4 [ Excircle ]

Let $A B C$ be an equilateral triangle. Three planes through the lines $A B, B C$, and $A C$ form an angle $\phi$ with the plane $A B C$ and intersect at point $D 1$. In addition, three planes through the same lines form an angle $2 \phi$ with the plane $A B C$ and intersect at point $D 2$. Find $\phi$, given that points $D 1$ and $D 2$ are at equal distances from the plane $A B C$.
Let's denote the side of the equilateral triangle $ABC$ by $a$. Then the radius of the inscribed circle of the triangle is $\frac{\sqrt{2}}{6}$, and the radius of the exscribed circle is $-\frac{\sqrt{2} \sqrt{3}}{2}$. The lateral faces of the triangular pyramid $ABCD_1$ form equal angles with the plane of the base $ABC$, so the height $D_1O_1$ of the pyramid passes through the center of the inscribed or the center of the exscribed circle of the triangle $ABC$. Similarly, the height $D_2O_2$ also passes through the center of the inscribed or the center of the exscribed circle of the triangle $ABC$. If $O_1$ and $O_2$ coincide with the center of the inscribed circle of the triangle $ABC$ or both $O_1$ and $O_2$ are the centers of the exscribed circles, then either

$$
D_1O_1=\frac{a \sqrt{3}}{6} \cdot \operatorname{tg} \phi < \frac{a \sqrt{3}}{6} \cdot \operatorname{tg} 2 \phi = D_2O_2
$$

or

$$
D_1O_1=\frac{a \sqrt{3}}{2} \cdot \operatorname{tg} \phi < \frac{a \sqrt{3}}{2} \cdot \operatorname{tg} 2 \phi = D_2O_2
$$

which contradicts the condition of the problem ($D_1O_1 = D_2O_2$). If $O_1$ is the center of the inscribed circle and $O_2$ is the center of the exscribed circle of the triangle $ABC$, then

$$
\begin{aligned}
D_1O_1= & \frac{a \sqrt{3}}{6} \cdot \operatorname{tg} \phi, D_2O_2=\frac{a \sqrt{3}}{2} \cdot \operatorname{tg} 2 \phi \\
& \frac{a \sqrt{3}}{6} \cdot \operatorname{tg} \phi = \frac{a \sqrt{3}}{2} \cdot \operatorname{tg} 2 \phi
\end{aligned}
$$

from which $\operatorname{tg} 2 \phi = -5$, which is impossible. If $O_1$ is the center of the exscribed circle and $O_2$ is the center of the inscribed circle of the triangle $ABC$, then

$$
\begin{gathered}
D_1O_1=\frac{a \sqrt{3}}{2} \cdot \operatorname{tg} \phi, D_2O_2=\frac{a \sqrt{3}}{6} \cdot \operatorname{tg} 2 \phi, \\
\frac{a \sqrt{3}}{2} \cdot \operatorname{tg} \phi = \frac{a \sqrt{3}}{6} \cdot \operatorname{tg} 2 \phi, \\
3 \operatorname{tg} \phi = \operatorname{tg} 2 \phi, 3 \operatorname{tg} \phi = \frac{2 \operatorname{tg}}{1-\operatorname{tr} \varphi},
\end{gathered}
$$

from which $\operatorname{tg} 2 \phi = \frac{1}{3}, \phi = 30^{\circ}$.

## Answer

$30^{\circ}$.