A new point on the nine-point circle III

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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 Wanders 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.

The fact that the common ratio of the slopes of BW,AW,CW is necessarily positive further confirms that these lines all make acute angles or all make obtuse angles with the positive x-axis.

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.

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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

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

  1. Let \triangle ABC be such that AB is parallel to the x-axis, and BC and CA 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
  2. Let m_{WA},m_{WB},m_{WC} be the slopes of the line segments from W to the vertices of \triangle ABC, and let m_{AB},m_{BC},m_{CA} 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.
  3. PROVE that BW^2+CW^2=(r^2+r^{-2})AW^2 for any \triangle ABC with side-slopes a,ar,ar^2.
  4. 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.
  5. 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}.