Let ABC be a triangle, let O be its circumcenter, and let H be its orthocenter.
Let P be a point on the segment OH.
Prove that
6r≤PA+PB+PC≤3R,
where r is the inradius and R the circumradius of triangle ABC.
Moderator edit: This is true only if the point P lies inside the triangle ABC. (Of course, this is always fulfilled if triangle ABC is acute-angled, since in this case the segment OH completely lies inside the triangle ABC; but if triangle ABC is obtuse-angled, then the condition about P lying inside the triangle ABC is really necessary.) inequalitiesgeometrycircumcircleinradiusfunctiontrigonometrygeometric transformation