12.310. On the surface of a sphere with radius $R$ lie all the vertices of an isosceles trapezoid, where the smaller base is equal to the lateral side, and the acute angle is $\alpha$. Find the distance from the center of the sphere to the plane of the trapezoid, if the larger base of the trapezoid is equal to the radius of the sphere.
## Solution.

Let the given isosceles trapezoid $ABCD, BC \| AD, AD=R, AB=BC=CD, \angle BAD=\alpha, 0<\alpha<\frac{\pi}{2}$, be inscribed in a section of a given sphere, $O_{1}$ - the center of the sphere, $O_{2}$ - the center of the section (Fig. 12.172). Then $O_{1} O_{2}$ is the required distance from the center of the sphere to the plane of the trapezoid.

$\angle BAC = \angle BCA$, since $AB = BC$ (Fig. 12.172, a), $\angle BCA = \angle CAD$, since $BC \| AD$. $\angle CAD = \angle BAC = \frac{\alpha}{2}, \angle ACD = \pi - (\angle CAD + \angle ADC) = \pi - \frac{3\alpha}{2}$.

The radius of the circle circumscribed around $\triangle ACD$:

$$
\begin{aligned}
& r = \frac{AD}{2 \sin \angle ACD} = \frac{R}{2 \sin \left(\pi - \frac{3\alpha}{2}\right)} = \frac{R}{2 \sin \frac{3\alpha}{2}} \\
& \text{In } \triangle O_{1} O_{2} A \left(\angle O_{1} O_{2} A = \frac{\pi}{2}\right):
\end{aligned}
$$

$$
\begin{aligned}
& O_{1} O_{2} = \sqrt{O_{1} A^{2} - O_{2} A^{2}} = \sqrt{R^{2} - r^{2}} = \sqrt{R^{2} - \frac{R^{2}}{4 \sin^{2} \frac{3\alpha}{2}}} = \frac{R}{\sin \frac{3\alpha}{2}} \sqrt{\sin^{2} \frac{3\alpha}{2} - \frac{1}{4}} = \\
& = \frac{R}{\sin \frac{3\alpha}{2}} \sqrt{\sin^{2} \frac{3\alpha}{2} - \sin^{2} \frac{\pi}{6}} = \frac{R}{\sin \frac{3\alpha}{2}} \sqrt{\frac{1 - \cos 3\alpha}{2} - \frac{1 - \cos \frac{\pi}{3}}{2}} = \\
& = \frac{R}{\sin \frac{3\alpha}{2}} \sqrt{\frac{1}{2} \left(\cos \frac{\pi}{3} - \cos 3\alpha\right)} = \frac{R}{\sin \frac{3\alpha}{2}} \sqrt{\sin \left(\frac{3\alpha}{2} + \frac{\pi}{2}\right) \sin \left(\frac{3\alpha}{2} - \frac{\pi}{6}\right)} \\
& \text{Answer: } \frac{R}{\sin \frac{3\alpha}{2}} \sqrt{\sin \left(\frac{3\alpha}{2} + \frac{\pi}{6}\right) \sin \left(\frac{3\alpha}{2} - \frac{\pi}{6}\right)} \\
& 12.311 .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1065.jpg?height=565&width=535&top_left_y=771&top_left_x=96)

Fig. 12.173

The angle between the generatrix and the base plane of the cone is $\alpha$. A sphere is inscribed in this cone. A tangent line is drawn to the circle of contact between the spherical and conical surfaces, and a plane is drawn through this line parallel to the height of the cone. Find the area of the section of the sphere by this plane.

## Solution.

Consider the axial section of the given set of bodies (Fig. 12.173): $\triangle ABC$ is the axial section of the cone, $BD$ is its height, $BD=H, \angle BAD=\alpha, O$ is the center of the sphere inscribed in the cone, $E$ is the point of tangency of the sphere and the generatrix $BC$ of the cone, $EF$ is the diameter of the section of the required area.

Draw the perpendicular $OK$ to $EF$. Then $EK$ is the radius of the section of the required area.

In $\triangle ADB \left(\angle ADB = 90^{\circ}\right): AD = BD \operatorname{ctg} \angle BAD = \frac{H}{\operatorname{tg} \alpha}$.

In $\triangle ADO \left(\angle ADO = 90^{\circ}\right)$:

$$
OD = AD \operatorname{tg} \angle OAD = \frac{H \operatorname{tg} \frac{\alpha}{2}}{\operatorname{tg} \alpha} = \frac{H \operatorname{tg} \frac{\alpha}{2} \left(1 - \operatorname{tg}^{2} \frac{\alpha}{2}\right)}{2 \operatorname{tg} \frac{\alpha}{2}} =
$$

$$
= \frac{H \left(1 - \frac{\sin^{2} \frac{\alpha}{2}}{\cos^{2} \frac{\alpha}{2}}\right)}{2} = \frac{H \left(\cos^{2} \frac{\alpha}{2} - \sin^{2} \frac{\alpha}{2}\right)}{2 \cos^{2} \frac{\alpha}{2}} = \frac{H \cos \alpha}{2 \cos^{2} \frac{\alpha}{2}}
$$

$\angle BOE = \angle BCD = \alpha$.

$\angle EOK = 90^{\circ} - \angle BOE = 90^{\circ} - \alpha$.

$$
OE = OD = \frac{H \cos \alpha}{2 \cos^{2} \frac{\alpha}{2}}
$$

In $\triangle OKE \left(\angle OKE = 90^{\circ}\right)$:

$$
EK = OE \sin \angle EOK = \frac{H \cos \alpha}{2 \cos^{2} \frac{\alpha}{2}} \sin \left(90^{\circ} - \alpha\right) = \frac{H \cos^{2} \alpha}{2 \cos^{2} \frac{\alpha}{2}}
$$

The area of the section $S = \pi \cdot EK^{2} = \frac{\pi H^{2} \cos^{4} \alpha}{4 \cos^{4} \frac{\alpha}{2}}$.

Answer: $\frac{\pi H^{2} \cos^{4} \alpha}{4 \cos^{4} \frac{\alpha}{2}}$.