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)



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)





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 Kosnita point coincides with vertex
- (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
- 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
.