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# Geometric sequence on a staircase

In case you’ve used a staircase before, you’ve “experienced” a special case of a geometric sequence as a consequence.

For a case study, consider the diagram below:

Join to , to , and to to form :

We’ll prove that the slopes of the sides of form a geometric sequence, if .

#### Example

A portion of a staircase is shown below:

Verify that the slopes of the sides of form a geometric sequence.

Easy-peasy.

The slopes of sides are , respectively. They form a geometric sequence with a common ratio .

(The fact that the slope of side is implies that it makes an angle of with the horizontal. In real life, however, the angle is usually around .)

#### Example

A portion of a staircase is shown below:

Verify that the slopes of the sides of form a geometric sequence.

Easy-peasy as before.

The slopes of sides are , respectively. They form a geometric sequence with a common ratio of .

(It shouldn’t come as a surprise that the common ratio is this time too. Indeed, it is almost always the case in a staircase, because the “rise” and “run” stay constant throughout the entire span of the staircase. For this reason, the geometric sequence obtained on a staircase is a special case.)

## Basic background

From Example 3 to Example 6 below, we lay the foundation for our staircase story. Don’t worry if it seems to “escalate” quite quickly; remember we’re on a staircase .

PROVE that if the slopes of the sides of a triangle form a geometric sequence with a positive common ratio, then the triangle contains an obtuse angle.

Actually, any triangle in which the three sides have positive slopes contains an obtuse angle (for now we’re content with slopes in geometric sequence).

Let be such that sides have slopes , respectively. If the common ratio is positive, then the angle at is obtuse. To see this, let the lengths of sides be , respectively. Then, as we’ll show in our next post, the following pseudo-pythagorean property holds:

Thus, we can use the cosine law to compute the cosine of the angle at :

The obtuse angle at is well-oriented in such a way that a special can be cut out from it.

In , let be the slopes of sides , respectively. Then a point divides in the ratio if and only if it shares the same -coordinate as vertex .

Thus, will be a vertical line. Take note of this.

The problem deals with coordinates, so let’s begin by specifying the coordinates of the vertices of the parent triangle . Assume are the vertices. Since the slopes of sides are , there are relationships among the coordinates:

(1)

Now, suppose a point divides side in the ratio , with . Using the formula for internal division of a line segment, we can write the -coordinate of point in terms of the -coordinates of and as follows:

Therefore, point shares the same -coordinate as vertex .

Conversely, suppose that point shares the same -coordinate as vertex . We proceed to find the -coordinate of . Since the slope of is and it goes through , its equation is . Using the fact that at point , we have:

Let’s manipulate slightly:

Thus, is the point . By examining its coordinates, we see that it actually divides side in the ratio .

In , let be the slopes of sides , respectively. Then a point divides in the ratio if and only if it shares the same -coordinate as vertex .

Thus, will be a horizontal line. Together with the previous example, the parent contains a special right triangle in which both legs and are parallel to the coordinate axes. Further, . The presence of this right triangle is key to our staircase story.

Let’s find the coordinate of point which divides in the ratio . Using equation (1) and the formula for internal division of a line segment:

Easier than thought. Conversely, suppose that shares the same -coordinate as vertex , so that . We’ll find its -coordinate, using the fact that it lies on side whose equation is :

These calculations show that is the point , and it divides side in the ratio if we inspect its coordinates.

In the above diagram, point divides in the ratio while point divides in the ratio .

#### (Main goal)

Let be a triangle. Then the following two statements are equivalent:

1. the slopes of the sides form a geometric sequence with positive common ratio;
2. there are internal points and on such that is vertical, is horizontal, and .

The implication follows from the three previous examples. For , consider the diagram below:

We’ll show that

(2)

Since , we have:

Because are co-linear, we have , and so . Using this in the previous equation, we have:

The latter means that either or . However, in our diagram we have and , thus ruling out the possibility of . We therefore settle for .

Returning to (2) and using some equality relations among the slopes of the co-linear points , we have:

This proves that the slopes of the sides of form a geometric progression. The common ratio is positive in view of the diagram.

In , let be the slopes of sides , respectively. For any positive integer , PROVE that there are points and on such that the slopes of the sides of form a geometric sequence .

Note that both Example 4 and Example 5 result from the special case , but the associated slopes do not form a geometric sequence. Also, need not be a positive integer; any positive number works as well.

We need to construct points and first.

Let be the point which divides side in the ratio . (The restriction that is essential to exclude the midpoint of .) Using equation (1) and the formula for internal division of a line segment, we can determine the coordinates of as follows:

So, is the point . Next, let’s construct point by dividing side in the ratio .

Thus, is the point . Together with equation (1), the vertices of are:

From these, we obtain as the slopes of sides , respectively. Note that yields as the slopes of — same slopes as those of the parent triangle .

The next example shows that if we exclude three points from , namely, the midpoint of , the point on that shares the same -coordinate as , and the point on that shares the same -coordinate as , then we obtain what may be regarded as a stronger version of Example 6.

For any , the following statements are equivalent:

1. the slopes of sides form a geometric sequence with positive common ratio;
2. for any internal point on , there is an internal point on such that and the slopes of form a geometric sequence.

Bear in mind the three excluded points on . Basically, these are the points where the slope of will be undefined, the slope of will be zero, and where . Naturally, these were excluded to ensure that we have proper slopes.

To show , let be the point which divides in the ratio , where is a positive integer. Choose as the point which divides in the ratio . Then . By Example 7, the slopes of the sides of form a geometric progression.

To show , take . Then has to be taken as , so that the equality is satisfied (both sides being zero).

PROVE that the two infinite geometric progressions

(3)

(4)

can be obtained from the same triangle.

You already know how to generate (4); if not, see here. So we focus on obtaining (3), following a similar procedure and using the same triangle that yields (4).

Begin with a triangle whose slopes are — like the one shown below:

where , , are the vertices.

For an integer , set . Then is an internal point on ; further, as and as . These limits ensure that always resides within side .

Next, let’s compute the slope of cevian for each :

Letting gives rise to the infinite geometric sequence in (3).

## That special right triangle

We conclude by isolating a unique property of the special right triangle that appears in staircases.

PROVE that a triangle is a right triangle with legs parallel to the coordinate axes if and only if the slopes of its three medians form a geometric sequence with common ratio .

First suppose that we have a right triangle with legs parallel to the coordinate axes. This direction of the proof was done in our previous post, but let’s go through it briefly. Fix the vertices at , , and ; a little modification will be needed for the general case.

Now calculate the slopes of medians :

These form a geometric progression with common ratio of .

Conversely, suppose that we have with vertices , , and in such a way that the slopes of the three medians form a geometric sequence with common ratio .

If we denote the slopes of the medians from vertices by respectively, then:

Suppose that the geometric sequence of median slopes is , then , and :

Clear fractions from the first equation above:

(5)

Now clear fractions from the second equation:

(6)

It turns out that can be eliminated if we perform the operation (5)-(6):

Since the slopes of the three medians form a geometric sequence, we can’t have . Otherwise this would result in an undefined slope for the median from vertex , namely , and a geometric sequence cannot contain such a term. So the only option is to take , whence . Our triangle now has one side parallel to the -axis.

Put in equation (5):

Since we already had , it’s impossible to have in addition, otherwise this would lead to , which is not allowed for a triangle. So we take , whence . Our triangle now has a side that is parallel to the -axis.

Therefore, has side parallel to the -axis, and side parallel to the -axis.

## Takeaway

We’ve used staircases in various cases and in various places — homes, airports, train stations, offices, etc. As we ascend our staircase, we “run” forward and “rise” higher, so our slopes are positive. In turn, the geometric sequence we describe has a positive common ratio — in most cases, .

Embellished with nice properties, triangles with slopes in geometric progression are established as objects to cherish and relish. And, they not only appeal in theory, they appear in real life too.

1. (Three out) Given with vertices , the slopes of the sides form a geometric sequence with common ratio . Find two points and on for which the slopes of the sides of do not form a geometric sequence.
(The third point is the midpoint of , namely . Apart from these three points, given any other point within , there is a corresponding point within for which the slopes of the sides of form a geometric sequence.)
2. Consider the diagram below:

in which and have the same -coordinates, and share the same -coordinates, and .

• If also , PROVE that the slopes of the sides of form a geometric sequence with a common ratio of ;
• Give an example to show that if and do not trisect , then the common ratio is no longer .
3. PROVE that if the slopes of the sides of a triangle are all positive (and non-zero), then the triangle contains an obtuse angle.
(Similarly, if the slopes are all negative, there’ll be an obtuse angle.)
4. In , let be the slopes of sides , respectively. If , PROVE that side is the longest side.
5. In , let be the slopes of sides , respectively. If , PROVE that side is NOT always the longest side.
6. PROVE that the two infinite geometric sequences and can be obtained from the same triangle, using slopes.
7. (Close enough) Let be a triangle in which sides have slopes , with and . Let and be points on which divide in the ratios and . PROVE that the area of is times the area of .
(In the limit , we have area of area of .)
8. Find coordinates for the vertices of a non-right triangle which contains a right triangle such that the area of is times the area of .
9. Let be such that the slopes of sides form the geometric sequence , with .
• Let be the point which divides in the ratio and let be the midpoint of and . PROVE that the slope of cevian is times the slope of .
• Let be the point which divides in the ratio and let be the midpoint of and . PROVE that the slope of cevian is times the slope of .
• Conclude that every triangle with slopes (where ) contains a sub-triangle with slopes , a geometric sequence with common ratio .
10. (Right inside) Let () be the slopes of sides in . PROVE that there are points and on for which .
11. Let () be the slopes of sides in . PROVE that there two points and on for which the slopes of form an arithmetic sequence with a common difference of , namely: .
12. Let () be the slopes of sides in . PROVE that there two points and on for which the slopes of form an arithmetic sequence with a common difference of , namely: .
13. Let , , be a right triangle in the first quadrant. Choose any vertex, say , and draw two cevians and to side .
• For a positive integer , let and divide side in the ratios and , respectively. PROVE that and have the same centroid
• Can this be extended to all triangles?
14. (Almost right) Let be the slopes of sides in . PROVE that there are points and on such that
(Compare this with exercise 10 above.)

## Thanksgiving

Today is a special day in this poster’s life. As such, he gives praise to the Owner of life.

Mind always goes back to that marked Thursday, June 14, 2018. In turn, it CAN`T hold back from giving thanks.

What happened was very basic (as simple as HTML), but its effect became “drastic” (actually, dramatic, dynamic), and so one’s gratitude is made public.