Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.8 AMC 10/12 Find the answer China

Suppose in the tetrahedron $ABCD$ , $AB = 1$ , $CD = \sqrt{3}$ , the distance and angle between the lines $AB$ and $CD$ are $2$ and $\frac{\pi}{3}$ respectively. Then the volume of the tetrahedron equals:

Pick one

Solution

As in the diagram, from point $C$ draw a line $CE$ such that it is equal and parallel to $AB$ . Construct a prism $ABF$ - $ECD$ with $\triangle CDE$ as base and $BC$ as a lateral edge. Denote $V_1$ as the volume of the tetrahedron and $V_2$ the volume of the prism, then $V_1 = \frac{1}{3} V_2$ .

Since $S_{\triangle CDE} = \frac{1}{2} CE \cdot CD \sin \angle ECD$ , and the common perpendicular line $MN$ of $AB$ and $CD$ is the height of the prism, then
$V_2 = \frac{1}{2} MN \cdot CE \cdot CD \sin \angle ECD = \frac{3}{2}.$
So $V_1 = \frac{1}{3} V_2 = \frac{1}{2}$ .

Answer: B.

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