Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Taiwan

Given an arbitrary triangle $\triangle ABC$ , let its circumcircle be $O_1$ , its nine-point circle be $O_2$ ; and let the circle with diameter the segment joining the orthocenter $H$ and the centroid $G$ of $\triangle ABC$ be $O_3$ . Prove that $O_1, O_2, O_3$ are coaxial. (That is, there exists a line such that every point on it has the same power with respect to all three circles. The power of a point with respect to a circle is the square of the distance from the point to the center of the circle minus the square of the radius of the circle.)

Remark: The nine-point circle of a triangle is the circle passing through the nine points consisting of the midpoints of the three sides, the feet of the three altitudes, and the midpoints of the segments joining the three vertices to the orthocenter.

Solution

Lemma: The orthocenter and the centroid are the two centers of similitude of $O_1, O_2$ .

Proof of Lemma: The nine-point circle passes through the midpoints of $\overline{HA}, \overline{HB}, \overline{HC}$ , so $H$ is one center of similitude.
The nine-point circle passes through the midpoints of $\overline{AB}, \overline{BC}, \overline{CA}$ (denoted respectively by $M, N, P$ ), and $\overline{AG} : \overline{GM} = 2 : 1$ , $\overline{BG} : \overline{GN} = 2 : 1$ , $\overline{CG} : \overline{GP} = 2 : 1$ , so $G$ is also a center of similitude.

Now return to the proof of the original problem. When $ABC$ is acute, let $X$ be the intersection of the radical axis of $O_1, O_2$ with $OH$ , let $x$ be the radius of $O_1$ , $y$ the radius of $O_2$ , and let $\overline{XO} = a, \overline{XG} = b, \overline{XN} = c, \overline{XH} = d$ .

$\text{We have } \frac{a-b}{b-c} = \frac{a-d}{c-d} = \frac{x}{y},\quad a^2 - x^2 = c^2 - y^2.$
$\text{From this we get } b = \frac{cx+ay}{x+y},\quad d = \frac{cx-ay}{x-y},$
$bd = \frac{c^2 x^2 - a^2 y^2}{x^2 - y^2} = \frac{(c^2 - y^2)(x^2 - y^2)}{x^2 - y^2} = c^2 - y^2.$

Hence the three circles are coaxial (the case of an obtuse angle is computed similarly).

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.