10,11

In the pyramid $ABCD$, the faces $ABC$ and $ADC$ are isosceles triangles with a common base $AC$. A sphere of radius $R$ with its center at point $O$, which lies on the face $ABC$, is tangent to all the edges of the pyramid $ABCD$. Find the lengths of the segments into which the points of tangency of the sphere divide the edges of the pyramid, and the volume of the pyramid $ABCD$, if the angle $ABC$ is equal to $2\alpha$. Find the value of the angle $ABC$ for which the volume of the pyramid will be the smallest. Find this smallest value of the volume of the pyramid $ABCD$.
Let the sphere touch the edges $A B, B C, A C, A D, D C, D B$ at points $M, N, E, K, L, F$ respectively. The segments of tangents drawn from one point to the sphere are equal. Denote

$$
B M=B N=B F=a, C E=C N=C L=b, D K=D L=D F=c,
$$

and since $A B=B C$, then $A M=A E=A K=b$. The section of the sphere by the plane $A B C$ is a circle with center $O$, inscribed in the isosceles triangle $A B C$, so $B O, A O$ and $C O$ are the angle bisectors of this triangle. Therefore,

$$
\angle O B A=\angle C B O=\alpha, \angle B C O=\frac{1}{2} \angle B C A=\frac{1}{2}\left(90^{\circ}-\angle C B O\right)=45^{\circ}-\frac{2}{2}
$$

The radius of the sphere, drawn to the point of tangency with the line, is perpendicular to this line, so $O F \perp B D$ and $O N \perp B C$. The right triangles $O F B$ and $O N B$ are equal by the leg ( $O F=O N=R$ ) and the common hypotenuse $O B$, so $\angle D B O=\angle C B O=\alpha$. From the right triangles $O N B$ and $O N C$ we find that

$$
\begin{aligned}
a=B N= & O N \operatorname{ctg} O B N=R \operatorname{ctg} \alpha, b=C N=O N \operatorname{ctg} O C N= \\
& =R \operatorname{ctg}\left(45^{\circ}-\frac{\mathbf{2}}{2}\right)=R \cdot \frac{\sin \left(90^{\beta}-\hat{r}_{2}\right\}}{1-\cos \left(90^{\beta}-\mathbf{r}_{2}\right)}=\frac{R\left(\operatorname { R e } \left(\boldsymbol{r}_{\mathbf{2}}\right.}{1-\sin (\mathbf{2} .}
\end{aligned}
$$

Let $D T=h$ be the height of the pyramid. From the equality of the lateral edges $D A$ and $D C$ it follows that the point $T$ is equidistant from the points $A$ and $C$, and since $A B=B C$, the point $T$ lies on the line $B E$ containing the height of the isosceles triangle $A B C$. From the right triangles $C T D$ and $B T D$ we get that

$$
B T=B D \cos \angle D B O=(a+c) \cos \alpha
$$

By the cosine theorem $C T 2=B T 2+B C 2-2 B T \cdot B C \cos \alpha=B T 2+(a+b) 2-2(a+b) B T \cos \alpha$,

$$
\begin{aligned}
& (b+c) 2-(a+c) 2=(a+b) 2-2(a+c)(a+b) \cos 2 \alpha, \\
& \text { from which we find that }
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-47.jpg?height=89&width=454&top_left_y=590&top_left_x=802)

Denote $\sin \alpha=t$ and find the minimum value of the function $f(t)=\frac{(1+t)^{2}}{l(2-t)}$ on the interval $(0 ; 1)$. The equation $f^{\prime}(t)=$ $\frac{2(1+t)(2-1)}{t^{2}(2-t)^{2}}=0$ has a unique solution $t 0=\frac{1}{2}$ on this interval. On the interval $\left(0 ; \frac{1}{2}\right)$ the function $f(t)$ is decreasing $\left(f^{\prime}(t)\right.$

$0\right)$. Therefore, at the point $t 0=\frac{1}{2}$ the function $f(t)$ reaches its minimum value on the interval $(0 ; 1)$. Then $\alpha=30^{\circ}$ and $\angle A B C=2 \alpha=60^{\circ}$. In this case,

$$
V_{\text {min }}=\frac{2}{3} R 3 \cdot f\left(\frac{1}{2}\right)=\frac{2}{3} R 3 \cdot \frac{\left(1+\frac{1}{2}\right)^{2}}{\frac{1}{2}\left(2-\frac{1}{2}\right)}=2 R 3
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-47.jpg?height=167&width=1370&top_left_y=1597&top_left_x=341)