$[$ [extension of a tetrahedron to a parallelepiped]

Segment $A B(A B=1)$, being a chord of a sphere with radius 1, is positioned at an angle of $60^{\circ}$ to the diameter $C D$ of this sphere. The distance from the end $C$ of the diameter to the nearest end $A$ of the chord $A B$ is $\sqrt{2}$. Find $B D$.

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Complete the tetrahedron $ABCD$ to a parallelepiped $AKBLNDMC (AN\|KD\|BM\|LC)$ by drawing pairs of parallel planes through its opposite edges. Let $O$ and $Q$ be the centers of the faces $NDMC$ and $AKBL$ respectively. Then $O$ is the center of a sphere, $KL=CD=2, OQ \perp AB, OA=OB=1$ as radii of the sphere, $KL \| CD, QB$ $=QA=\frac{1}{2}, QL=QK=1$. Let $\angle BQL=60^{\circ}$. Then by the cosine theorem,

\[
BL=\sqrt{QB^2+QL^2-2 QB \cdot QL \cos 60^{\circ}}=\sqrt{\frac{1}{4}+1-2 \cdot \frac{1}{2} \cdot \frac{1}{2}}=\frac{1}{2} \sqrt{3}
\]

Therefore, triangle $BQL$ is a right triangle, $\angle QBL=90^{\circ}$. This means $AB \perp BL$, and since $BM \| OQ$ and $OQ \perp AB$, then $AB \perp BM$, hence the line $AB$ is perpendicular to the plane of the face $BMLC$. Then $AB \perp BC$, so in the right triangle $ABC$, the hypotenuse $AC$ is greater than the leg $BC$. But according to the condition $AC < BC$, so $\angle BQL$ $=120^{\circ}$. In this case $\angle BQK=60^{\circ}$. Then

\[
BK=\sqrt{QB^2+QK^2-2 QB \cdot QK \cos 60^{\circ}}=\sqrt{\frac{1}{4}+1-2 \cdot \frac{1}{2} \cdot \frac{1}{2}}=\frac{1}{2} \sqrt{3}
\]

\[
OQ^2=OA^2-QA^2=1-\frac{1}{4}=\frac{3}{4}
\]

Thus, $AN=CL=OQ=\frac{1}{2} \sqrt{3}$. In the parallelogram $ALCN$

\[
NL^2+AC^2=2AN^2+2AL^2
\]

Therefore,

\[
\begin{aligned}
& BD^2=NL^2=2AN^2+2AL^2-AC^2= \\
& =2AN^2+2BK^2-AC^2=\frac{3}{2}+\frac{3}{2}-2=1 .
\end{aligned}
\]

Thus, $BD=1$.

## Answer

1.00