IMO 1973 SL 1

Origin: BUL

Problem

Let a tetrahedron ABCD be inscribed in a sphere S. Find the locus of points P inside the sphere S for which the equality AP PA1

• BP PB1
• CP PC1
• DP PD1 = 4 holds, where A1, B1, C1, and D1 are the intersection points of S with the lines AP, BP, CP, and DP, respectively.

Solution

The condition of the point P can be written in the form AP 2 AP \cdotPA1 + BP 2 BP \cdotPB1 + CP 2 CP \cdotPC1 + DP 2 DP \cdotPD1 = 4. All the four denominators are equal to R2 −OP 2, i.e., to the power of P with respect to S. Thus the condition becomes AP 2 + BP 2 + CP 2 + DP 2 = 4(R2 −OP 2). (1) Let M and N be the midpoints of segments AB and CD respectively, and G the midpoint of MN, or the centroid of ABCD. By Stewart’s formula, an arbitrary point P satisfies AP 2 + BP 2 + CP 2 + DP 2 = 2MP 2 + 2NP 2 + 1 2AB2 + 1 2CD2 = 4GP 2 + MN 2 + 1 2(AB2 + CD2). Particularly, for P \equivO we get 4R2 = 4OG2 + MN 2 + 1 2(AB2 + CD2), and the above equality becomes AP 2 + BP 2 + CP 2 + DP 2 = 4GP 2 + 4R2 −4OG2. Therefore (1) is equivalent to OG2 = OP 2 +GP 2 ⇔\angleOPG = 90◦. Hence the locus of points P is the sphere with diameter OG. Now the converse is easy.