A regular tetrahedron has four faces, each of which is an equilateral triangle - Leaving Cert Mathematics - Question 9 - 2014

From the triangle OAB (where O is the circumcenter), we know ( |AOB| = 120^\circ ). By the Sine Rule, [ |OA| = \frac{2a \sin 30^\circ}{\sin 120^\circ} = \frac{2a \cdot \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{2a}{\sqrt{3}}. ] Thus, the radius of the cylinder is ( \frac{2a}{\sqrt{3}} ).

We drop a vertical line from vertex D of the tetrahedron to O, forming a right triangle. Let h be the height of the cylinder: [ h^2 + \left( \frac{2a}{\sqrt{3}} \right)^2 = (2a)^2. ] Solving this gives us [ h^2 = 4a^2 - \frac{4a^2}{3} = \frac{8a^2}{3} \Rightarrow h = \sqrt{\frac{8}{3}} a = \frac{2a\sqrt{2}}{\sqrt{3}}. ]

The volume ( V ) of the cylinder can be determined using the formula: [ V = \pi r^2 h. ] Substituting the radius and height, we get: [ V = \pi \left(\frac{2a}{\sqrt{3}}\right)^2 \left(\frac{2a\sqrt{2}}{\sqrt{3}}\right) = \pi \cdot \frac{4a^2}{3} \cdot \frac{2a\sqrt{2}}{\sqrt{3}} = \frac{8a^3\sqrt{2}}{3\sqrt{3}} \cdot \pi. ]

Further simplifying gives us: [ V = \pi \cdot \frac{8a^3 \sqrt{2}}{3\sqrt{3}} = \frac{8}{9} \sqrt{6} \pi a^3. ] This is the required volume of the cylindrical container.