A square-based right pyramid has its lateral faces forming a $60^{\circ}$ angle with the base. The plane that bisects the angle between the base and one of the lateral faces divides the pyramid into two parts. Determine the ratio of the volumes of the two resulting solids.
Let the base of the pyramid be the square $ABCD$, and its fifth vertex be $E$. The side face $AED$ and the angle bisector of the base intersect the edges $EB$ and $EC$ at points $F$ and $G$, respectively. Let $H$ and $I$ be the midpoints of the edges $AD$ and $BC$, respectively, and $J$ be the intersection point of the angle bisector plane and the segment $EI$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_febab40a3eaffda39bfcg-1.jpg?height=518&width=501&top_left_y=283&top_left_x=796)

The angle bisector plane cuts the pyramid $ABCDE$ into the pyramid $AFGDE$. First, we determine the volume of this pyramid. Let the side length of the square $ABCD$ be $a$. Since the side faces form a $60^\circ$ angle with the base, $\angle EHI = \angle EIH = 60^\circ$, so the triangle $EHI$ is equilateral, with side length $HI = AB = a$. Since $AFGD$ is the angle bisector plane, $\angle JHI = 30^\circ$, which means $HJ$ is the altitude of the triangle $EHI$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_febab40a3eaffda39bfcg-1.jpg?height=403&width=439&top_left_y=1029&top_left_x=824)

Figure 2

The point $J$ bisects the segment $EI$, so the segment $FG$, which is parallel to $BC$ and passes through $J$, is the midline of the triangle $EBC$, with length $FG = \frac{BC}{2} = \frac{a}{2}$. The quadrilateral $ADGF$ is a trapezoid with bases $a$ and $\frac{a}{2}$, and height $HJ = \frac{a \sqrt{3}}{2}$. Therefore, its area is:

$$
\frac{1}{2} \left(a + \frac{a}{2}\right) \cdot \frac{a \sqrt{3}}{2} = \frac{a^2 \cdot 3 \sqrt{3}}{8}
$$

$EJ$ is perpendicular to the plane $ADFG$ (since $EJ \perp FG$ and $EJ \perp HJ$), so the height of the pyramid $AFGDE$ from $E$ is $EJ = \frac{a}{2}$. Thus, the volume of the pyramid is:

$$
V_{AFGDE} = \frac{1}{3} \cdot \frac{a^2 \cdot 3 \sqrt{3}}{8} \cdot \frac{a}{2} = \frac{a^3 \sqrt{3}}{16}
$$

The height of the pyramid $ABCDE$ from $E$ is equal to the height of the equilateral triangle $EHI$ from $E$, so the volume of this pyramid is:

$$
V_{ABCDE} = \frac{1}{3} \cdot a^2 \cdot \frac{a \cdot \sqrt{3}}{2} = \frac{a^3 \cdot \sqrt{3}}{6}
$$

The volume of the solid $ABCDGF$ is the difference between the volumes of the two pyramids:

$$
V_{ABCDGF} = V_{ABCDE} - V_{AFGDE} = \frac{a^3 \cdot \sqrt{3}}{6} - \frac{a^3 \cdot \sqrt{3}}{16} = \frac{a^3 \cdot 5 \sqrt{3}}{48}
$$

The ratio of the volumes of the two solids determined by the angle bisector plane is:

$$
\frac{V_{AFGDE}}{V_{ABCDGF}} = \frac{\frac{a^3 \cdot \sqrt{3}}{16}}{\frac{a^3 \cdot 5 \sqrt{3}}{48}} = \frac{3}{5}
$$

Based on the work of Zoltán Jombik (Esztergom, P. Temesvári Ferences Gymnasium, 1st year).