8,9

In a regular triangular pyramid $A B C D$, the side of the base $A B C$ is $12, \angle A D B=2 \operatorname{arctg} \frac{3}{4}$. In triangle $A B D$, the bisector $B A 1$ is drawn, and in triangle $B C D$, the median $B C 1$ and the altitude $C B 1$ are drawn. Find: 1) the volume of the pyramid $A 1 B 1 C 1 D ; 2)$ the area of the projection of triangle $A 1 B 1 C 1$ on the plane $A B C$.
Let $M$ be the midpoint of $AB$ (Fig.1), $H$ be the center of the equilateral triangle $ABC$. Denote $\angle ADB = \phi$. Then

$$
\begin{aligned}
& AH = \frac{2}{3} CM = \frac{2}{3} \cdot \frac{12 \sqrt{3}}{2} = 4 \sqrt{3}, \angle ADM = \frac{\varphi}{2}, \operatorname{tg} \frac{\varphi}{2} = \frac{3}{4} \text{, }
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-43.jpg?height=108&width=1140&top_left_y=1195&top_left_x=459)

From the right triangles $AMD$ and $AHD$, we find that

$$
AD = \frac{AM}{\sin \frac{4}{2}} = \frac{6}{3} = 10, \quad DH = \sqrt{AD^2 - AH^2} = \sqrt{100 - 48} = 2 \sqrt{13}
$$

Then

$$
V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot DH = \frac{1}{3} \cdot \frac{12^2 \sqrt{3}}{4} \cdot 2 \sqrt{13} = 24 \sqrt{39}
$$

By the angle bisector theorem in the triangle, $\frac{DA_1}{A_1A} = \frac{DB}{BA} = \frac{10}{12} = \frac{5}{6}$, so $\frac{DA_1}{DA} = \frac{5}{11}$. From the right triangle $CB_1D$, we find that $\frac{DB_1}{DC} = \cos \phi = \frac{7}{25}$, so $\frac{DB_1}{DB} = \frac{DB_1}{DC} = \frac{7}{25}$, and since $\frac{DC_1}{DC} = \frac{1}{2}$, then

$$
V_{A_1B_1C_1D} = \frac{DA_1}{DA} \cdot \frac{DB_1}{DB} \cdot \frac{DC_1}{DC} V_{ABCD} = \frac{5}{11} \cdot \frac{7}{25} \cdot \frac{1}{2} \cdot 24 \sqrt{39} = \frac{84 \sqrt{39}}{55}
$$

Let $A_2$, $B_2$, and $C_2$ be the orthogonal projections of points $A_1$, $B_1$, and $C_1$ onto the plane $ABC$ (Fig.2). Points $A_2$, $B_2$, and $C_2$ lie on segments $HA$, $HB$, and $HC$ respectively, and

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-43.jpg?height=543&width=798&top_left_y=2004&top_left_x=630)

Therefore,

$$
S_{\triangle A_2B_2C_2} = S_{\triangle A_2HB_2} + S_{\triangle B_2HC_2} + S_{\triangle A_2HC_2} = \frac{5}{11} \cdot \frac{7}{25} \cdot \frac{1}{33} S_{\triangle ABC} + \frac{7}{25} \cdot \frac{1}{2} \cdot \frac{1}{3} S_{\triangle ABC} + \frac{5}{11} \cdot \frac{1}{2} \cdot \frac{1}{3} S_{\triangle ABC} =
$$

$$
= \left( \frac{5}{11} \cdot \frac{7}{25} + \frac{7}{25} \cdot \frac{1}{2} + \frac{5}{11} \cdot \frac{1}{2} \right) \cdot \frac{1}{3} \cdot S_{\triangle ABC} = \left( \frac{5}{11} \cdot \frac{7}{25} + \frac{7}{25} \cdot \frac{1}{2} + \frac{5}{11} \cdot \frac{1}{2} \right) \cdot \frac{1}{3} \cdot \frac{12^2 \sqrt{3}}{4} = \frac{1632 \sqrt{3}}{275}
$$.

## Answer

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