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# properties of regular tetrahedron

A regular tetrahedron may be formed such that each of its edges is a diagonal of a face of a cube; then the tetrahedron has been inscribed in the cube.

It’s apparent that a plane passing through the midpoints of three parallel edges of the cube cuts the regular tetrahedron into two congruent pentahedrons and that the intersection figure is a square, the midpoint $M$ of which is the centroid of the tetrahedron.

The angles between the four half-lines from the centroid $M$ of the regular tetrahedron to the vertices are $2\arctan\!{\sqrt{2}}$ ($\approx 109^{\circ}$), which is equal the angle between the four covalent bonds of a carbon atom. A half of this angle, $\alpha$, can be found from the right triangle in the below figure, where the catheti are $\frac{s}{\sqrt{2}}$ and $\frac{s}{2}$.

One can consider the regular tetrahedron as a cone. Let its edge be $a$ and its height $h$. Because of symmetry, a height line intersects the corresponding base triangle in the centroid of this equilateral triangle. Thus we have (see the below diagram) the rectangular triangle with hypotenuse $a$, one cathetus $h$ and the other cathetus $\frac{2}{3}\!\cdot\!\frac{a\sqrt{3}}{2}=\frac{a}{\sqrt{3}}$ (i.e. $\frac{2}{3}$ of the median $\frac{a\sqrt{3}}{2}$ of the equilateral triangle — see the common point of triangle medians). The Pythagorean theorem then gives

$h\;=\;\sqrt{a^{2}-\left(\frac{a}{\sqrt{3}}\right)^{2}}\;=\;\frac{a\sqrt{6}}{3}.$ |

Consequently, the height of the regular tetrahedron is $\displaystyle\frac{a\sqrt{6}}{3}$.

Since the area of the base triangle is $\frac{a^{2}\sqrt{3}}{4}$, the volume (one third of the product of the base and the height) of the regular tetrahedron is $\displaystyle\frac{a^{3}\sqrt{2}}{12}$.

## Mathematics Subject Classification

51E99*no label found*

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