[ Regular tetrahedron ]

[ Vector product ]

Find the radius of the sphere circumscribed around a regular tetrahedron with edge $a$.

#
First method.

Let $ABCD$ be a regular tetrahedron with edge $a$ (Fig.1), $M$ be the center of the face $ABC$, and $R$ be the desired radius of the circumscribed sphere. Since $DM$ is the height of the tetrahedron, the angle between the edge $AD$ and the plane of the face $ABC$ is the angle $\angle DAM$. Let's denote it as $\alpha$. In the right triangle $DAM$, it is known that $AD = a$, $AM = \frac{a \sqrt{3}}{3}$. Therefore,

$$
\cos \alpha = \cos \angle DAM = \frac{AM}{AD} = \frac{\frac{a \sqrt{3}}{3}}{a} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}, \sin \alpha = \sqrt{\frac{2}{3}}
$$

We draw a section of the tetrahedron by a plane passing through the edge $AD$ and the midpoint $L$ of the opposite edge $BC$ (Fig.2). We obtain the triangle $ALD$, the vertices $A$ and $D$ of which lie on a circle of radius $R$ with the center on the height $DM$. Extend $AL$ to intersect the circle at point $E$. The isosceles triangle $ADE$ is inscribed in a circle of radius $R$, so

$$
R = \frac{DE}{2 \sin \angle DAE} = \frac{AD}{2 \sin \angle 2} = \frac{a}{2 \sqrt{\frac{2}{3}}} = \frac{a \sqrt{6}}{4}.
$$

## Second method.

Let $ABCD$ be a regular tetrahedron with edge $a$, and $R$ be the desired radius of the circumscribed sphere. The center of the sphere circumscribed around a regular tetrahedron lies on each of the four heights of the tetrahedron. The heights of a regular tetrahedron are its medians, and the medians of any triangular pyramid intersect at one point and are divided by it in the ratio 3:1, counting from the vertex. Therefore, the center of the circumscribed sphere coincides with the point of intersection of the heights of the regular tetrahedron, and the radius is $\frac{3}{4}$ of the height of the tetrahedron. Therefore,

$$
R = \frac{3}{4} a \sqrt{\frac{2}{3}} = \frac{a \sqrt{6}}{4}
$$

## Answer

$\frac{a \sqrt{6}}{4}$.