Nine-Point Circle in the Complex Plane

The theorem

Theorem

Let $O$ and $H$ be the circumcenter and orthocenter of a triangle $ABC$ , and let its circumradius be $R$ . The three side midpoints, the three altitude feet, and the midpoints of $AH,BH,CH$ lie on a circle whose center is the midpoint of $OH$ and whose radius is $R/2$ .

Triangle ABC with its three altitudes, circumcircle, and nine-point circle; the side midpoints, altitude feet, and vertex–orthocenter midpoints lie on the smaller circle, whose center N is the midpoint of OH. — Figure 1. The homothety centered at H with ratio 1/2 sends the circumcircle to the nine-point circle.

The unit-circle model

Place the circumcircle at

\Gamma:\quad |z|=1

and take its center $O$ as the origin. Identify the Euclidean vector plane with the complex plane, and let the complex coordinates of $A,B,C$ be $a,b,c$ . Then

|a|=|b|=|c|=1.

Under this identification, the usual circumcenter-origin vector formula for the orthocenter gives

h=a+b+c.

Three formulas, one circle

Let $N$ be the point with coordinate

n=\frac h2=\frac{a+b+c}{2},

so $N$ is the midpoint of $OH$ . Let $e_a$ , $m_a$ , and $d_a$ denote, respectively, the midpoint of $AH$ , the midpoint of $BC$ , and the foot of the altitude from $A$ . The first two formulas are immediate. In the unit-circle model, the standard formula for the foot of the altitude from $A$ gives the third:

\begin{aligned} e_a&=\frac{a+h}{2}=\frac{2a+b+c}{2},\\ m_a&=\frac{b+c}{2},\\ d_a&=\frac12\left(a+b+c-\bar a bc\right). \end{aligned}

Therefore

e_a-n=\frac a2, \qquad m_a-n=-\frac a2, \qquad d_a-n=-\frac12\bar a bc.

Since $|a|=|b|=|c|=1$ ,

|e_a-n| =|m_a-n| =|d_a-n| =\frac12.

The cyclic formulas give the other six points. Thus all nine points lie on the circle

\mathcal N:\quad |z-n|=\frac12,

centered at $N$ with radius $1/2$ .

The homothety

Now define

T(w)=\frac{h+w}{2}.

Since $T(w)=n+w/2$ ,

T(\Gamma) =\left\{n+\frac w2:|w|=1\right\} =\mathcal N.

In particular, the three triples

\begin{aligned} &(a,b,c),\\ &(-a,-b,-c),\\ &\left(-\bar a bc,-a\bar b c,-ab\bar c\right) \end{aligned}

on $\Gamma$ are sent respectively to the vertex–orthocenter midpoints, the side midpoints, and the altitude feet.

Finally,

T(w)-h=\frac{w-h}{2},

so $T$ is the homothety centered at $H$ with ratio $1/2$ . Returning from the unit-circle normalization, the radius of $\mathcal N$ is $R/2$ .

The theorem

The unit-circle model

Three formulas, one circle

The homothety

Related notes and diagrams