IMO 1973 SL 9

Origin: ROM

Problem

Let Ox, Oy, Oz be three rays, and G a point inside the trihe- dron Oxyz. Consider all planes passing through G and cutting Ox, Oy, Oz at points A, B, C, respectively. How is the plane to be placed in order to yield a tetrahedron OABC with minimal perimeter?

Solution

Let a, b, c be vectors going along Ox, Oy, Oz, respectively, such that −−\to OG = a + b + c. Now let A \inOx, B \inOy, C \inOz and let −\to OA = \alphaa, −−\to OB = \betab, −−\to OC = \gammac, where \alpha, \beta, \gamma > 0. Point G belongs to a plane ABC with A \inOx, B \inOy, C \inOz if and only if there exist positive real numbers \lambda, µ, \nu with sum 1 such that \lambda−\to OA+µ−−\to OB+\nu−−\to OC = −−\to OG, which is equivalent to \lambda\alpha = µ\beta = \nu\gamma = 1. Such \lambda, µ, \nu exist if and only if \alpha, \beta, \gamma > 0 and \alpha + 1 \beta + 1 \gamma = 1. Since the volume of OABC is proportional to the product \alpha\beta\gamma, it is minimized when 1 \alpha \cdot 1 \beta \cdot 1 \gamma is maximized, which occurs when \alpha = \beta = \gamma = 3 and G is the centroid of \triangleABC.