3450.21 – Tetrahedron Inscribed in a Cube

Tags:Problem Set 14Algebra/TrigonometryGeometry-Euclidean

Four of the eight vertexes of a cube are the vertexes of a regular tetrahedron as in the figure below. Relax for a moment and enjoy how neatly the regular tetrahedron fits inside the cube. The other four vertexes of the cube, incidentally, form a second, inter-penetrating tetrahedron.

Find the ratio of the surface area of the cube to the surface area of the tetrahedron.


Let the edge of the cube be xx. Then SCSC, the surface area of the cube, is 6x26x^2.

The faces of the tetrahedron are equilateral triangles whose edges are diagonals of the cube's square faces. If yy is the edge length of the tetrahedron, then y=x2y = x\sqrt{2} since the diagonal of a square's face is the hypotenuse of a 45-45-90 triangle. Now the area of an equilateral triangle of edge yy is y23/4=x23/2y^2 \sqrt{3}/4 = x^2 \sqrt{3}/2 so STST, the surface area of the tetrahedron, is ST=4x23/2=2x23.ST = 4 x^2 \sqrt{3}/2 = 2 x^2 \sqrt{3}.

Thus the desired ratio of areas is

SCST=6x22x23=33=3.\frac{SC}{ST} = \frac{6x^2}{2 x^2 \sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}.

Students who like making models might make a nice 3-d model of this figure.

(N.b. the 45-45-90 right triangle is sometimes known as the western, or cowboy's, triangle. Why?)