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# Two propositions on geometric slopes

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 .

However, the slopes of the sides of , namely do not form a geometric progression. This doesn’t contradict the equivalent statements in our first proposition; it simply illustrates the fact that a hypothesis cannot be dismissed, else things can go amiss.

#### 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 .

Combining our first and second propositions, we obtain the following three equivalent statements for with vertices at :

(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!

1. (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.
2. (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.)
3. (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 .
4. (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.)
5. 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.)
6. 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.)
7. has vertices at . Find three different pairs for which the slopes of the sides form a geometric progression with a common ratio of .
8. (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.)
9. 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.)
10. 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 sides form a geometric progression with a common ratio of .)