Redirecting right bisectors

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Among the four classical triangle centers — centroid, circumcenter, incenter, orthocenter — only the circumcenter is usually constructed from lines that do not originate from any of the vertices of the parent triangle. It turns out that we can actually alter this tradition, and that’s what we’re after in today’s discussion. We’ll construct lines that originate from each vertex of the parent triangle whose point of concurrence is the circumcenter.

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Radial slopes

In \triangle ABC, let m_1,m_2,m_3 be the slopes of sides AB,BC,CA in that order. Let O be the circumcenter. Then the slope of radius OB can be given by:

(1)   \begin{equation*} m_{OB}=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)} \end{equation*}

Note the position of m_3, the slope of the side opposite vertex B. It’s the key to the “cyclic permutations” that yield similar expressions for the slopes of the radii through vertices A and C:

(2)   \begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)} \end{split} \end{equation*}

Easy to remember as they all radiate simplicity.

Consider \triangle ABC with vertices at A(3,12), B(0,0), C(6,6). Its circumcenter is O\left(-\frac{1}{2},\frac{13}{2}\right). Verify that the slopes of the radii OA, OB, OC satisfy equations (1,2).

Using the given coordinates:

  • the slope of OA is \frac{12-13/2}{3+1/2}=\frac{11}{7}
  • the slope of OB is \frac{0-13/2}{0+1/2}=-13
  • the slope of OC is \frac{6-13/2}{6+1/2}=-\frac{1}{13}

In order to use equations (1,2), write the side-slopes as m_1=4, m_2=1, m_3=-2 for AB, BC, CA.

    \begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ &=\frac{(1-(-2)\times 4)+(-2+4)\times 1}{(1-(-2)\times 4)\times 1-(-2+4)}\\ &=\frac{11}{7}\\ m_{OB}&=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)}\\ &=\frac{(1-4\times 1)+(4+1)\times(-2)}{(1-4\times 1)\times(-2)-(4+1)}\\ &=-13\\ m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{(1-1\times (-2))+(1+(-2))\times 4}{(1-1\times (-2))\times 4-(1+(-2))}\\ &=-\frac{1}{13} \end{split} \end{equation*}

We obtain the same results.

Suppose that the slopes of the sides of a triangle are 1,2,3. PROVE that the circumcenter shares the same y-coordinate with one of the vertices.

As easy as 1,2,3. Let \triangle ABC be such that the slopes of sides AB,BC,CA are m_1=1,m_2=2,m_3=3, respectively. By the second equation in (2), the slope of radius OC is:

    \begin{equation*} \begin{split} m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{(1-2\times 3)+(2+3)(1)}{(1-2\times 3)(1)-(2+3))}\\ &=0 \end{split} \end{equation*}

Thus, vertex C shares the same y-coordinate as the circumcenter. C for common. C for cool.

Similarly, if the slopes of the sides of a triangle are -1,-2,-3, then the circumcenter shares the same y-coordinate with one of the vertices. See the exercises for a more general pattern, and see below for a sample picture:

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Notice the horizontal radius OC, as well as our favourite point W on the nine-point circle. W shares the same x-coordinate with the nine-point center N in this case.

Suppose that the slopes of the sides of a triangle are -1,2,3. PROVE that the circumcenter shares the same x-coordinate with one of the vertices.

Let \triangle ABC be such that the slopes of sides AB,BC,CA are m_1=-1, m_2=2, m_3=3, respectively. By the second equation in (2), the slope of radius OC is:

    \begin{equation*} \begin{split} m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{(1-2\times 3)+(2+3)(-1)}{(1-2\times 3)(-1)-(2+3))}\\ &=\frac{-10}{0} \end{split} \end{equation*}

Thus, vertex C has the same x-coordinate as the circumcenter. Here C stands for cauton — d0n’t divide by zer0.

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Notice the vertical radius OC. Notice also the point W on the nine-point circle; it has the same y-coordinate as the nine-point center N in this case.

Suppose that the slopes of the sides of a triangle are -1, -2, 3. PROVE that the circumcenter shares an x-coordinate with a vertex and also shares a y-coordinate with another vertex.

Pick. This. One. Out.

Put m_1=-1, m_2=-2, and m_3=3 for the slopes of sides AB, BC, CA. Consider the slopes of radii OA and OB:

    \begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ &=\frac{(1+3)+(2)(-2)}{(1+3)(-2)-(-2)}\\ &=0\\ m_{OB}&=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)}\\ &=\frac{(1-2)+(-3)(3)}{(1-2)(3)-(-3)}\\ &=\frac{-10}{0} \end{split} \end{equation*}

This confirms that radius OA is horizontal, while radius OB is vertical. See the exercises for a more general pattern.

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The point W lies on the right bisector of AB in this case. Can you see why we named W the point that Wanders here and there on the nine-point circle?

\triangle ABC has vertices at A(0,6), B(0,0), C(3,0). Find the slopes of the three radii through vertices A,B,C.

It’s essential to consider an instance in which one of the slopes of the parent triangle is not well-behaved, as is the case here. We have m_1=\infty, m_2=0, m_3=-2. Notice the slight manoeuvres in the calculations:

    \begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ &=\frac{1-m_1m_3}{-m_1-m_3}\\ &=\frac{\frac{1}{m_1}-m_3}{-1-\frac{m_3}{m_1}}\\ &=\frac{0-m_3}{-1-0}\\ &=m_3=-2\\ m_{OB}&=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)}\\ &=\frac{1+m_1m_3}{m_3-m_1}\\ &=-m_3=2\\ m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{1+m_1m_3}{m_1-m_3}\\ &=m_3=-2 \end{split} \end{equation*}

And we obtain a characterization of right triangles with legs parallel to the coordinate axes: a right triangle has legs parallel to the coordinate axes if and only if two of its radii have opposite slopes.

If the slopes of the parent triangle form a geometric progression of the form \frac{1}{r},\pm 1, r (where r\neq 0,\pm 1), PROVE that the slopes of the radii form another geometric progression of the form \frac{1}{R},\mp 1, R.

It’s really, really hard to overlook geometric progressions. Let m_1=\frac{1}{r}, m_2=1, m_3=r be the side-slopes. Then:

  • slope of radius OA is: \frac{(1-1)+\left(r+\frac{1}{r}\right)(1)}{(1-1)(1)-\left(r+\frac{1}{r}\right)}=-1
  • slope of radius OB is: \frac{\left(1-\frac{1}{r}\right)+\left(\frac{1}{r}+1\right)(r)}{\left(1-\frac{1}{r}\right)(r)-\left(\frac{1}{r}+1\right)}=\frac{r^2+2r-1}{r^2-2r-1}
  • slope of radius OC is: \frac{(1-r)+\left(1+r\right)\left(\frac{1}{r}\right)}{(1-r)\left(\frac{1}{r}\right)-(1+r)}=\frac{r^2-2r-1}{r^2+2r-1}

So we obtain a geometric progression of the form \frac{1}{R},-1,R, where R=\frac{r^2+2r-1}{r^2-2r-1} or R=\frac{r^2-2r-1}{r^2+2r-1}. We must impose another restriction on r, namely:

    \[r^2+2r-1\neq 0\implies r\neq -1\pm\sqrt{2},\quad\textrm{and}\scriptstyle\quad r^2-2r-1\neq 0\implies r\neq 1\pm\sqrt{2}\]

The new common ratio -R can equal the original common ratio r when

    \[-\left(\frac{r^2+2r-1}{r^2-2r-1}\right)=r\scriptstyle\implies r^3+3r^2-3r-1=0\implies (r-1)(r^2+4r+1)=0\]

Whenever that special quadratic r^2+4r+1 is sighted, get excited: it signals the presence of an equilateral triangle with slopes in geometric progression. An equilateral triangle. Slopes in geometric progression. Pick them out.

If the slopes of the sides of an equilateral triangle form a geometric progression, then so do the slopes of the three radii from the vertices. Furthermore, the two three-term geometric progressions have the same common ratio (but are different because of the \pm 1 in the middle terms).

Radical strategy

Not quite radical per se, but still fundamentally different from — and eventually consistent with — the traditional method for finding the circumcenter. Our technique is merely intended to demonstrate what’s possible in principle; for practical computational purposes it’s inconvenent, and so not recommended.

Determine the circumcenter of \triangle ABC having vertices at A(2,6), B(0,0), C(8,0) by finding the point of concurrence of the three lines through A, B, C having slopes given by equations (1,2).

Here m_1=3, m_2=0, and m_3=-1 as per the slopes of AB, BC, CA. Using equations (1,2), we obtain the slopes of the three lines (not proper to call them radii yet):

    \[m_{OA}=-2,~m_{OB}=\frac{1}{2},~m_{OC}=-\frac{1}{2}\]

and hence the equations of the three lines:

(3)   \begin{equation*} \begin{split} \textrm{line through }A:~  y-6&=-2(x-2)\\ y&=-2x+10\\ \textrm{line through }B:~y-0&=\frac{1}{2}\left(x-0\right)\\ y&=\frac{1}{2}x\\ \textrm{line through }C:~y-0&=-\frac{1}{2}\left(x-8\right)\\ y&=-\frac{1}{2}x+4 \end{split} \end{equation*}

Consider the equations of the lines through B and C:

    \[\frac{1}{2}x=-\frac{1}{2}x+4\implies x=4\implies y=2\]

The pair (4,2) also satisfies the equation of the line through A:

    \[y=-2x+10,\quad 2=-2\times 4+10\]

Thus, the three equations are consistent and the circumcenter is therefore located at (4,2).

Don’t use ths method in real life.

Determine the circumcenter of \triangle ABC having vertices at A(2,6), B(-2,4), C(0,0) by finding the point of concurrence of the three radii OA, OB, OC.

Here we’re jumping the gun by calling radii, but don’t mind. We have m_1=\frac{1}{2}, m_2=-2, and m_3=3 as per the slopes of AB, BC, CA. Using equations (1,2), we obtain the slopes of the three radii:

    \[m_{OA}=3,~m_{OB}=-\frac{1}{3},~m_{OC}=3\]

and hence the equations of the three radii:

(4)   \begin{equation*} \begin{split} \textrm{radius OA: }  y-6&=3(x-2)\\ y&=3x\\ \textrm{radius OB: }y-4&=-\frac{1}{3}\left(x--2\right)\\ y&=-\frac{1}{3}x+\frac{10}{3}\\ \textrm{radius OC: }y-0&=3\left(x-0\right)\\ y&=3x \end{split} \end{equation*}

Consider the equations of radii OB and OC:

    \[3x=-\frac{1}{3}x+\frac{10}{3}\implies 9x=-x+10\implies x=1\implies y=3\]

The pair (1,3) also satisfies the equation of radius OA:

    \[y=3x,\quad 3=1\times 3\]

Thus, the three equations are consistent and the circumcenter is therefore located at (1,3).

Notice that (1,3) is the midpoint of AC. So the parent triangle is right-angled.

Right scenario

Today’s discussion bodes well for right triangles.

Suppose that m_1m_2=-1. PROVE that the slopes of the radii OA and OC are equal.

Let m_1m_2=-1. According to equation (2):

    \begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ &=\frac{1-m_3m_1+m_2m_3+m_1m_2}{m_2-m_1m_2m_3-m_1-m_3}\\ &=\frac{1-m_3m_1+m_2m_3-1}{m_2+m_3-m_1-m_3}\\ &=\frac{m_3(m_2-m_1)}{m_2-m_1}\\ &=m_3\\ m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{1-m_2m_3+m_1m_2+m_1m_3}{m_1-m_1m_2m_3-m_2-m_3}\\ &=\frac{1-m_2m_3-1+m_1m_3}{m_1+m_3-m_2-m_3}\\ &=\frac{m_3(m_1-m_2)}{m_1-m_2}\\ &=m_3 \end{split} \end{equation*}

Can you see how this verifies that the circumcenter of a right triangle must lie on the hypotenuse? The slope way.

We encountered a right triangle in example 5.

PROVE that if the slopes of two radii are equal, then the triangle contains 90^{\circ}.

Suppose that the slope of radius OA is equal to the slope of radius OB. By equations (1,2):

    \begin{equation*} \begin{split} \frac{(1-m_1m_3)+(m_1+m_3)m_2}{(1-m_1m_3)m_2-(m_1+m_3)}&=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)}\\ &\vdots\cdots\vdots\ddots\\ 2(m_1^2+1)(m_3-m_2)(m_2m_3+1)&=0\\ \implies m_2m_3+1&=0 \end{split} \end{equation*}

A couple of steps were skipped, but you get the gist.

Takeaway

The three statements below are equivalent for any triangle:

  • the triangle is a right triangle
  • the slopes of two radii are equal
  • one radius and one side have the same slope.

Tasks

  1. Find a triangle ABC in which the slopes of its three radii are equal to the negatives of the slopes of its three sides.
  2. Give an example of a right triangle having one radius horizontal and another radius vertical.
  3. Let \triangle ABC be an arbitrary triangle with circumcenter O.
    • If side BC has slope m=1, PROVE that the slopes of the radii OB and OC are reciprocals of each other.
    • If side BC has slope m=0, PROVE that the slopes of the radii OB and OC are negatives of each other.
  4. In \triangle ABC, Let m_1,m_2,m_3 be the slopes of sides AB,BC,CA, and let the circumcenter be O.
    • PROVE that there’s a horizontal radius if m_3=\frac{m_1m_2-1}{m_1+m_2}. Which of the radii OA, OB, OC will it be?
    • PROVE that there’s a vertical radius if m_3=\frac{m_1+m_2}{1-m_1m_2}. Which of the radii OA,OB,OC will it be?
    • Deduce that a circumcenter cannot coincide with a vertex, unless the triangle is a single point.
  5. (Pattern recognition) Suppose that the set \{m_1,m_2,m_3\} of slopes produce a horizontal radius, in light of the previous exercise. PROVE that each of the following sets of slopes also produce a horizontal radius:
    • \{-m_1,-m_2,-m_3\}
    • \left\{-\frac{1}{m_1},-\frac{1}{m_2},m_3\right\}
    • \left\{\frac{1}{m_1},\frac{1}{m_2},-m_3\right\}
  6. (Perpendicular radii) Suppose that radii OA and OB are perpendicular. As usual, let m_1, m_2, m_3 be the slopes of sides AB, BC, CA.
    • PROVE that \left(1+m_1^2\right)\Big(\left(1+m_2m_3\right)^2-\left(m_2-m_3\right)^2\Big)=0
    • Deduce that m_2=\frac{1+m_3}{1-m_3} or m_2=\frac{m_3-1}{m_3+1}
    • If m_1=\pm 1, deduce that one of the radii OA or OB is horizontal and the other vertical.
  7. (Principal result) Here’s how to obtain the main result of our discussion. In \triangle ABC, let m_1,m_2,m_3 be the slopes of sides AB,BC,CA, and let the circumcenter be O. Suppose that \angle OBC=\alpha.
    • If O is inside the parent triangle, PROVE that \angle BAC is \frac{\pi}{2}-\alpha
    • If O is outside the parent triangle, PROVE that \angle BAC is \frac{\pi}{2}+\alpha
    • Derive equation (1).
  8. In \triangle ABC, Let m_1,m_2,m_3 be the slopes of sides AB,BC,CA.
    • Suppose that m_1+m_2=0. PROVE that the slope of the radius OB is \frac{1}{m_3}. In particular, if m_3=0, deduce that the diameter through vertex B is vertical.
    • Suppose that m_1m_2=1. PROVE that the slope of the radius OB is -m_3. In particular, if m_3=0, deduce that the diameter through vertex B is horizontal.
  9. PROVE that any triangle with side-slopes -3,2,7 contains:
    • a horizontal radius
    • a vertical median. (A vertical median always shows up when the side-slopes form an arithmetic progression, like -3,2,7.)
  10. PROVE a result similar to that of exercise 5 above in the case of vertical radius.

A variant of Kosnita’s theorem

Let G be the centroid of \triangle ABC. Denote by G_a,G_b,G_c the centroids of triangles BCG,CAG,ABG, respectively. Then the lines AG_a,BG_b,CG_c are concurrent at G, the centroid of the parent triangle.

Further, let O be the circumcenter of \triangle ABC. Denote by O_a,O_b,O_c the circumcenters of triangles BCO,CAO,ABO, in that order. According to Kosnita’s theorem, the lines AO_a,BO_b,CO_c are concurrent at a point called the Kosnita point.

Now let W be that point that wanders here and there on the nine-point circle of \triangle ABC (coordinates given by (1) or (2) or (3) or (4)). Denote by W_a,W_b,W_c the “W-centers” of triangles BCW,CAW,ABW respectively. In this post we show that the lines AW_a,BW_b,CW_c are concurrent, if the slopes of the parent triangle form a geometric progression.

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(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*}

(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*}

(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*}

(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*}

A(x_1,y_1),~B(x_2,y_2),C(x_3,y_3) are the coordinates of the vertices, while m_1,m_2,m_3 are the slopes of sides AB,BC,CA. Equations (2),(3),(4) are all equivalent (to (1)) representations of point W.

Concurrence

Throughout, W will denote the “W-center” of \triangle ABC, while W_a,W_b,W_c represent the “W-centers” of triangles BCW,CAW,ABW.

Express the slopes of AW_c and WW_c in terms of the slopes of the parent triangle ABC.

Let the slopes of AB,BC,CA be m_1,m_2,m_3 in that order. Since W_c is constructed from \triangle ABW, the slope of AW_c is the product of the slopes of AB and AW, divided by the negative of the slope of BW (the way it works in general, as shown in this post, is as follows: take the product of the slopes of the two sides that originate from the reference vertex and then divide by the negative of the slope of the opposite side).

(5)   \begin{equation*} \begin{split} \textrm{slope of}~AW_c&=\frac{m_1\times\left(\frac{m_1m_3}{-m_2}\right)}{-\left(\frac{m_1m_2}{-m_3}\right)}\\ &=-m_1\left(\frac{m_3}{m_2}\right)^2 \end{split} \end{equation*}

Similarly, the slope of WW_c is the product of the slopes of AW and BW, divided by the negative of the slope of AB:

(6)   \begin{equation*} \begin{split} \textrm{slope of}~WW_c&=\frac{\left(\frac{m_1m_3}{-m_2}\right)\times\left(\frac{m_1m_2}{-m_3}\right)}{-m_1}\\ &=-m_1\\ \end{split} \end{equation*}

Express the slopes of AW_b and WW_b in terms of the slopes of the parent triangle ABC.

As before, let m_1,m_2,m_3 be the slopes of sides AB,BC,CA respectively. Since W_b is constructed from \triangle CAW, the slope of AW_b is the product of the slopes of CA and AW, divided by the negative of the slope of CW:

(7)   \begin{equation*} \begin{split} \textrm{slope of}~AW_b&=\frac{m_3\times\left(\frac{m_1m_3}{-m_2}\right)}{-\left(\frac{m_2m_3}{-m_1}\right)}\\ &=-m_3\left(\frac{m_1}{m_2}\right)^2 \end{split} \end{equation*}

Similarly, the slope of WW_b is the product of the slopes of AW and CW, divided by the negative of the slope of CA:

(8)   \begin{equation*} \begin{split} \textrm{slope of}~WW_b&=\frac{\left(\frac{m_1m_3}{-m_2}\right)\times\left(\frac{m_2m_3}{-m_1}\right)}{-m_3}\\ &=-m_3\\ \end{split} \end{equation*}

Suppose that the slopes of sides AB,BC,CA are a,ar,ar^2. PROVE that the quadrilateral AW_cWW_b is a parallelogram.

Set m_1=a,m_2=ar,m_3=ar^2. From equation (5):

    \begin{equation*} \begin{split} \textrm{slope of}~AW_c&=-m_1\left(\frac{m_3}{m_2}\right)^2\\ &=-a\left(\frac{ar^2}{ar}\right)^2\\ &=-ar^2 \end{split} \end{equation*}

From equation (6):

    \begin{equation*} \begin{split} \textrm{slope of}~WW_c&=-m_1\\ &=-a \end{split} \end{equation*}

From equation (7):

    \begin{equation*} \begin{split} \textrm{slope of}~AW_b&=-m_3\left(\frac{m_1}{m_2}\right)^2\\ &=-ar^2\left(\frac{a}{ar}\right)^2\\ &=-a \end{split} \end{equation*}

From equation (8):

    \begin{equation*} \begin{split} \textrm{slope of}~WW_b&=-m_3\\ &=-ar^2 \end{split} \end{equation*}

Thus, AW_c is parallel to WW_b, and WW_c is parallel to AW_b. There the parallelogram AW_cWW_b goes.

PROVE that W and W_a both lie on the median through vertex A, if the slopes of sides AB,BC,CA are a,ar,ar^2 in that order.

We’ve already seen that W lies on the median through vertex A in example 6 here.

To see that W_a also lies on the median through vertex A, recall that W_a was constructed from \triangle BCW. As such, the slope of the line WW_a is the product of the slopes of BW and CW divided by the negative of the slope of BC:

    \begin{equation*} \begin{split} \textrm{slope of}~WW_a&=\frac{\left(\frac{m_1m_2}{-m_3}\right)\times\left(\frac{m_2m_3}{-m_1}\right)}{-m_2}\\ &=-m_2\\ &=-ar \end{split} \end{equation*}

The slope of the median through vertex A is -ar, the slope of AW is also -ar, and the slope of WW_a is again -ar. Thus, the points A,W,W_a are co-linear together with the midpoint of BC.

(Main goal)

Suppose that the slopes of sides AB,BC,CA form a geometric progression a,ar,ar^2. PROVE that the lines AW_a, BW_b, CW_c are concurrent.

First consider the segment W_bW_c. In the parallelogram constructed in example 3, W_bW_c and AW are diagonals, so their midpoints coincide. Next, draw the lines BW_b and CW_c and extend till they intersect at \overline{W}, say. We claim that the line AW_a also passes through \overline{W}. This follows because the midpoint of W_bW_c lies on the median through A — as do W and W_a — so the line AW_a will go through \overline{W} (in fact, it is a median — extended if necessary — in \triangle W_bW_c\overline{W}).

In any \triangle ABC, PROVE that \angle W_aWW_b=\angle C or \angle W_aWW_b=\pi-\angle C, \angle W_bWW_c=\angle A or \angle W_bWW_c=\pi-\angle A, \angle W_cWW_a=\angle B or \angle W_cWW_a=\pi-\angle B .

As usual, let the slopes be m_1,m_2,m_3 for sides AB,BC,CA. Consider \angle W_bWW_c for example. In equation (6) we saw that the slope of WW_c=-m_1, the negative of the slope of side AB. In equation (8) we saw that the slope of WW_b=-m_3, the negative of the slope of AC. Thus, the angle between WW_b and WW_c is either angle A itself, or its supplement.

Coordinates

In the case of triangles with slopes in geometric progression a,ar,ar^2, we’re able to explicitly determine the coordinates of the point of concurrence of the lines AW_a, BW_b, CW_c:

(9)   \begin{equation*} \begin{split} x&=x_3+\frac{(r^4-2r^3+r^2-1)(y_2-y_3)}{ar(r^5+r^4-3r^3+3r^2-r-1)}\\ y&=\frac{(r^5-r^3+2r^2-r)y_2+(r^4-2r^3+r^2-1)y_3}{r^5+r^4-3r^3+3r^2-r-1} \end{split} \end{equation*}

How does one identify x_3,y_2,y_3 from a given triangle? Well, easy: just arrange the slopes in such a way that they appear in the format a,ar,ar^2 for sides AB,BC,CA respectively. Then (x_3,y_3) are the coordinates of vertex C and y_2 is the y-coordinate of B. (Notice how the coefficients of y_2 and y_3 in the second equation in (9) add up to the denominator. So in a sense, they’re “weighted”.)

Given \triangle ABC with vertices at A(0,0), B(3,3), C(1.5,6), find the point where the lines AW_a, BW_b, CW_c concur.

The slopes of sides AB,BC,CA are 1,-2,4 in that order; they form a geometric progression of the form a,ar,ar^2. As specified in equation (9), we identify x_3=1.5, y_3=6, and y_2=3.

    \begin{equation*} \begin{split} x&=x_3+\frac{(r^4-2r^3+r^2-1)(y_2-y_3)}{ar(r^5+r^4-3r^3+3r^2-r-1)}\\ &=1.5+\frac{(16+16+4-1)(3-6)}{(-2)(-32+16+24+12+2-1)}\\ &=1.5+\frac{-105}{-42}\\ &=4\\ y&=\frac{(r^5-r^3+2r^2-r)y_2+(r^4-2r^3+r^2-1)y_3}{r^5+r^4-3r^3+3r^2-r-1}\\ &=\frac{(-32+8+8+2)(3)+(16+16+4-1)(6)}{-32+16+24+12+2-1}\\ &=\frac{-42+210}{21}\\ &=8 \end{split} \end{equation*}

The point of concurrence is at \overline{W}(4,8) as shown below:

Rendered by QuickLaTeX.com

Notice point W on the nine-point circle of the parent triangle ABC. Points W_c, W_a, W_b also lie on the nine-point circles of triangles ABW, BCW, CAW.

In the exercises we ask you to derive the following equation:

(10)   \begin{equation*} \scriptstyle AW^2+BW^2+CW^2=\frac{1}{2}\left(\frac{m_2^2+m_3^2}{(m_2-m_3)^2}AB^2+\frac{m_3^2+m_1^2}{(m_3-m_1)^2}BC^2+\frac{m_1^2+m_2^2}{(m_1-m_2)^2}CA^2\right) \end{equation*}

Don’t be intimidated by equation (10), especially with the slope terms m_1,m_2,m_3 that appear there; its derivation can be accomplished without coordinates. In fact, triangles ABW, BCW, CAW always have their slopes in geometric progressions — irrespective of what happens in the parent triangle ABC (exception: when one side of the parent triangle is parallel to the x or y axis) — and so a certain approximate pythagorean identity can be applied to derive equation (10).

In \triangle ABC, let a,ar,ar^2 be the slopes of sides AB,BC,CA. PROVE that AW^2+BW^2+CW^2=\left(\frac{r^4+r^2+1}{r^4-2r^2+1}\right)BC^2.

The side-lengths AB,BC,CA satisfy an approximate pythagorean identity:

    \[AB^2+CA^2=\frac{r^2+1}{(r+1)^2}BC^2\]

Using equation (10) with m_1=a, m_2=ar, and m_3=ar^2:

    \begin{equation*} \begin{split} AW^2+BW^2+CW^2&=\scriptstyle \frac{1}{2}\left(\frac{m_2^2+m_3^2}{(m_2-m_3)^2}AB^2+\frac{m_3^2+m_1^2}{(m_3-m_1)^2}BC^2+\frac{m_1^2+m_2^2}{(m_1-m_2)^2}CA^2\right)\\ &=\scriptstyle \frac{1}{2}\left(\frac{(ar^2)^2+(ar)^2}{(ar^2-ar)^2}AB^2+\frac{(ar^2)^2+a^2}{(ar^2-a)^2}BC^2+\frac{a^2+(ar)^2}{(a-ar)^2}CA^2\right)\\ &=\scriptstyle\frac{1}{2}\left(\frac{r^2+1}{(r-1)^2}AB^2+\frac{r^4+1}{(r^2-1)^2}BC^2+\frac{1+r^2}{(1-r)^2}CA^2\right)\\ &=\frac{1}{2}\left(\frac{r^2+1}{(r-1)^2}(AB^2+CA^2)+\frac{r^4+1}{(r^2-1)^2}BC^2\right)\\ &=\frac{1}{2}\left(\frac{r^2+1}{(r-1)^2}\frac{r^2+1}{(r+1)^2}BC^2+\frac{r^4+1}{(r^2-1)^2}BC^2\right)\\ &=\frac{1}{2}\left(\frac{(r^2+1)^2+r^4+1}{(r^2-1)^2}\right)BC^2\\ &=\frac{1}{2}\left(\frac{2(r^4+r^2+1)}{(r^2-1)^2}\right)BC^2\\ &=\frac{r^4+r^2+1}{r^4-2r^2+1}BC^2 \end{split} \end{equation*}

In \triangle ABC, let a,ar,ar^2 be the slopes of sides AB,BC,CA. PROVE that AW=\left|\frac{r}{r^2-1}\right| BC.

We had, from the preceding example, that:

    \[AW^2+BW^2+CW^2=\left(\frac{r^4+r^2+1}{r^4-2r^2+1}\right)BC^2\]

In a previous post we asked you to prove that

    \[BW^2+CW^2=(r^2+r^{-2})AW^2\]

Combine these two equations:

    \begin{equation*} \begin{split} AW^2+(r^2+r^{-2})AW^2&=\left(\frac{r^4+r^2+1}{r^4-2r^2+1}\right)BC^2\\ \left(\frac{r^4+r^2+1}{r^2}\right)AW^2&=\left(\frac{r^4+r^2+1}{r^4-2r^2+1}\right)BC^2\\ \therefore AW^2&=\frac{r^2}{(r^2-1)^2}BC^2\\ AW&=\left|\frac{r}{r^2-1}\right| BC \end{split} \end{equation*}

Can you guess a value of r for which AW=BC? There it goes — it’s that golden ratio thing.

Coincidence

And a cauton.

Consider \triangle ABC with vertices at A(2.5,10), B(0,0), C(3,6). PROVE that the lines AW_a, BW_b, CW_c concur at (3,6).

Observe that the slopes of sides AB,BC,CA are 4,2,-8; they do not form a geometric progression (even when re-arranged as 2,4,-8). So this example suggests that there are other instances of concurrence of the lines AW_a, BW_b, CW_c beyond geometric progressions.

  • the “W-center” of the parent triangle is W(2,2) — obtained by solving the consistent equations y=x, y=4x-6, y=16x-30
  • the “W-center” of \triangle ABW is W_c\left(8/3,-2/3\rght) — obtained by solving the consistent system y=-\frac{1}{4}x, y=-4x+10, y=-64x+170. Using W_c\left(8/3,-2/3\rght) and C(3,6), we obtain the equation \boxed{y=20x-54} as the equation of line CW_c
  • the “W-center” of \triangle BCW is W_a\left(4,-2) — obtained by solving the consistent system y=-\frac{1}{2}x, y=-2x+6, y=-8x+30. Using W_a\left(4,-2) and A(2.5,10), we obtain the equation \boxed{y=-8x+30} for the line AW_a — incidentally, this is also the equation of line AC
  • the “W-center” of \triangle CAW is W_b\left(7/3,14/3\rght) — obtained by solving the consistent system y=2x, y=8x-14, y=32x-70. Using W_b\left(7/3,14/3\rght) and B(0,0), we obtain the equation \boxed{y=2x} — which, incidentally, is the equation of line BC. Together with the preceding incidence, we obtain a coincidence
  • the equations of lines AW_a, BW_b, CW_c are y=-8x+30, y=2x, y=20x-54. These three lines concur at (3,6) — coordinates of vertex C. C for coincidence. C for caution.

Takeaway

In \triangle ABC, let a,ar,ar^2 be the slopes of sides AB,BC,CA. Let W be that point on the nine-point circle whose coordinates are given by equation (1). Then the four statements below are equivalent:

  • AW^2+BW^2+CW^2=4BC^2
  • r^4-3r^2+1=0
  • r^2-r-1=0 or r^2+r+1=0
  • AW=BC

You can see the golden ratio popping up in the third statement above. That seemingly ubiquitous golden ratio thing is increasingly becoming conspicuous in our theory.

Tasks

  1. Find a triangle ABC and a point W on its nine-point circle such that AW=BC.
  2. In \triangle ABC, Let m_1,m_2,m_3 be the slopes of sides AB,BC,CA. Let W be the point whose coordinates were given in equation (1). PROVE that:
    • AW^2+BW^2=\frac{m_2^2+m_3^2}{(m_2-m_3)^2}AB^2
    • BW^2+CW^2=\frac{m_3^2+m_1^2}{(m_3-m_1)^2}BC^2
    • CW^2+AW^2=\frac{m_1^2+m_2^2}{(m_1-m_2)^2}CA^2
    • \scriptstyle AW^2+BW^2+CW^2=\frac{1}{2}\left(\frac{m_2^2+m_3^2}{(m_2-m_3)^2}AB^2+\frac{m_3^2+m_1^2}{(m_3-m_1)^2}BC^2+\frac{m_1^2+m_2^2}{(m_1-m_2)^2}CA^2\right)
  3. (Coordinates) Use equation (2) to prove that the “W-center” of \triangle ABW is \left(\frac{m_2^2x_2-m_3^2x_1}{m_2^2-m_3^2},\frac{m_2^2y_1-m_3^2y_2}{m_2^2-m_3^2}\right).
  4. (Congruence) In \triangle ABC, let A',B',C' be points that are diametrically opposite vertices A,B,C, respectively. PROVE that:
    • \triangle A'B'C' is congruent to \triangle ABC
    • the “W” center of \triangle ABC and the “W-center” of \triangle A'B'C' have their midpoint at the circumcenter of the parent \triangle ABC.
      (In general, it’s a similar scenario with the orthocenter and co — all due to a certain homothecy centered at the circumcenter \cdots.)
  5. (Coincidence) Let \triangle ABC be such that AB is parallel to the x-axis, and BC and CA have reciprocal slopes. PROVE that:
    • the Kosnita point coincides with vertex C
    • the point D diametrically opposite this Kosnita point forms an isosceles trapezoid ABCD in conjuction with the other points.
  6. (Concurrent coincidence) \triangle ABC has vertices at A(5,20), B(0,0), C(6,12). PROVE that:
    • W is the point (4,4)
    • W_a is the point (8,-4)
    • W_b is the point \left(\frac{14}{3},\frac{28}{3}\right)
    • W_c is the point \left(\frac{32}{15},-\frac{8}{15}\right)
    • the point of concurrence of the lines AW_a, BW_b, CW_c is (6,12).
  7. In any triangle ABC, PROVE that:
    • the slope of W_aW_b is the negative of the slope of CW
    • the slope of W_bW_c is the negative of the slope of AW
    • the slope of W_cW_a is the negative of the slope of BW
  8. Given a triangle ABC with side-slopes m_1,m_2,m_3, find a triangle with side-slopes \frac{m_1m_2}{m_3}, \frac{m_2m_3}{m_1}, \frac{m_3m_1}{m_2}.
    (Hint: Consider \triangle W_aW_bW_c.)
  9. If the slopes of sides AB,BC,CA form a geometric progression a,ar,ar^2, PROVE that W_bW_c is parallel to side BC.
  10. PROVE that \frac{r^4+r^2+1}{r^4-2r^2+1}=\frac{r^6-1}{\left(r^2-1\right)^3}.