From the midpoint of the height of a regular triangular pyramid, perpendiculars are dropped to the lateral edge and to the lateral face. These perpendiculars are equal to $a$ and $b$ respectively. Find the volume of the pyramid. Does the problem have a solution for any $a$ and $b$?
Drop perpendiculars $H P$ and $H Q$ from the base $H$ of the height $D H$ of the given regular triangular pyramid $A B C D$ to the lateral edge $C D$ and the lateral face $A B D$, respectively. Then $H P=2 a, H Q=2 b$. Let $\alpha$ and $\beta$ be the angles between the lateral edge and the plane of the base $A B C$, and the plane of the lateral face and the plane of the base $A B C$, respectively. Let $x$ be the side length of the base of the pyramid, and $M$ be the midpoint of $A B$. From the right triangles $H P C$ and $H Q M$, we find that

$$
\begin{gathered}
C H=H P / \sin \angle D C H, M H=H Q \cdot \sin \angle D M H, \text { or } \\
x \sqrt{3} / 3=2 a / \sin \alpha, x \sqrt{3} / 6=2 b / \sin \beta,
\end{gathered}
$$

from which

$$
\begin{gathered}
\sin \beta=(2 b / a) \cdot \sin \alpha, \sin ^{2} \beta=\left(4 b^{2} / a^{2}\right) \cdot \sin ^{2} \alpha \\
1 / \sin ^{2} \alpha=\left(4 b^{2} / a^{2}\right) / \sin ^{2} \beta
\end{gathered}
$$

From the right triangles $D H C$ and $D H M$, we find that

$$
D H=C H \cdot \operatorname{tg} \alpha, D H=M H \cdot \operatorname{tg} \beta,
$$

from which

$$
\begin{gathered}
\operatorname{tg} \beta=(C H / M H) \cdot \operatorname{tg} \alpha=2 \cdot \operatorname{tg} \alpha, \operatorname{ctg} \alpha=2 \cdot \operatorname{ctg} \beta \\
\operatorname{ctg}^{2} \alpha=4 \cdot \operatorname{ctg}^{2} \beta, 1 / \sin ^{2} \alpha-1=4\left(1 / \sin ^{2} \beta-1\right) \\
\left(4 b^{2} / a^{2}\right) / \sin ^{2} \beta-1=4\left(1 / \sin ^{2} \beta-1\right) \\
\left(4 b^{2} / a^{2}\right)-\sin ^{2} \beta=4\left(1-\sin ^{2} \beta\right) \\
3 \cdot \sin ^{2} \beta=4\left(1-b^{2} / a^{2}\right)=4\left(a^{2}-b^{2}\right) / a^{2}
\end{gathered}
$$

From which

$$
\begin{gathered}
\sin \beta=2 \sqrt{a^{2}-b^{2}} /(a \sqrt{3}) \\
\sin \alpha=(a /(2 b)) \cdot \sin \beta=\sqrt{a^{2}-b^{2}} /(b \sqrt{3}) \\
\cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\sqrt{1-4\left(a^{2}-b^{2}\right) /\left(3 b^{2}\right)}=
\end{gathered}
$$

$=\sqrt{4 b^{2}-a^{2}} /(b \sqrt{3})$,

$x \sqrt{a^{2}-b^{2}}$,

$x=6 a b / \sqrt{a^{2}-b^{2}}$,

$D H=H P / \cos \angle D H P=2 a / \cos \alpha=2 a b \sqrt{4 b^{2}-a^{2}}$.

Therefore,

$$
\begin{aligned}
& V(A B C D)=\frac{1}{3} S(A B C) \cdot D H=\frac{1}{3}\left(x^{2} \sqrt{4 b^{2}-a^{2}}\right)= \\
& =\frac{1}{3}\left(36 a^{2} b^{2} /\left(a^{2}-b^{2}\right)\right)\left(\sqrt{4 b^{2}-a^{2}}\right)= \\
& \quad=18 a^{3} b^{3} /\left(\left(a^{2}-b^{2}\right) \sqrt{4 b^{2}-a^{2}}\right)
\end{aligned}
$$