A variant of Kosnita’s theorem – High School Geometry

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.

(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:

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

Find a triangle $ABC$ and a point $W$ on its nine-point circle such that $AW=BC$ .
In , Let be the slopes of sides . Let be the point whose coordinates were given in equation (1). PROVE that:
- $AW^2+BW^2=\frac{m_2^2+m_3^2}{(m_2-m_3)^2}AB^2$
- $BW^2+CW^2=\frac{m_3^2+m_1^2}{(m_3-m_1)^2}BC^2$
- $CW^2+AW^2=\frac{m_1^2+m_2^2}{(m_1-m_2)^2}CA^2$
- $\scriptstyle AW^2+BW^2+CW^2=\frac{1}{2}\left(\frac{m_2^2+m_3^2}{(m_2-m_3)^2}AB^2+\frac{m_3^2+m_1^2}{(m_3-m_1)^2}BC^2+\frac{m_1^2+m_2^2}{(m_1-m_2)^2}CA^2\right)$
(Coordinates) Use equation (2) to prove that the “W-center” of $\triangle ABW$ is $\left(\frac{m_2^2x_2-m_3^2x_1}{m_2^2-m_3^2},\frac{m_2^2y_1-m_3^2y_2}{m_2^2-m_3^2}\right).$
(Congruence) In , let be points that are diametrically opposite vertices , 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 $\cdots$ .)
(Coincidence) Let be such that is parallel to the -axis, and and 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.
(Concurrent coincidence) has vertices at , , . 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 $(6,12)$ .
In any triangle , 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 $BW$
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 $\triangle W_aW_bW_c$ .)
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 $BC$ .
PROVE that $\frac{r^4+r^2+1}{r^4-2r^2+1}=\frac{r^6-1}{\left(r^2-1\right)^3}$ .