IMO 1967 LL ITA26

Origin: ITA

Problem

Let ABCD be a regular tetrahedron. To an arbitrary point M on one edge, say CD, corresponds the point P = P(M), which is the intersection of two lines AH and BK, drawn from A orthogonally to BM and from B orthogonally to AM. What is the locus of P as M varies?

Solution

Let L be the midpoint of the edge AB. Since P is the orthocenter of \triangleABM and ML is its altitude, P lies on ML and therefore belongs to the triangular area LCD. Moreover, from the similarity of triangles ALP and MLB we have LP \cdotLM = LA\cdotLB = a2/4, where a is the side length of tetrahedron ABCD. It easily follows that the locus of P is the image of the segment CD under the inversion of the plane LCD with center L and radius a/2. This locus is the arc of a circle with center L and endpoints at the orthocenters of triangles ABC and ABD.