $O$ lies on the line $DM$, where $M$ is the center of the base $ABC$, and since points $A$ and $D$ lie on the sphere, $O$ also lies on the perpendicular bisector of the side $AD$ of the triangle $ADM$. Denote $AB=BC=AC=a, \angle ADM=\phi$. If $K$ is the midpoint of $AD$, then

$$
\begin{aligned}
& \cos \phi=\frac{DK}{DO}=\frac{h}{R}=\frac{b}{2R}, DM=AD \cos \phi=b \cdot \frac{b}{2R}=\frac{b^2}{2R}, \\
& \frac{a}{\sqrt{3}}=AM=AD \sin \phi=b \sqrt{1-\cos^2 \varphi}=b \sqrt{1-\frac{b^2}{4R^2}}=b \frac{\sqrt{4R^2-b^2}}{2R}, \\
& a=\frac{b \sqrt{3} \sqrt{4R^2-b^2}}{2R}
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
& V_{ABCD}=\frac{1}{3} S_{\triangle ABC} \cdot DM=\frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot DM= \\
& =\frac{1}{3}\left(\frac{b \sqrt{3} \cdot \sqrt{4R^2-b^2}}{2R}\right)^2 \cdot \frac{\sqrt{3}}{4} \cdot \frac{b^2}{2R} \cdot \frac{b^4 \sqrt{3(4R^2-b^2)}}{32R^3}
\end{aligned}
$$

Find the volume of a regular triangular pyramid with lateral edge $b$ and the area $Q$ of a lateral face.

## Solution

Let $M$ be the center of the base $ABC$ of the regular triangular pyramid $ABCD$ with lateral edges $DA=DB=DC=b$, and let $K$ be the midpoint of $AB$. Denote $AB=BC=AC=a$. Then

$$
\begin{gathered}
AM=\frac{a}{\sqrt{3}}, DK=\sqrt{AD^2-AK^2}=\sqrt{b^2-\frac{a^2}{4}} \\
Q=S_{\triangle ABD}=\frac{1}{2} AB \cdot DK=\frac{1}{2} a \sqrt{b^2-\frac{a^2}{4}}=\frac{1}{4} a \sqrt{4b^2-a^2} \\
4Q=a \sqrt{4b^2-a^2}, 16Q^2=4a^2b^2-a^4, a^4-4a^2b^2+16Q^2=0 .
\end{gathered}
$$

From the obtained equation, we find that

$$
a^2=2b^2 \pm \sqrt{4b^2-16Q^2}
$$

Therefore,

$$
\begin{gathered}
DM=\sqrt{AM}=\sqrt{b^2-b^2+\frac{a^2}{3}} \\
\quad=\sqrt{b^2 \mp 2 \sqrt{b^2-4Q^2}} \sqrt{3}
\end{gathered}
$$

Thus,

$$
\begin{gathered}
V_{ABCD}=\frac{1}{3} S_{\triangle ABC} \cdot DM=\frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot DM \\
=\frac{\sqrt{3}}{12} \cdot\left(2b^2 \pm \sqrt{4b^2-16Q^2}\right) \\
=\frac{1}{12}\left(b^2 \pm \sqrt{b^2-4Q^2}\right) \sqrt{b^2 \mp 2 \sqrt{b^2-4Q^2}}
\end{gathered}
$$

## Answer

$\frac{1}{12}\left(b^2 \pm \sqrt{b^2-4Q^2}\right) \sqrt{b^2 \mp 2 \sqrt{b^2-4Q^2}}$ (if $2Q<b^2<\sqrt{\frac{19}{3}}$, then both solutions are valid).

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