Library / Saudi Arabian Mathematical Competitions /284 of 395

Saudi Arabian Mathematical Competitions, 2015

Geometry Difficulty 6.3 National Olympiad Prove it Saudi Arabia

Let $BD$ and $CE$ be altitudes of an arbitrary scalene triangle $ABC$ with orthocenter $H$ and circumcenter $O$ . Let $M$ and $N$ be the midpoints of sides $AB$ , respectively $AC$ , and $P$ the intersection point of lines $MN$ and $DE$ . Prove that lines $AP$ and $OH$ are perpendicular.

Solution

Because $\angle ADH = \angle HEA = 90^{\circ}$ , quadrilateral $AEHD$ is cyclic with $AH$ a diameter of its circumcircle $\omega_{H}$ .

Because $\angle ANO = \angle OMA = 90^{\circ}$ , quadrilateral $AMON$ is cyclic with $AO$ a diameter of its circumcircle $\omega_{O}$ .

Let $\omega_{9}$ be the nine point circle of the triangle $ABC$ . It is the circumcircle of quadrilateral $DEMN$ .

Because the circles $\omega_{H}$ and $\omega_{9}$ intersect at $D$ and $E$ , the line $DE$ is radical axis of $\omega_{H}$ and $\omega_{9}$ . Because the circles $\omega_{O}$ and $\omega_{9}$ intersect at $M$ and $N$ , the line $MN$ is radical axis of $\omega_{O}$ and $\omega_{9}$ . We deduce that $P$ is the radical center of the circles $\omega_{H}, \omega_{O}$ and $\omega_{9}$ , and therefore, the line $AP$ is the radical axis of the circles $\omega_{H}$ and $\omega_{O}$ , which is perpendicular to the line $H' O'$ , where $H'$ is the center of $\omega_{H}$ and $O'$ the center of $\omega_{O}$ . But $H'$ is the midpoint of $AH$ and $O'$ is the midpoint of $AO$ . We deduce that $AP$ is perpendicular to $HO$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.