IMO 1995 SL G6

Origin: USA | Category: Geometry

Problem

Let A1A2A3A4 be a tetrahedron, G its centroid, and A′ 1, A′ 2, A′ 3, and A′ 4 the points where the circumsphere of A1A2A3A4 in- tersects GA1, GA2, GA3, and GA4, respectively. Prove that GA1 \cdot GA2 \cdot GA3 \cdot GA4 \leqGA′ 1 \cdot GA′ 2 \cdot GA′ 3 \cdot GA′ and GA′ + GA′ + GA′ + GA′ \leq GA1 + GA2 + GA3 + GA4 .

Solution

Let O be the circumcenter and R the circumradius of A1A2A3A4. We have OA2 i = (−−\to OG + (−−\to OAi −−−\to OG))2 = OG2 + GA2 i + 2−−\to OG \cdot −−\to GAi. Summing up these equalities for i = 1, 2, 3, 4 and using that 4 i=1 −−\to GAi = −\to0 , we obtain  i=1 OA2 i = 4OG2 +  i=1 GA2 i ⇐⇒  i=1 GA2 i = 4(R2 −OG2). (1) Now we have that the potential of G with respect to the sphere equals GAi \cdot GA′ i = R2 −OG2. Plugging in these expressions for GA′ i, we reduce the inequalities we must prove to GA1 \cdot GA2 \cdot GA3 \cdot GA4 \leq(R2 −OG2)2 (2) and (R2 −OG2)  i=1 GAi \geq  i=1 GAi. (3)

Inequality (2) immediately follows from (1) and the quadratic-geometric mean inequality for GAi. Since from the Cauchy–Schwarz inequality we have 4 i=1 GA4 i \geq1 4 i=1 GAi and 4 i=1 GAi 4 i=1 GAi

\geq16, inequality (3) follows from (1) and from

 i=1 GA2 i

 i=1 GAi

\geq1

 i=1 GAi  i=1 GAi

\geq4  i=1 GAi.