A new point on the nine-point circle III

Look at the diagram above, where a number of things appear to be happening. The red circle that goes through points $H,U,V,X$ is our focus for now. The center $W$ of this circle is itself a point on the nine-point circle of the parent $\triangle ABC$ . We’ll examine more of the properties of $W$ in this post, especially when the side-slopes of $\triangle ABC$ form a geometric progression.

Simplified coordinates

If $\triangle ABC$ has vertices at $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ , then the coordinates of point $W$ are given by:

(1) $\begin{equation*} \begin{split} x&=\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ y&=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)} \end{split} \end{equation*}$

Compare (1) with its equivalent, simplified version:

(2) $\begin{equation*} x=\frac{m_2x_2-m_3x_1}{m_2-m_3},~y=\frac{m_2y_1-m_3y_2}{m_2-m_3} \end{equation*}$

Here, $m_1,m_2,m_3$ are the slopes of sides $AB,BC,CA$ , respectively. Notice that $m_1$ and the coordinates of $C(x_3,y_3)$ are explicitly “missing” in equation (2). One implication is that the calculation of $W$ requires only two sides and the slopes of those two sides. So (2) can take another form:

(3) $\begin{equation*} x=\frac{m_1x_1-m_2x_3}{m_1-m_2},~y=\frac{m_1y_3-m_2y_1}{m_1-m_2} \end{equation*}$

Or this other form:

(4) $\begin{equation*} x=\frac{m_3x_3-m_1x_2}{m_3-m_1},~y=\frac{m_3y_2-m_1y_3}{m_3-m_1} \end{equation*}$

Ample freedom. Simple concept.

Find the coordinates of point $W$

for $\triangle ABC$

with vertices located at $A(1,4)$

, $B(0,0)$

, $C(3,6)$

Set $(x_1,y_1)=(1,4)$ , $(x_2,y_2)=(0,0)$ , and $(x_3,y_3)=(3,6)$ . The slopes of sides $AB,BC$ , and $CA$ are then $m_1=4$ , $m_2=2$ , and $m_3=1$ , respectively.

Using equation (2):

$\begin{equation*} \begin{split} x&=\frac{m_2x_2-m_3x_1}{m_2-m_3},~y=\frac{m_2y_1-m_3y_2}{m_2-m_3}\\ &=\frac{2(0)-1(1)}{2-1}=-1,~y=\frac{2(4)-1(0)}{2-1}=8 \end{split} \end{equation*}$

Using equation (3):

$\begin{equation*} \begin{split} x&=\frac{m_1x_1-m_2x_3}{m_1-m_2},~y=\frac{m_1y_3-m_2y_1}{m_1-m_2}\\ &=\frac{4(1)-2(3)}{4-2}=-1,~y=\frac{4(6)-2(4)}{4-2}=8 \end{split} \end{equation*}$

Using equation (4):

$\begin{equation*} \begin{split} x&=\frac{m_3x_3-m_1x_2}{m_3-m_1},~y=\frac{m_3y_2-m_1y_3}{m_3-m_1}\\ &=\frac{4(0)-1(3)}{4-1}=-1,~y=\frac{4(6)-1(0)}{4-1}=8 \end{split} \end{equation*}$

In each case we obtain $W(-1,8)$ . It’s that point that $W$ anders here and there on the nine-point circle.

If $\triangle ABC$

has one side parallel to the $x$ -axis, PROVE that $W$

coincides with the foot of an altitude.

The converse is also true. Fix the vertices at $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ . Let the slopes of $AB,BC,CA$ be $m_1,m_2,m_3$ and suppose that $m_2=0$ . Using equation (2):

$\begin{equation*} \begin{split} x&=\frac{m_2x_2-m_3x_1}{m_2-m_3},~y=\frac{m_2y_1-m_3y_2}{m_2-m_3}\\ &=\frac{0-m_3x_1}{0-m_3}=x_1,~y=\frac{0-m_3y_2}{0-m_3}=y_2 \end{split} \end{equation*}$

Thus, $W$ is the point $(x_1,y_2=y_3)$ , which is the foot of the altitude from vertex $A$ .

If $\triangle ABC$

has two sides with opposite slopes, PROVE that $W$

is the midpoint of the third side.

The converse is also true. Fix the vertices at $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ . Let the slopes of $AB,BC,CA$ be $m_1,m_2,m_3$ and suppose that $m_2=-m_3$ . Using equation (2):

$\begin{equation*} \begin{split} x&=\frac{m_2x_2-m_3x_1}{m_2-m_3}\\ &=\frac{m_2x_2+m_2x_1}{m_2+m_2}\\ &=\frac{x_2+x_1}{2}\\ y&=\frac{m_2y_1-m_3y_2}{m_2-m_3}\\ &=\frac{m_2y_1+m_2y_2}{m_2+m_2}\\ &=\frac{y_1+y_2}{2} \end{split} \end{equation*}$

Thus, $W$ is the point $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ , which is the midpoint of $AB$ .

If the slopes of the legs of a right triangle are $\pm 1$

, PROVE that $W$

coincides with the circumcenter of the triangle.

This follows from example 3, since $\pm 1$ are opposites.

Special case

Most of our points behave extremely well if the side-slopes of the parent triangle form a geometric progression. Expectedly, our favorite point $W$ is not an exception.

If the side-slopes of $\triangle ABC$

form a geometric progression with common ratio $r$

, PROVE that the slopes of the line segments $BW,AW,CW$

form a geometric progression with common ratio $r^2$

Let the slopes of sides $AB,BC,CA$ be $m_1=a$ , $m_2=ar$ , $m_3=ar^2$ . From this post, the slopes from $W$ to vertices $A,B,C$ are:

(5) $\begin{equation*} \begin{split} m_{WA}&=\frac{m_1m_3}{-m_2},~m_{WB}=\frac{m_1m_2}{-m_3},~m_{WC}=\frac{m_2m_3}{-m_1}\\ m_{WA}&=\frac{a\times ar^2}{-ar}\\ &=-ar\\ &\cdots\\ m_{WB}&=\frac{a\times ar}{-ar^2}\\ &=-\frac{a}{r}\\ &\cdots\\ m_{WC}&=\frac{ar\times ar^2}{-a}\\ &=-ar^3 \end{split} \end{equation*}$

Thus, $m_{WB},m_{WA},m_{WC}$ correspond to the geometric progression $-\frac{a}{r},-ar,-ar^3$ with common ratio $r^2$ .

If the side-slopes of $AB,BC,CA$

form a geometric progression $a,ar,ar^2$

in that order, PROVE that $W$

lies on the median through vertex $A$

(hence is co-linear with the centroid, the midpoint of $BC$

, and vertex $A$

From the preceding example, the slope from $W$ to vertex $A$ is $-ar$ . From this post, the slope from vertex $A$ to the midpoint of side $BC$ is also $-ar$ . Thus, $W$ lies on the median through vertex $A$ .

If the slopes of the sides $AB,BC,CA$

of $\triangle ABC$

form a geometric progression $a,ar,ar^2$

in that order, PROVE that the area of $\triangle ABW$

is equal to the area of $\triangle ACW$

As noted in example 6 above, the point $W$ is on the median through vertex $A$ . Suppose that $W$ lies outside the triangle (usually when $r> 0$ ). Let $M$ be the midpoint of side $BC$ . Then the area of $\triangle BMW$ is equal to the area of $MCW$ (since $WM$ is a median in $\triangle WBC$ ). But then the area of $\triangle ABM$ is also equal to the area of $\triangle AMC$ (since $AM$ is a median in $\triangle ABC$ ). Together we have that the area of $\triangle ABW$ is equal to the area of $\triangle ACW$ . Same conclusion when $W$ is inside the parent triangle.

In the diagram above, point $W$ is inside the given triangle. The areas of triangles $ABW$ and $ACW$ are equal.

Sample calculations

Since $W$ lies on the median through vertex $A$ , it is natural to ask what ratio both it and the centroid $G$ divide the entire median length $AM$ .

(6) $\begin{equation*} \frac{GW}{AW}=\frac{r^2+r+1}{3r},~\frac{MW}{GW}=\frac{3(r^2+1)}{2(r^2+r+1)}},~\frac{AM}{AW}=\frac{(r-1)^2}{2r} \end{equation*}$

As you already know, the relations in (6) hold when the slopes of the parent triangle form a geometric progression. Absolute values may be necessary in some cases.

Let $\triangle ABC$

be equilateral with sides $AB,BC,CA$

having slopes $a,ar,ar^2$

. PROVE that $W$

is midway between vertex $A$

and centroid $G$

The common ratio $r$ of the slopes $a,ar,ar^2$ of an equilateral triangle satisfies the quadratic equation

(7) $\begin{equation*} r^2+4r+1=0 \end{equation*}$

Use equation (7) and the first ratio in equation (6) to obtain

$\frac{GW}{AW}=\left|\frac{r^2+r+1}{3r}\right |=\left|\frac{-3r}{3r}\right |=1$

Thus, $AW=GW$ and so $W$ is the midpoint of $A$ and $G$ . In addition, since the centroid divides median $AM$ in the ratio $1:2$ , it follows that $AW=WG=GM$ in this case.

Let $\triangle ABC$

be right isosceles with sides $AB,BC,CA$

having slopes $a,ar,ar^2$

. PROVE that $3GW=2AW$

Use the first ratio in equation (6) and the fact that the common ratio of the slopes $a,ar,ar^2$ of a right isosceles triangle satisfies the quadratic equation

(8) $\begin{equation*} r^2+3r+1=0 \end{equation*}$

and obtain

$\frac{GW}{AW}=\left|\frac{r^2+r+1}{3r}\right |=\left|\frac{-2r}{3r}\right |=\frac{2}{3}$

Thus, $2AW=3GW$ .

Suppose that the slopes of sides $AB,BC,CA$

form a geometric progression $a,ar,ar^2$

. If $r> 0$

, PROVE that the point $W$

lies outside the triangle.

We have:

$\frac{AM}{MW}=\frac{(r-1)^2}{r^2+1}< 1~~\textrm{when}~~r> 0$

Thus, the line segment $MW$ is longer than the median $AM$ . (If $r< 0$ , we can’t conclude that $W$ lies inside the triangle.)

Takeaway

In $\triangle ABC$ , let $G$ be the centroid, $M$ the midpoint of side $BC$ , and $W$ as given by equation (1). If the slopes of sides $AB,BC,CA$ form a geometric progression $a,ar,ar^2,$ then the following statements are equivalent:

$q=-19$ , where $q$ is the sum of quotients of slopes defined here
$r^2+3r+1=0$ or $r^2+7r+1=0$
$AB^2+CA^2=3BC^2$ or $AB^2+CA^2=\frac{7}{5}BC^2$
$m_{AB}^2+m_{CA}^2=\frac{7}{5}m_{BC}^2$ or $m_{AB}^2+m_{CA}^2=3m_{BC}^2$
$r=-\phi^2$ or $r=-\phi^4$ or $r=-\frac{1}{\phi^2}$ or $r=-\frac{1}{\phi^4}$ , where $\phi$ is the golden ratio
$\frac{AW}{GW}=\frac{3}{2}$ or $\frac{AW}{GW}=2$
$\frac{AM}{AW}=\frac{5}{2}$ or $\frac{AM}{AW}=\frac{9}{2}$
$\frac{AM}{WM}=\frac{5}{3}$ or $\frac{AM}{WM}=3$ .

Eight easy equivalences. A triangle that satisfies any of these (together with $ar^2=-1$ or $a^2r^3=-1$ ) will be a right isosceles triangle (or its “look-alike”).

Tasks

Let be such that is parallel to the -axis, and and have reciprocal slopes. PROVE that:
- the orthocenter $H$ is a reflection of vertex $C$ across the $x$ -axis
- $W$ is the foot of the altitude from vertex $C$
- $W$ is the midpoint of $C$ and $H$
Let be the slopes of the line segments from to the vertices of , and let be the side-slopes.
- PROVE that $m_{WA}\times m_{WB}\times m_{WC}=-\left(m_{AB}\times m_{BC}\times m_{CA}\right)$ .
- If $\triangle ABC$ is right-angled, deduce that the product $m_{WA}\times m_{WB}\times m_{WC}$ equals the slope of the hypotenuse.
PROVE that $BW^2+CW^2=(r^2+r^{-2})AW^2$ for any $\triangle ABC$ with side-slopes $a,ar,ar^2$ .
Find coordinates for the vertices of $\triangle ABC$ having the property that a point $W$ on its nine-point circle is twice as far from vertex $A$ as it is from the centroid $G$ .
Suppose that the slopes of sides $AB,BC,CA$ are $a,ar,ar^2$ , in that order. Let $W$ be as given in equation (1), and let $M$ be the midpoint of $BC$ . PROVE that $\frac{AW}{MW}=\frac{2r}{r^2+1}$ .