Let be the *centroid* of Denote by the centroids of triangles , respectively. Then the lines are concurrent at , the centroid of the parent triangle.

Further, let be the *circumcenter* of . Denote by the circumcenters of triangles , in that order. According to Kosnita’s theorem, the lines are concurrent at a point called the Kosnita point.

Now let be that point that wanders here and there on the *nine-point circle* of (coordinates given by (1) or (2) or (3) or (4)). Denote by the “-centers” of triangles respectively. In this post we show that the lines are concurrent, if the slopes of the parent triangle form a *geometric progression*.

(1)

(2)

(3)

(4)

are the coordinates of the vertices, while are the slopes of sides . Equations (2),(3),(4) are all equivalent (to (1)) representations of point .

## Concurrence

Throughout, will denote the “W-center” of , while represent the “W-centers” of triangles .

Let the slopes of be in that order. Since is constructed from , the slope of is the product of the slopes of and , divided by the __negative__ of the slope of (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)

Similarly, the slope of is the product of the slopes of and , divided by the negative of the slope of :

(6)

As before, let be the slopes of sides respectively. Since is constructed from , the slope of is the product of the slopes of and , divided by the __negative__ of the slope of :

(7)

Similarly, the slope of is the product of the slopes of and , divided by the negative of the slope of :

(8)

*parallelogram*.

Set . From equation (5):

From equation (6):

From equation (7):

From equation (8):

Thus, is parallel to , and is parallel to . There the parallelogram goes.

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

To see that also lies on the median through vertex , recall that was constructed from . As such, the slope of the line is the product of the slopes of and divided by the *negative* of the slope of :

The slope of the median through vertex is , the slope of is also , and the slope of is again . Thus, the points are co-linear together with the midpoint of .

#### (Main goal)

*geometric progression*PROVE that the lines , , are

*concurrent*.

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

As usual, let the slopes be for sides . Consider for example. In equation (6) we saw that the slope of , the negative of the slope of side . In equation (8) we saw that the slope of , the negative of the slope of . Thus, the angle between and is either angle itself, or its supplement.

## Coordinates

In the case of triangles with slopes in geometric progression , we’re able to explicitly determine the coordinates of the point of concurrence of the lines , , :

(9)

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

The slopes of sides are in that order; they form a geometric progression of the form . As specified in equation (9), we identify , , and .

The point of concurrence is at as shown below:

Notice point on the *nine-point circle* of the parent triangle . Points , , also lie on the *nine-point circles* of triangles , , .

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

(10)

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

The side-lengths satisfy an approximate pythagorean identity:

Using equation (10) with , , and :

We had, from the preceding example, that:

In a previous post we asked you to prove that

Combine these two equations:

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

## Coincidence

And a cauton.

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

- the “W-center” of the parent triangle is — obtained by solving the consistent equations , ,
- the “W-center” of is — obtained by solving the consistent system , , . Using and , we obtain the equation as the equation of line
- the “W-center” of is — obtained by solving the consistent system , , . Using and , we obtain the equation for the line — incidentally, this is also the equation of line
- the “W-center” of is — obtained by solving the consistent system , , . Using and , we obtain the equation — which, incidentally, is the equation of line . Together with the preceding incidence, we obtain a coincidence
- the equations of lines , , are , , . These three lines concur at — coordinates of vertex . C for coincidence. C for caution.

## Takeaway

In , let be the slopes of sides . Let be that point on the *nine-point circle* whose coordinates are given by equation (1). Then the four statements below are *equivalent*:

- or

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 and a point on its
*nine-point circle*such that . - In , Let be the slopes of sides . Let be the point whose coordinates were given in equation (1). PROVE that:
- (Coordinates) Use equation (2) to prove that the “W-center” of is
- (Congruence) In , let be points that are
*diametrically opposite*vertices , respectively. PROVE that:- is congruent to
- the “W” center of and the “W-center” of have their midpoint at the circumcenter of the parent

(In general, it’s a similar scenario with the orthocenter and co — all due to a certain homothecy centered at the circumcenter .)

- (Coincidence) Let be such that is parallel to the -axis, and and have
*reciprocal*slopes. PROVE that:- the
*Kosnita point*coincides with vertex - the point diametrically opposite this Kosnita point forms an isosceles trapezoid in conjuction with the other points.

- the
- (Concurrent coincidence) has vertices at , , . PROVE that:
- is the point
- is the point
- is the point
- is the point
- the point of concurrence of the lines , , is .

- In any triangle , PROVE that:
- the slope of is the negative of the slope of
- the slope of is the negative of the slope of
- the slope of is the negative of the slope of

- Given a triangle with side-slopes , find a triangle with side-slopes , , .

(Hint: Consider .) - If the slopes of sides form a geometric progression , PROVE that is parallel to side .
- PROVE that .