IMO 1970 SL 3

Origin: BUL

Problem

In the tetrahedron SABC the angle BSC is a right angle, and the projection of the vertex S to the plane ABC is the intersection of the altitudes of the triangle ABC. Let z be the radius of the inscribed circle of the triangle ABC. Prove that SA2 + SB2 + SC2 \geq18z2.

Solution

We shall use the following lemma Lemma. If an altitude of a tetrahedron passes through the orthocenter of the opposite side, then each of the other altitudes possesses the same property. Proof. Denote the tetrahedron by SABC and let a = BC, b = CA, c = AB, m = SA, n = SB, p = SC. It is enough to prove that an altitude passes through the orthocenter of the opposite side if and only if a2 + m2 = b2 + n2 = c2 + p2. Suppose that the foot S′ of the altitude from S is the orthocenter of ABC. Then SS′ \perpABC ⇒SB2 −SC2 = S′B2 −S′C2. But from AS′ \perpBC it follows that AB2 −AC2 = S′B2 −S′C2. From these two equalities it can be concluded that n2 −p2 = c2 −b2, or equivalently, n2 + b2 = c2 + p2. Analogously, a2 + m2 = n2 + b2, so we have proved the first part of the equivalence. Now suppose that a2 + m2 = b2 + n2 = c2 + p2. Defining S′ as before, we get n2 −p2 = S′B2 −S′C2. From the condition n2 −p2 = c2 −b2 (⇔b2 + n2 = c2 + p2) we conclude that AS′ \perpBC. In the same way CS′ \perpAB, which proves that S′ is the orthocenter of \triangleABC. The lemma is thus proven. Now using the lemma it is easy to see that if one of the angles at S is right, than so are the others. Indeed, suppose that \angleASB = \pi/2. From the lemma we have that the altitude from C passes through the orthocenter of \triangleASB, which is S, so CS \perpASB and \angleCSA = \angleCSB = \pi/2. Therefore m2 + n2 = c2, n2 + p2 = a2, and p2 + m2 = b2, so it follows that m2 + n2 + p2 = (a2 + b2 + c2)/2. By the inequality between the arithmetic and quadric means, we have that (a2 + b2 + c2)/2 \geq2s2/3, where s denotes the semiperimeter of \triangleABC. It remains to be shown that 2s2/3 \geq18r2. Since S\triangleABC = sr, this is equivalent to 2s4/3 \geq 18S2 ABC = 18s(s −a)(s −b)(s −c) by Heron’s formula. This reduces to s3 \geq27(s−a)(s−b)(s−c), which is an obvious consequence of the AM–GM mean inequality. Remark. In the place of the lemma one could prove that the opposite edges of the tetrahedron are mutually perpendicular and proceed in the same way.