The base of the pyramid is a triangle with sides 5, 12, and 13, and its height forms equal angles with the heights of the lateral faces (dropped from the same vertex), which are not less than $30^{\circ}$. What is the maximum volume that such a pyramid can have?
Let $A 1, B 1$ and $C 1$ be the bases of the perpendiculars dropped from the base $O$ of the height $D O$ of the pyramid $A B C D$ to the sides $B C, A C$ and $A B$ of the base $A B C$, respectively, and $B C=12, A C=5, A B=13$. By the theorem of three perpendiculars, $D B 1 \perp A C, D C 1 \perp A B$ and $D A 1 \perp B C$. Therefore, $D B 1, D C 1$ and $D A 1$ are the heights of the lateral faces of the pyramid. According to the problem, $\angle O D B 1=\angle O D C 1=\angle O D A 1$. Right triangles $O D B 1, O D C 1$ and $O D A 1$ are equal by the leg and the adjacent acute angle, so $O B 1=O C 1=O A 1$, i.e., point $O$ is equidistant from the lines on which the sides of triangle $A B C$ lie. Therefore, $O$ is either the center of the inscribed circle of this triangle (Fig.1) or the center of its exscribed circle (Fig.2). Let $\angle O D B 1=\angle O D C 1=\angle O D A 1=\alpha \geqslant 30^{\circ}$. Note that triangle $A B C$ is a right triangle $(A C 2+B C 2=25+144=169=132=A B 2)$, and $\angle A C B=90^{\circ}$. Let $r$ be the radius of its inscribed circle, and $r_{a}, r_{b}$ and $r_{c}$ be the radii of the exscribed circles touching the sides $B C, A C$ and $A B$ respectively, $S$ be the area of triangle $A B C$, and $p$ be its semiperimeter. Then

$$
\begin{gathered}
S=\frac{1}{2} A C \cdot B C=\frac{1}{2} \cdot 5 \cdot 12=30 \\
p=\frac{1}{2}(A C+B C+A B)=\frac{1}{2}(5+12+13)=15, r=\frac{S}{p}=\frac{30}{15}=2 \\
r_{a}=\frac{S}{p-B C}=\frac{30}{3}=10, r_{b}=\frac{S}{p-A C}=\frac{30}{10}=3, r_{c}=\frac{S}{p-B C}=\frac{30}{2}=15
\end{gathered}
$$

If $h, h_{a}, h_{b}$ and $h_{c}$ are the heights of the pyramid corresponding to each of the considered cases, then

$$
h=r \operatorname{ctg} \alpha, h_{a}=r_{a} \operatorname{ctg} \alpha, h_{b}=r_{b} \operatorname{ctg} \alpha, h_{c}=r_{c} \operatorname{ctg} \alpha
$$

Since in each of these cases the area of the base of the pyramid is the same, the volume of the pyramid is maximal if its height is maximal. In turn, the maximum height corresponds to the maximum of the four found radii, i.e., $r_{c}=15$. Therefore,

$$
V_{A B C D}=\frac{1}{3} S_{\triangle A B C} \cdot D O=\frac{1}{3} S \cdot h_{c}=\frac{1}{3} \cdot 30 \cdot 15 \operatorname{ctg} \alpha=150 \operatorname{ctg} \alpha
$$

and since $\alpha \geqslant 30^{\circ}$, then $\operatorname{ctg} \alpha \leqslant \operatorname{ctg} 30^{\circ}=\sqrt{3}$. Therefore, $V_{A B C D} \leqslant 150 \sqrt{3}$, and equality is achieved if the radius is $r_{c}$ and $\alpha=30^{\circ}$.

## Answer

$150 \sqrt{3}$.