[ Tangents to Spheres ]

Given four points $A, B, C, D$, not lying in the same plane. A sphere is tangent to the lines $A B$ and $A D$ at point $A$, and to the lines $B C$ and $C D$ at point $C$. Find the area of the sphere, given that $A B=1, B D=2, \angle A B C = \angle B A D = 90^{\circ}$.
Pass a plane through points $A, C$ and the center of the sphere.

## Solution

Let $O$ be the center of the sphere, $r$ be its radius. Consider the tetrahedron $A B C D$. Since the sphere touches its edges $A B$ and $A D$ at point $A$, the line $O A$ is perpendicular to two intersecting lines $A B$ and $A D$ in the plane of face $A B D$, so the line $O A$ is perpendicular to this plane. Therefore, the sphere touches the plane of face $A B D$ at point $A$.

Similarly, we can prove that the sphere touches the plane of face $B C D$ at point $C$. Thus, the sphere is inscribed in the dihedral angle formed by the planes of faces $A B D$ and $A B C$.

Let the plane passing through the intersecting lines $O A$ and $O C$ intersect the edge $B D$ of the specified dihedral angle at point $K$. Since $O A$ is perpendicular to the plane of face $A B D$, and $O C$ is perpendicular to the plane of face $B C D$, the line $B D$ is perpendicular to two intersecting lines $O A$ and $O C$ in the plane $A K C$. Therefore, $B D \perp A K$ and $B D \perp C K$, i.e., $A K$ and $C K$ are altitudes of triangles $A B D$ and $C B D$.

In the right triangle $A B D$, it is known that $B D=2$ and $A B=1$, and $B D$ is the hypotenuse, so

$$
\angle A D B=30^{\circ}, A D=\sqrt{3}, A K=\frac{1}{2} A D=\sqrt{3} / 2
$$

By the theorem of the equality of segments of tangents drawn from a point to a sphere, $B C=B A=1, K C=K A=\sqrt{3}/2$. Thus, the problem reduces to finding the radius $r$ of the circle that touches the sides $K A$ and $K C$ of the isosceles triangle $A K B$ at points $A$ and $B$, respectively, and

$$
K A=K C=\sqrt{A B^{2}+B C^{2}}=\sqrt{1+1}=\sqrt{2}.
$$

Let $K M$ be the altitude of triangle $A K C$. Then

$$
A M=C M=\sqrt{A K^{2}-A M^{2}}=\sqrt{(\sqrt{3} / 2)^{2}-(\sqrt{2} / 2)^{2}}=1 / 2.
$$

Denote $\angle O K C=\alpha$. From the right triangles $A K M$ and $A K O$, we find that

$$
r=A O=A K \cdot \operatorname{tg} \alpha=A K \cdot A M / K M=(\sqrt{3} / 2) \cdot(\sqrt{2} / 2) /(1 / 2)=\sqrt{6} / 2
$$

Therefore, the surface area of the sphere is $4 \pi r^{2}=6 \pi$.

## Answer

$6 \pi$.

Submit a comment