Basically, geometric slopes involve three numbers that are in geometric progression — like
— but at the same time are slopes of the sides of a triangle. Note that this is not a standard terminology; we’re using it here to keep the title short.
We present two “propositions” that describe precisely when the slopes of the sides of a triangle form a geometric progression.
Notation and characterization
In , let’s denote the slopes of the three medians from vertices
by
, respectively. Similarly, the slopes of sides
will be denoted by
.
If , PROVE that
, where
has vertices at
.
is the slope of the line segment joining
to the midpoint of
;
is the slope of side
. So we have:
Now, if , write
. Clear fractions:
Expand and combine like terms:
Obtain , from which
can be isolated as:
Example 2 (First proposition)
For with vertices at
and non-zero side slopes, PROVE that the slopes of sides
form a geometric progression if, and only if,
.
In other words, once one of the medians has a slope that is the negative of the slope of the side that contains its “foot”, then the slopes of the sides of the triangle form a geometric progression.
The non-zero requirement for the slopes of the sides is crucial. See Example 3 below. Also, click here for a pdf version of some of the equations in this post.
Suppose that ; we’ll show that the slopes of sides
(namely
) form a geometric progression. It suffices to show that
.
Since , we can use
from Example 1 and obtain
:
, and so the slopes
form a geometric progression. The converse was proved in Example 8 of our post on January 28, 2020.
Consider the right triangle with vertices at
. PROVE that the slope of the median from vertex
is the negative of the slope of the hypotenuse
.
Using the given coordinates and the mid-point formula, the mid-point of the hypotenuse is
and so the slope of the median from vertex is
Now, the slope of the hypotenuse is
which is the negative of the slope of the median from vertex .


Example 4 (Second proposition)
In , PROVE that
if, and only if,
, where
is the length of the median from vertex
and the triangle’s vertices are located at
.
Consequently, we obtain a second characterization of slopes in geometric progression.
First suppose that . Since
.
Now, the length of the median from vertex
is the distance from
to the midpoint of
:
A little care should be taken pertaining the sign of , so its absolute value should be used instead.
Conversely, suppose that . Usually,
, in view of the given coordinates. Together with the assumption of the converse, this amounts to:
Extracting square roots, we get . Since the slope of a median cannot be equal to the slope of the side it meets, we must then take the negative square root, and so
. In terms of our notation, this translates to
, or
.


(that is, the slope of the median from vertex
is the negative of the slope of side
);
the (non-zero) slopes of sides form a geometric progression;
the length of the median from vertex is
.
Applying the first proposition
Our first application goes to the right triangle considered in Example 3; it will be the basis of a future post — where we aim to unveil a hidden characteristic in every triangle.
PROVE that the right triangle with vertices at
contains a “sub-triangle” whose side slopes form a geometric progression with a common ratio of
.
In view of Example 3, we know that the median from vertex has a slope that’s the negative of the slope of the hypotenuse
. Consider the midpoint of median from
to
, located at
, say.
The triangle with vertices is the desired “sub-triangle”. Indeed, the slopes of its sides are:
This verifies that the slopes of the “sub-triangle” form a geometric progression with a common ratio of
.
PROVE that if the slopes of the sides of a triangle form a geometric progression (with common ratio ), then so do the slopes of the three medians, provided
(or
).
In Example 7 and Example 8 below, we explain what underlies the restriction .
If has sides
with slopes
in geometric progression, we’ll prove that the median slopes
form a new geometric progression.
Let the coordinates of be
. We have
and
From our first proposition, being in geometric progression means that
, and so
. That is,
. From Example 1,
.
So, the median slopes form a geometric progression, with
as the geometric mean.
Suppose that the slopes of the sides of a triangle form a geometric progression with a common ratio of . PROVE that the triangle contains a horizontal median.
We’ve given a proof of this before by explicitly determining the coordinates of the vertices in terms of the common ratio ; here we present a direct approach that avoids row reduction (linear algebra).
Assume the geometric progression of the slopes is , where
are the triangle’s vertices. Since the common ratio is
, we have:
Isolate from the first equation:
Also, from the second equation, isolating gives
Equate the two expressions obtained for :
Thus, the slope of the median from vertex is zero, that is,
, giving a horizontal median.
Suppose that the slopes of the sides of a triangle form a geometric progression with a common ratio of . PROVE that the triangle contains a vertical median.
As in Example 7, the fact the geometric progression has a common ratio of means we can write:
We obtain , from which
Also, from , isolating
gives
Equate the two expressions obtained for :
Therefore, the slope of the median from vertex , given by
, is undefined, so this median is vertical.
Suppose that is isosceles and that the slope of the median from the “apex”
is
. PROVE that the slopes of its sides form a geometric progression.
This is based on the fact that the slope of the median from the “apex” of an isosceles triangle is the negative reciprocal of the slope of the base. In particular, if , then
, giving
. By our first proposition, this means the slopes of sides
form a geometric progression. Similarly, if
, then
, also giving
, and we again obtain a geometric progression for the slopes of sides
.
A numerical problem
has vertices at
. Determine suitable values of
and
for which the slopes of sides
form a geometric progression.
Using our first proposition, we can tackle this problem by somehow “forcing” a median from side to have a slope that’s the negative of side
‘s slope. (Note that we can also use Exercise 1 below, which may be easier.)
Let’s find the equation of this “forced” median. The midpoint of is
, and its slope is
. So we want a median through
with slope
. Its equation is:
Thus, any point on the line — except the trio
— will do for the coordinates of vertex
. Why exclude these three? First, the point
is the midpoint of
, so we can’t form a triangle with it. Next, the point
shares the same
-coordinate with vertex
, which will make the slope of
zero if
is chosen as the point
. Similarly, we can’t take
for
because then
will have the same
-coordinate with vertex
, resulting in an undefined slope.
With the exclusions out of the way, we can take any other point, say , as
. Then
has vertices at
, and the slopes of sides
become
respectively. They form a geometric progression with a common ratio of (or
if we reverse the numbers).
A few other points on the line that can be chosen in place of
— and the corresponding geometric progressions for the slopes of the sides — are listed below:
Observe that the choice yields the same set of geometric progression as our initial choice
.
As in, this whole thing is COOL!
Takeaway
In the domain of triangles, the GOAT (Greatest Of All Triangles) is the triangle whose slopes form a geometric progression. No jokes.
That told, our first proposition is the basis for our future application. In particular, what we did for the right triangle in Example 5 will be done for any triangle. As in, A-N-Y!
Tasks
- (Generating formula) Let
be the coordinates of the vertices of
. Require that
and that
. PROVE that the slopes of sides
form a geometric progression if, and only if,
Thus, in order to obtain coordinates for the vertices of a triangle whose slopes form a geometric progression, begin by selecting “any” five numbers — ensuring that no two
s are equal and no two
s are equal — then use the above relation to determine the value of the remaining coordinate.
- (Special ratio) Let
be the coordinates of the vertices of
. PROVE that:
if, and only if,
(that is, the medians through
and
are perpendicular)
- if the slopes
form a geometric progression, then
if, and only if,
(that is, the median through
is vertical)
- if the slopes
form a geometric progression and
, then the median through
is horizontal
- if the slopes
form a geometric progression and
, then the common ratio of the geometric progression is
(If the slopes of the sides of a triangle form a geometric progression, no other value for the common ratio yields the relation; only
does.)
- (Half right) Let
be a triangle in which the slopes
form a geometric progression. PROVE that the acute angle between the median from
and side
is
if, and only if,
or
.
- (Pseudo perpendicularity) Suppose that the slopes
of the sides
of
form a geometric progression with common ratio
. Let the lengths of sides
be
, respectively. PROVE that the following pseudo-pythagorean property holds:
.
(If we let, then we obtain a full pythagorean relationship. However,
is unacceptable for a geometric progression. Note that this doesn’t imply the impossibility of having right triangles with slopes in geometric progression; there are actually many of them — with an appropriately chosen first term and a negative common ratio. Now, if we let
, we get
, a peach.)
- Suppose that the slopes of a right triangle form a geometric progression. PROVE that the slope of the hypotenuse cannot be the geometric mean of the progression.
(For a change, attempt this exercise without recourse to perpendicular slopes. Use our second proposition (Example 4) and the fact that the length of the median to the hypotenuse is half of the length of the hypotenuse.) - Let
be the coordinates of the vertices of an equilateral
. If the slopes of sides
form a geometric progression, prove that
, or
(Note the order of the slopes.) has vertices at
. Find three different pairs
for which the slopes of the sides form a geometric progression with a common ratio of
.
- (Three possibilities) Let
be the coordinates of the vertices of a triangle. If the slopes of the sides form a geometric progression, PROVE that at least one of these three is true:
;
;
.
(Use the generating formula.)
- PROVE that every right triangle
with legs
and
on — or parallel to — the coordinate axes contains a “sub-triangle”
that has the following property: the length of the median from vertex
is one-fourth the length of the hypotenuse
; that is,
.
(For instance, consider the “sub-triangle” we constructed in Example 5.) - Find suitable choice of coordinates for the vertices of
that has the property that
, where
is the length of the median from vertex
and
is the length of side
.
(Consider the case where the slopes of sidesform a geometric progression with a common ratio of
.)