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Prove that HA'+HB'+HC'<2R

Source: Canadian Repêchage 2013: Problem 8

May 20, 2014
geometrycircumcircletrigonometrygeometry proposed

Problem Statement

Let ABC\triangle ABC be an acute-angled triangle with orthocentre HH and circumcentre OO. Let RR be the radius of the circumcircle. \begin{align*} \text{Let }\mathit{A'}\text{ be the point on }\mathit{AO}\text{ (extended if necessary) for which }\mathit{HA'}\perp\mathit{AO}. \\ \text{Let }\mathit{B'}\text{ be the point on }\mathit{BO}\text{ (extended if necessary) for which }\mathit{HB'}\perp\mathit{BO}. \\ \text{Let }\mathit{C'}\text{ be the point on }\mathit{CO}\text{ (extended if necessary) for which }\mathit{HC'}\perp\mathit{CO}.\end{align*} Prove that HA+HB+HC<2RHA'+HB'+HC'<2R
(Note: The orthocentre of a triangle is the intersection of the three altitudes of the triangle. The circumcircle of a triangle is the circle passing through the triangle’s three vertices. The circummcentre is the centre of the circumcircle.)