Construction of a Triangle from Circumcenter, Orthocenter and Incenter

Jack D'Aurizio
25 March, 2016

By Euler's formula, $OI^2=R^2-2Rr,\;$ hence it is not difficult to draw both the incircle and the circumcircle. Otherwise, we may exploit the fact that the orthocenter is the incenter of the orthic triangle and go this way: the intersection $A'\;$ between the angle bisector $AI\;$ and the circumcircle $(ABC)\;$ is the midpoint of the arc $BC,\;$ hence $BC\;$ is perpendicular to $OA'.\;$ We may take a point $P\;$ on $(ABC)\;$ and its symmetric with respect to $OA',\;$ $P',\;$ then considering the perpendicular to $IP\;$ through $P\;$ and the perpendicular to $IP'\;$ through $P',\;$ meeting at $Q.\;$ If we take $R\;$ as the intersection between $AA'\;$ and the parallel to $OA'\;$ through $Q,\;$ the circle having $IR\;$ as a diameter intersects $(ABC)\;$ at $B,C.\;$

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