IMO 1966 · Problem 3

Prove: The sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.

Official text of the problem. Official problem paper (PDF)

Solution

Let the vertices be \(A_1,A_2,A_3,A_4\), the circumcenter be \(O\), the circumradius be \(R>0\), and the common edge length be \(a\). Take \(O\) as the origin and put \[ \mathbf v_i=\overrightarrow{OA_i}. \]

First we show that \[ \mathbf v_1+\mathbf v_2+\mathbf v_3+\mathbf v_4=\mathbf0. \] Indeed, \(\|\mathbf v_i\|=R\), and for \(i\ne j\), \[ a^2=\|\mathbf v_i-\mathbf v_j\|^2 =2R^2-2\mathbf v_i\cdot\mathbf v_j, \] so \(\mathbf v_i\cdot\mathbf v_j=R^2-a^2/2\). Therefore, writing \(\mathbf s=\sum_{i=1}^4\mathbf v_i\), we have \[ \mathbf s\cdot\mathbf v_i =R^2+3\left(R^2-\frac{a^2}{2}\right) \] for every \(i\). Thus \(\mathbf s\) is perpendicular to each of \(\mathbf v_1-\mathbf v_4,\mathbf v_2-\mathbf v_4,\mathbf v_3-\mathbf v_4\). These vectors span space because the tetrahedron’s vertices are not coplanar. Hence \(\mathbf s=\mathbf0\).

Now let \(P\) be any point in space, with position vector \(\mathbf p\). By the Cauchy–Schwarz inequality, \[ R\,PA_i =\|\mathbf v_i\|\,\|\mathbf v_i-\mathbf p\| \ge \mathbf v_i\cdot(\mathbf v_i-\mathbf p) =R^2-\mathbf v_i\cdot\mathbf p. \] Summing and using \(\sum_i\mathbf v_i=\mathbf0\), we obtain \[ R\sum_{i=1}^4 PA_i\ge 4R^2, \qquad\text{so}\qquad \sum_{i=1}^4 PA_i\ge 4R=\sum_{i=1}^4 OA_i. \]

It remains to establish strictness when \(P\ne O\). Equality in the sum requires equality in each Cauchy–Schwarz inequality, so \[ \mathbf v_i-\mathbf p=\lambda_i\mathbf v_i \quad\text{for some }\lambda_i\ge0 \] for every \(i\). Consequently, \(\mathbf p\) is a scalar multiple of every \(\mathbf v_i\). If \(\mathbf p\ne\mathbf0\), all four vertices would lie on the line through \(O\) in direction \(\mathbf p\), which is impossible. Thus equality requires \(P=O\).

Therefore every other point \(P\) satisfies \[ \boxed{\sum_{i=1}^4 PA_i>\sum_{i=1}^4 OA_i}. \]

The solution was checked against John Scholes' IMO solution archive (Kalva).