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

  1. 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.
  2. Let \triangle ABC be such that side BC is parallel to the x-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.)
  3. 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.
  4. 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.
  5. In \triangle ABC, let m_1,m_2,m_3 be the slopes of sides AB,BC,CA. 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.)