IMO 1970 SL 5

Origin: CZS

Problem

Let M be an interior point of the tetrahedron ABCD. Prove that −−\to MA vol(MBCD) + −−\to MB vol(MACD) +−−\to MC vol(MABD) + −−\to MD vol(MABC) = 0 (vol(PQRS) denotes the volume of the tetrahedron PQRS).

Solution

Denote respectively by A1, B1, C1 and D1 the points of intersection of the lines AM, BM, CM, and DM with the opposite sides of the tetrahe- dron. Since vol(MBCD) = vol(ABCD)−−−\to MA1/−−\to AA1, the relation we have to prove is equivalent to −−\to MA \cdot −−−\to MA1 −−\to AA1

• −−\to MB \cdot −−−\to MB1 −−\to BB1
• −−\to MC \cdot −−−\to MC1 −−\to CC1
• −−\to MD \cdot −−−\to MD1 −−−\to DD1 = 0. (1) There exist unique real numbers \alpha, \beta, \gamma, and \delta such that \alpha+\beta +\gamma +\delta = 1 and for every point O in space −−\to OM = \alpha−\to OA + \beta−−\to OB + \gamma−−\to OC + \delta−−\to OD. (2) (This follows easily from −−\to OM = −\to OA+−−\to AM = −\to OA+k−−\to AB +l−\to AC +m−−\to AD = −−\to AB + k(−−\to OB −−\to OA) + l(−−\to OC −−\to OA) + m(−−\to OD −−\to OA) for some k, l, m \inR.) Further, from the condition that A1 belongs to the plane BCD we obtain for every O in space the following equality for some \beta′, \gamma′, \delta′: −−\to OA1 = \beta′−−\to OB + \gamma′−−\to OC + \delta′−−\to OD. (3) However, for \lambda = −−−\to MA1/−−\to AA1, −−\to OM = \lambda−\to OA+(1−\lambda)−−\to OA1; hence substituting (2) and (3) in this expression and equating coefficients for −\to OA we obtain \lambda = −−−\to MA1/−−\to AA1 = \alpha. Analogously, \beta = −−−\to MB1/−−\to BB1, \gamma = −−−\to MC1/−−\to CC1, and \delta = −−−\to MD1/−−−\to DD1; hence (1) follows immediately for O = M. Remark. The statement of the problem actually follows from the fact that M is the center of mass of the system with masses vol(MBCD), vol(MACD), vol(MABD), vol(MABC) at A, B, C, D respectively. Our proof is actually a formal verification of this fact.