MathDB
Geometrical inequality, similar to treegoner's one

Source: Moldavian Olympiad

November 6, 2004
inequalitiesgeometrycircumcircleinradiusfunctiontrigonometrygeometric transformation

Problem Statement

Let ABCABC be a triangle, let OO be its circumcenter, and let HH be its orthocenter. Let PP be a point on the segment OHOH. Prove that 6rPA+PB+PC3R6r\leq PA+PB+PC\leq 3R, where rr is the inradius and RR the circumradius of triangle ABCABC. Moderator edit: This is true only if the point PP lies inside the triangle ABCABC. (Of course, this is always fulfilled if triangle ABCABC is acute-angled, since in this case the segment OHOH completely lies inside the triangle ABCABC; but if triangle ABCABC is obtuse-angled, then the condition about PP lying inside the triangle ABCABC is really necessary.)