IMO 1972 SL 5

Origin: GDR

Problem

Prove the following assertion: The four altitudes of a tetrahe- dron ABCD intersect in a point if and only if AB2 + CD2 = BC2 + AD2 = CA2 + BD2.

Solution

Lemma. If X, Y, Z, T are points in space, then the lines XZ and Y T are perpendicular if and only if XY 2 + ZT 2 = Y Z2 + TX2. Proof. Consider the plane \pi through XZ parallel to Y T. If Y ′, T ′ are the feet of the perpendiculars to \pi from Y, T respectively, then XY 2 + ZT 2 = XY ′2 + ZT ′2 + 2Y Y ′2, and Y Z2 + TX2 = Y ′Z2 + T ′X2 + 2Y Y ′2. Since by the Pythagorean theorem XY ′2 +ZT ′2 = Y ′Z2 +T ′X2, i.e., XY ′2−Y ′Z2 = XT ′2−T ′Z2, if and only if Y ′T ′ \perpXZ, the statement follows. Assume that the four altitudes intersect in a point P. Then we have DP \perp ABC ⇒DP \perpAB and CP \perpABD ⇒CP \perpAB, which implies that CDP \perpAB, and CD \perpAB. By the lemma, AC2 + BD2 = AD2 + BC2. Using the same procedure we obtain the relation AD2 + BC2 = AB2 + CD2. Conversely, assume that AB2 + CD2 = AC2 + BD2 = AD2 + BC2. The lemma implies that AB \perpCD, AC \perpBD, AD \perpBC. Let \pi be the plane containing CD that is perpendicular to AB, and let hD be the altitude from D to ABC. Since \pi \perpAB, we have \pi \perpABC ⇒hD \subset\pi and \pi \perpABD ⇒hC \subset\pi. The altitudes hD and hC are not parallel; thus they have an intersection point PCD. Analogously, hB \caphC = {PBC} and hB \caphD = {PBD}, where both these points belong to \pi. On the other hand, hB doesn’t belong to \pi; otherwise, it would be perpendicular to both ACD and AB \subset\pi, i.e. AB \subsetACD, which is impossible. Hence, hB can have at most one common point with \pi, implying PBD = PCD. Analogously, PAB = PBD = PCD = PABCD.