2006/07 AHSMC Part 1 Problem 15

Let the base of the pyramid be the equilateral triangle $ABC$. Let $D$ be the midpoint of $BC$. Then $AD = \sqrt{AB^2 - BD^2} = \dfrac{\sqrt{3}}{2}$. Let $G$ be the centre of $ABC$. Then $G$ lies on $AD$ and $AG = \dfrac{2AD}{3} = \dfrac{\sqrt{3}}{3}$. Let $V$ be the fourth vertex of the pyramid and $O$ be the centre of the sphere passing through all four vertices. Then $O$ lies on $VG$ and let $r = VO = OA$ be the radius of the sphere. Since $VG - VO = OG = \sqrt{OA^2 - AG^2}$, we have $1 - r = \sqrt{r^2 - \dfrac{1}{3}}$. Squaring both sides, we have $1 - 2r + r^2 = r^2 - \dfrac{1}{3}$. It follows that $r = \dfrac{2}{3}$. The answer is (e).