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:

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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}.