12.447 In a regular truncated triangular pyramid, two spheres are inscribed; one touches all its faces, the other touches all its edges. Find the sine of the angle between the lateral edge and the plane of the base.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-316.jpg?height=289&width=390&top_left_y=1247&top_left_x=73)

Let $A B C A_{1} B_{1} C_{1}$ be the given regular truncated pyramid, $P$ and $P_{1}$ be the centers of its bases $A B C$ and $A_{1} B_{1} C_{1}$, $b$ and $a$ be the sides of the bases, $O$ be the center of the inscribed sphere, and $Q$ be the center of the sphere that touches all the edges of the pyramid.

Let $K$ be the projection of point $Q$ onto the lateral face $A A_{1} B_{1} B$, and $E_{1}, E_{2}, E_{3}, E_{4}$ be the points of tangency of the sphere $Q$ with the edges of this face. Then $K$ is the center of the circle with radius $E_{1} K$, inscribed in the isosceles trapezoid $A A_{1} B_{1} B$ (equal inclined lines $Q E_{i}$ have equal projections $\left.E_{i} K, i=1,2,3,4\right)$.

$$
\begin{aligned}
& \angle A_{1} K A=90^{\circ}(\text { theorem } 9) \Rightarrow E_{1} K^{2}=E_{1} A_{1} \cdot E_{1} A(\text { theorem } 5) \Rightarrow \\
& E_{1} K^{2}=A_{1} E_{2} \cdot A E_{4}=\frac{a \cdot b}{4}(\text { theorem } 3) \Rightarrow E_{2} E_{4}{ }^{2}=a b .
\end{aligned}
$$

Moreover, $A_{1} A=\frac{a+b}{2}$ (theorem 2).

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-317.jpg?height=321&width=274&top_left_y=140&top_left_x=221)

Now let $O M$ be the radius of the sphere $O$, drawn to the point of tangency $M$ with the face $A A_{1} B_{1} B$, which belongs to $E_{2} E_{4}$. By analogy with the previous reasoning, we have:

$O M^{2}=E_{2} P_{1} \cdot E_{4} P=\frac{a b}{12}, P_{1} P^{2}=4 \cdot O M^{2}=\frac{a b}{3} ;$

$E_{2} E_{4}=E_{2} P_{1}+E_{4} P=\frac{a+b}{2 \sqrt{3}} \Rightarrow \frac{a+b}{2 \sqrt{3}}=\sqrt{a b}$.

From this, $\sin \angle A_{1} A P=\frac{P_{1} P}{A_{1} A}=\sqrt{\frac{a b}{3}} ; \frac{a+b}{2}=\frac{1}{3}$.

Answer: $\frac{1}{3}$.