Nine-Point Circle

Motivation

I saw someone illustrating his/her solution with a “superior triangle”.

This reminds me the homothety about the centroid of factor −1/2.

The above picture \usetikzlibrary{calc} for computing coordinates from those of existing points. (A)!.25!(B) means $(A)+.25[(B)-(A)]$.

\begin{tikzpicture}[scale=2]
\coordinate (D) at (-0.7,1);
\coordinate (E) at (-1,0);
\coordinate (F) at (1,0);
\coordinate (A) at ($(E)!.5!(F)$);
\coordinate (B) at ($(F)!.5!(D)$);
\coordinate (C) at ($(D)!.5!(E)$);
\coordinate (G) at ($(D)!.5!(E)!1/3!(F)$);

\draw (A) -- (B) -- (C) -- cycle;
\draw (D) -- (E) -- (F) -- cycle;
\begin{scriptsize}
\fill (G) circle (0.5pt) node [left=2pt,anchor=north]{$G$};
\end{scriptsize}
\draw[->,-latex,dashed] (D) -- (A);
\draw[->,-latex,dashed] (E) -- (B);
\draw[->,-latex,dashed] (F) -- (C);
\end{tikzpicture}

Previous post

From the homothety between the nine-point circle and circumcircle about orthocenter with a factor of 2, we see that the nine-point center is the mid-point of orthocenter and circumcenter.

Motivation

Someone on Discord asked about the existence of the nine-point circle. It’s well-known that that can be proved by homothety.

Little reminder about homothety

Homothety preserves angles (and thus parallel lines). Homothetic polygons are similar, so the ratio of the corresponding sides is the same. Considering the radii of a circle under a homothety, we see that a homothety maps a circle to another circle.

Notation

• H: orthocenter
• G: centroid
• O: circumcenter
• ω: circumcircle
• H_A: feet of altitude with respect to A.
• M_A: midpoint of side a.
• E_A: Euler point with respect to A. (i.e. midpoint of A and H)

Problem

The second proof for nine-point circle on AoPS starts with a proved fact that the reflection of H about a and M_A lie on ω.