Look at the diagram above, where a number of things appear to be happening. The red circle that goes through points is our focus for now. The center
of this circle is itself a point on the nine-point circle of the parent
. We’ll examine more of the properties of
in this post, especially when the side-slopes of
form a geometric progression.
Simplified coordinates
If has vertices at
,
,
, then the coordinates of point
are given by:
(1)
Compare (1) with its equivalent, simplified version:
(2)
Here, are the slopes of sides
, respectively. Notice that
and the coordinates of
are explicitly “missing” in equation (2). One implication is that the calculation of
requires only two sides and the slopes of those two sides. So (2) can take another form:
(3)
(4)
Ample freedom. Simple concept.





Set ,
, and
. The slopes of sides
, and
are then
,
, and
, respectively.
Using equation (2):
Using equation (3):
Using equation (4):
In each case we obtain . It’s that point that
anders here and there on the nine-point circle.



The converse is also true. Fix the vertices at ,
,
. Let the slopes of
be
and suppose that
. Using equation (2):
Thus, is the point
, which is the foot of the altitude from vertex
.


The converse is also true. Fix the vertices at ,
,
. Let the slopes of
be
and suppose that
. Using equation (2):
Thus, is the point
, which is the midpoint of
.


This follows from example 3, since are opposites.
Special case
Most of our points behave extremely well if the side-slopes of the parent triangle form a geometric progression. Expectedly, our favorite point is not an exception.




Let the slopes of sides be
,
,
. From this post, the slopes from
to vertices
are:
(5)
Thus, correspond to the geometric progression
with common ratio
.






From the preceding example, the slope from to vertex
is
. From this post, the slope from vertex
to the midpoint of side
is also
. Thus,
lies on the median through vertex
.





As noted in example 6 above, the point is on the median through vertex
. Suppose that
lies outside the triangle (usually when
). Let
be the midpoint of side
. Then the area of
is equal to the area of
(since
is a median in
). But then the area of
is also equal to the area of
(since
is a median in
). Together we have that the area of
is equal to the area of
. Same conclusion when
is inside the parent triangle.
In the diagram above, point is inside the given triangle. The areas of triangles
and
are equal.
Sample calculations
Since lies on the median through vertex
, it is natural to ask what ratio both it and the centroid
divide the entire median length
.
(6)
As you already know, the relations in (6) hold when the slopes of the parent triangle form a geometric progression. Absolute values may be necessary in some cases.






The common ratio of the slopes
of an equilateral triangle satisfies the quadratic equation
(7)
Use equation (7) and the first ratio in equation (6) to obtain
Thus, and so
is the midpoint of
and
. In addition, since the centroid divides median
in the ratio
, it follows that
in this case.




Use the first ratio in equation (6) and the fact that the common ratio of the slopes of a right isosceles triangle satisfies the quadratic equation
(8)
and obtain
Thus, .




We have:
Thus, the line segment is longer than the median
. (If
, we can’t conclude that
lies inside the triangle.)
Takeaway
In , let
be the centroid,
the midpoint of side
, and
as given by equation (1). If the slopes of sides
form a geometric progression
then the following statements are equivalent:
, where
is the sum of quotients of slopes defined here
or
or
or
or
or
or
, where
is the golden ratio
or
or
or
.
Eight easy equivalences. A triangle that satisfies any of these (together with or
) will be a right isosceles triangle (or its “look-alike”).
Tasks
- Let
be such that
is parallel to the
-axis, and
and
have reciprocal slopes. PROVE that:
- the orthocenter
is a reflection of vertex
across the
-axis
is the foot of the altitude from vertex
is the midpoint of
and
- the orthocenter
- Let
be the slopes of the line segments from
to the vertices of
, and let
be the side-slopes.
- PROVE that
.
- If
is right-angled, deduce that the product
equals the slope of the hypotenuse.
- PROVE that
- PROVE that
for any
with side-slopes
.
- Find coordinates for the vertices of
having the property that a point
on its nine-point circle is twice as far from vertex
as it is from the centroid
.
- Suppose that the slopes of sides
are
, in that order. Let
be as given in equation (1), and let
be the midpoint of
. PROVE that
.