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Problem of the Week

Problem E and Solution

All Around the Cube


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.

A cube inscribed in a sphere with centre O. A line segment from the centre O to a vertex of the cube has length 6.

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 \(O\).


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 cube lying flat on one of its faces. Vertices A, B, and C are on the bottom face of the cube. AC is a diagonal of this face. Vertex D is on the top face of the cube above vertex C. AD is a diagonal of the cube which passes through the interior of the cube.

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 \(x^3=(4\sqrt{3})^3=192\sqrt{3} \mbox{ cm}^3\).