A new point on the nine-point circle II

Take vertex $A$ of triangle $ABC$ , together with the orthocenter $H$ of the triangle, and our new point $W$ with coordinates

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

then the resulting triangle $AHW$ has the following property: the slopes of its sides form a geometric progression. Same with vertices $B$ and $C$ as well as the corresponding triangles $BHW$ and $CHW$ (examples 6 and 10). This construction works for all triangles, except those triangles with one side parallel to the $x$ or $y$ axis.

Rigid transversals

If we draw three line segments from $W$ to each of the three vertices $A,B,C$ of any triangle $ABC$ , then the line segments $AW,BW,CW$ either all make acute angles with the $x$ -axis or all make obtuse angles with the $x$ -axis. See example 2 below.

In $\triangle ABC$ , let $m_1,m_2,m_3$ be the slopes of sides $AB,BC,CA$ , respectively. PROVE that the slopes of the line segments $AW, BW, CW$ are $\frac{m_1m_3}{-m_2}$ , $\frac{m_1m_2}{-m_3}$ , $\frac{m_2m_3}{-m_1}$ , in that order.

Let the coordinates of the vertices be $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_3)$ . With $W$ as in equation (1), we can calculate the slope of $AW$ (those of $BW$ and $CW$ are similar).

$\begin{equation*} \begin{split} \Delta x&= \frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}-x_1\\ &=\frac{(x_1-x_2)(y_2-y_3)(x_3-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ \Delta y&=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}-y_1\\ &=\frac{(y_1-y_2)(x_2-x_3)(y_1-y_3)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ \therefore \textrm{slope of AW}&=\frac{\Delta y}{\Delta x}\\ &=\frac{(y_1-y_2)(x_2-x_3)(y_1-y_3)}{(x_1-x_2)(y_2-y_3)(x_3-x_1)}\\ &=\frac{y_1-y_2}{x_1-x_2}\times\frac{y_1-y_3}{x_3-x_1}\times\frac{x_2-x_3}{y_2-y_3}\\ &=m_1\times (-m_3)\times\frac{1}{m_2}\\ &=\frac{m_1m_3}{-m_2} \end{split} \end{equation*}$

The placement of the negative sign beside $m_2$ was deliberate: to make it easy to remember the formula for the slope of $AW$ as well as those of $BW$ and $CW$ . Here’s how it works: the slope of $AW$ is the product of the slopes of the two sides ( $AB$ and $AC$ ) that originate from vertex $A$ divided by the negative of the slope of the side opposite vertex $A$ (side $BC$ ).

In any $\triangle ABC$ , PROVE that the line segments $AW,BW,CW$ either all make acute angles with the positive $x$ -axis or all make obtuse angles with the positive $x$ -axis.

The slopes from $W$ to $A,B,C$ are given by:

(2) $\begin{equation*} \frac{m_1m_3}{-m_2},~\frac{m_1m_2}{-m_3},~\frac{m_2m_3}{-m_1} \end{equation*}$

Re-write the terms in (2) as

(3) $\begin{equation*} \frac{m_1m_2m_3}{-m_2^2},~\frac{m_1m_2m_3}{-m_3^2},~\frac{m_1m_2m_3}{-m_1^2} \end{equation*}$

and consider four cases:

Case I: $m_1,m_2,m_3$ are all positive. In this case, the product $m_1m_2m_3$ is also positive and so each term in equation (3) is negative.
Case II: $m_1,m_2,m_3$ are all negative. In this case, the product $m_1m_2m_3$ is also negative and so each term in equation (3) is positive.
Case III: two of $m_1,m_2,m_3$ are positive while the third is negative; specifically, let $m_1,m_2$ be positive while $m_3$ is negative. In this case, the product $m_1m_2m_3$ is negative, and so the terms in equation (3) are all positive.
Case IV: one of $m_1,m_2,m_3$ is positive while the other two are negative; specifically, let $m_1$ be positive while $m_2$ and $m_3$ are negative. In this case, the product $m_1m_2m_3$ will be positive and so the terms in equation (3) will be negative.

In $\triangle ABC$ , let $m_1,m_2,m_3$ be the slopes of sides $AB,BC,CA$ , respectively. PROVE that the line segment $AW$ is perpendicular to side $BC$ if and only if $m_1m_3=1$ .

$AW$ can also be perpendicular to $AB$ or $AC$ , but requiring it to be perpendicular to $BC$ is more convenient.

First suppose that $AW$ is perpendicular to $BC$ . Since the slope of $AW$ is $\frac{m_1m_3}{-m_2}$ and the slope of $BC$ is $m_2$ , we have:

$\frac{m_1m_3}{-m_2}\times m_2=-1\implies m_1m_3=1.$

Conversely, suppose that $m_1m_3=1$ . Let’s simplify the slope of $AW$ :

$m_{AW}=\frac{m_1m_3}{-m_2}=\frac{1}{-m_2}.$

This shows that $AW$ is perpendicular to $BC$ .

(The case where $m_1m_3=1$ and $m_2=\pm 1$ is a special case. The case where $m_1m_3=1$ and $m_2=0$ is an extremely pleasant case.)

In $\triangle ABC$ , let $m_1,m_2,m_3$ be the slopes of sides $AB,BC,CA$ , respectively. PROVE that $AW$ is parallel to $BW$ if and only if $m_2=-m_3$ (in other words, the slopes of sides $BC$ and $CA$ are negatives of each other).

First suppose that $AW$ is parallel to $BW$ . Set their slopes equal:

$\frac{m_1m_3}{-m_2}=\frac{m_1m_2}{-m_3}\implies m_2^2=m_3^2\implies m_2=\pm m_3.$

However, since we can’t have $m_2=m_3$ in a triangle, we take $m_2=-m_3$ .

Conversely, suppose that $m_2=-m_3$ and re-calculate the slopes of $AW$ and $BW$ :

$m_{AW}=\frac{m_1m_3}{-m_2}=\frac{m_1m_3}{m_3}=m_1,~~m_{BW}=\frac{m_1m_2}{-m_3}=\frac{m_1m_2}{m_2}=m_1.$

What this means is that $AW$ and $BW$ are parallel to side $AB$ (in fact, $W$ is the midpoint of $AB$ under this condition).

Let $H$ be the orthocenter of $\triangle ABC$ , and let $W$ be the point with coordinates in equation (1). PROVE that the slope of $HW$ is the negative reciprocal of the product of the slopes of sides $AB,BC,CA$ .

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 its orthocenter are given by:

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

These are very much the same as the coordinates of point $W$ , except for the extra term $(y_1-y_2)(y_2-y_3)(y_3-y_1)$ in the numerator of the $x$ -coordinate and the extra term $-(x_1-x_2)(x_2-x_3)(x_3-x_1)$ in the $y$ -coordinate. It follows that the slope of the line segment $HW$ is:

(5) $\begin{equation*} m_{HW}=\frac{-(x_1-x_2)(x_2-x_3)(x_3-x_1)}{(y_1-y_2)(y_2-y_3)(y_3-y_1)}=\frac{-1}{m_1m_2m_3} \end{equation*}$

where $m_1,m_2,m_3$ are the slopes of sides $AB,BC,CA$ .

(Main goal)

For any triangle $ABC$ with non-zero side-slopes, PROVE that the three triangles $AHW$ , $BHW$ , and $CHW$ have their side-slopes in geometric progressions.

We do this for triangle $AHW$ . The same argument can be adapted for the other two triangles.

Let the slopes of sides $AB,BC,CA$ be $m_1,m_2,m_3$ . Assume all non-zero.

The slope from vertex $A$ to the orthocenter $H$ is $\frac{-1}{m_2}$ (by the definition of the altitude from vertex $A$ ). The slope of the line segment $WA$ is $\frac{m_1m_3}{-m_2}$ (by example 1), and the slope of the line segment $HW$ is $\frac{-1}{m_1m_2m_3}$ (by example 5). Since

$\frac{m_1m_3}{-m_2}\times\frac{-1}{m_1m_2m_3}=\frac{1}{m_2^2}=\left(\frac{-1}{m_2}\right)^2,$

the slopes of sides $WA,AH,HW$ form a geometric progression.

Right triangles

In $\triangle ABC$ , let $\angle C=90^{\circ}$ . PROVE that the slope of $CW$ is the reciprocal of the slope of the hypotenuse $AB$ .

Let $m_1,m_2,m_3$ be the slopes of sides $AB,BC,CA$ . By example 1, the slope of the line segment $CW$ is $\frac{m_2m_3}{-m_1}$ . Since the given triangle is right-angled at $C$ , sides $BC$ and $CA$ are perpendicular, and so $m_2m_3=-1$ , which in turn simplifies the slope of the line segment $CW$ :

$\frac{m_2m_3}{-m_1}=\frac{1}{m_1},~m_1\neq 0.$

Alternatively, one can also deduce this from example 5.

(The three triangles $AHW,BHW,CHW$ reduces to only two for a right triangle, since its orthocenter coincides with a vertex. But we do get a special quadrilateral.)

Related trapezium

Let $\triangle ABC$ be a right triangle in which the hypotenuse has slope $\pm 1$ and the legs are not parallel to the coordinate axes. PROVE that the quadrilateral $ABCW$ is a trapezium.

Excluding the case where the legs are parallel to the coordinate axes is essential; if not, $W$ coincides with the orthocenter, which is one of the vertices of a right triangle.

Suppose we have $\angle C=90^{\circ}$ and $m_{AB}=1$ . According to example 7, the slope of $CW$ will be $1$ as well, meaning that $CW$ is parallel to $AB$ . This gives the required trapezium $ABCW$ . Similarly, if the slope of the hypotenuse $AB$ is $-1$ (that is $m_{AB}=-1$ ), then the slope of $CW$ is again $-1$ , and this gives a trapezium $ABCW$ .

(Note that the slopes of $AW$ and $BW$ are reciprocals of each other under the above condition. Together with the slope of $CW$ as $\pm 1$ , we get that the slopes of the line segments $AW,CW,BW$ form a geometric progression, even though the slopes of the sides of the parent triangle $ABC$ do not form a geometric progression.)

Calculate the coordinates of point $W$ for $\triangle ABC$ with vertices at $A(0,0)$ , $B(10,10)$ , $C(4,12)$ . Deduce that $ABCW$ is a trapezium.

Observe that $\triangle ABC$ is right-angled at $C$ . Using equation (1) with $x_1=0,y_1=0$ , $x_2=10,y_2=10$ , and $x_3=4,y_3=12$ , we get:

$\begin{equation*} \begin{split} x&=\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{0+100(-4)+48(10)}{0+10(12)+4(-10)}\\ &=1\\ y&=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{0+100(12)+48(-10)}{0+10(12)+4(-10)}\\ &=9 \end{split} \end{equation*}$

Thus, $W$ is located at $(1,9)$ in the Cartesian plane. The slope of $CW$ is $\frac{12-9}{4-1}=1$ , and the slope of $AB$ is $1$ , so the two line segments are parallel. The slope of $AW$ is $9$ , while the slope of $BC$ is $-\frac{1}{3}$ . Therefore, we obtain a trapezium.

(Notice that the slopes of $BW, CW, AW$ form a geometric progression, despite the fact the the slopes of the parent triangle $ABC$ do not form a geometric progression.)

Calculate the coordinates of $W$ for $\triangle ABC$ with vertices at $A(0,0)$ , $B(6,12)$ , and $C(5,20)$ . Deduce that $AWBC$ is a trapezium. Verify also that the slopes of the sides of triangles $AHW$ , $BHW$ , and $CHW$ form geometric progressions.

Unlike in the preceding example, the given triangle $ABC$ is not right-angled this time. So this example is saying that we can still obtain this special trapezium even when the starting triangle is not right-angled. Using equation (1) with $x_1=0,y_1=0$ , $x_2=6,y_2=12$ , and $x_3=5,y_3=20$ , we get:

$\begin{equation*} \begin{split} x&=\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{0+72(-5)+100(6)}{0+6(20)+5(-12)}\\ &=4\\ y&=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{0+72(20)+100(-12)}{0+6(20)+5(-12)}\\ &=4 \end{split} \end{equation*}$

Thus, $W$ is the point $(4,4)$ . With this, the slope of $BW$ is $\frac{12-4}{6-4}=4$ . Since the slope of $AC$ is also $4$ , it follows that $AC$ is parallel to $BW$ . The slope of $AW$ is $1$ and the slope of $BC$ is $-8$ . So we obtain a trapezium $AWBC$ .

Behind the scences, we calculated the orthocenter of triangle $ABC$ ; it is the point $H\left(36,\frac{9}{2}\right)$ . Together with $W(4,4)$ and the given vertices $A(0,0)$ , $B(6,12)$ , $C(5,20)$ , we have the following slopes:

$\scriptstyle m_{HW}=\frac{1}{64},~m_{AW}=1,~m_{BW}=4,~m_{CW}=16,~m_{HA}=\frac{1}{8},~m_{HB}=-\frac{1}{4},~m_{HC}=-\frac{1}{2}.$

The slopes of the sides of $\triangle AHW$ are then $\frac{1}{64},\frac{1}{8},1$ ; the slopes of the sides of $\triangle BHW$ are $\frac{1}{64},-\frac{1}{4},4$ ; the slopes of the sides of $\triangle CHW$ are $\frac{1}{64},-\frac{1}{2},16$ . These form geometric progressions with common ratios $8,-16,-32$ , respectively.

(Notice that the slopes of $AW, BW, CW$ form a geometric progression, despite the fact the the slopes of the parent triangle $ABC$ do not form a geometric progression.)

Takeaway

Let $W$ be the point on the nine-point circle of $\triangle ABC$ whose coordinates were given in equation (1). Let $H$ be the orthocenter of $\triangle ABC$ . Then the following statements are equivalent:

the slopes of sides $AB$ and $AC$ are reciprocals of each other
the line segment $WA$ is perpendicular to side $BC$
$W$ is the midpoint of the line segment $AH$ .

You’ll be getting used to equivalent statements by now.

Tasks

For $\triangle ABC$ with vertices at $A(-3,6)$ , $B(0,0)$ , $C(2,16)$ , calculate the slopes of line segments $AW$ , $BW$ , and $CW$ . Verify that $AW$ is perpendicular to $AB$ .
Let be such that side is parallel to the -axis. PROVE that:
- $W$ coincides with the foot of the altitude from vertex $A$
- none of the triangles $AWH,BWH,CWH$ has slopes in geometric progression.
  (This is the only exception to the main point in today’s discussion.)
In $\triangle ABC$ , let $W$ be as given in equation (1). PROVE that the slopes of $AW$ and $BW$ are reciprocals of each other if and only if the slope of side $AB$ is $\pm 1.$
PROVE if the slopes of the sides of a triangle form a geometric progression with common ratio $r$ , then the slopes of $BW,AW,CW$ form a geometric progression with common ratio $r^2$ .
In , let be the slopes of sides . PROVE that the following statements are equivalent:
- $WA$ is parallel to $BC$
- $m_2^2+m_1m_3=0$
  (Under this condition, the slopes of line segments $BW,AW,CW$ form a geometric progression.)