Problem of the Week

Problem E and Solution

All Around the Cube

Problem

A cube is said to be inscribed in a sphere when all the vertices of the cube are on the surface of the sphere. In the diagram below, the cube is inscribed in the sphere with centre $$O$$. If the radius of the sphere is 6 cm, determine the volume of the cube.

Note:
In solving the above problem, it may be helpful to use the fact that if a cube is inscribed in a sphere with centre $$O$$, then the cube will also have centre $$O$$.

Solution

Label four of the vertices of the cube $$A$$, $$B$$, $$C$$, $$D$$, as shown in the diagram. Let $$x$$ represent the side length of the cube. Then $$AB=BC=CD=x$$.

The diagonals of a cube intersect at a point such that the distance from the intersection point to each vertex is equal. Since each vertex of the cube is on the sphere, the diagonal of the cube, $$AD$$, is equal in length to the diameter of the sphere. Therefore, $$AD=2(6)=12 \mbox{ cm}$$.
Each face of a cube is a square, so $$\angle ABC=90^{\circ}$$. Using the Pythagorean Theorem in $$\triangle ABC$$, $AC^2=AB^2+BC^2=x^2+x^2=2x^2$
In a cube the sides are perpendicular to the base. In particular, $$DC$$ is perpendicular to the base and it follows that $$DC \perp AC$$. Therefore $$\triangle DCA$$ is a right-angled triangle. Using the Pythagorean Theorem in $$\triangle DCA$$, $AD^2=AC^2+CD^2=2x^2+x^2=3x^2$ But $$AD=12$$, so $$AD^2=144$$. Then, \begin{aligned} 3x^2&=&144\\ x^2&=&48\\ x&=&4\sqrt{3}, \quad\text{ since x>0}\end{aligned} Therefore, the volume of the cube is $$x^3=(4\sqrt{3})^3=192\sqrt{3} \mbox{ cm}^3$$.