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:
- the slopes of the sides form a geometric sequence with positive common ratio;
- 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:
(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:
- the slopes of sides
form a geometric sequence with positive common ratio;
- 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.
Tasks
- (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.)
- 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
.
- If also
- 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.) - In
, let
be the slopes of sides
, respectively. If
, PROVE that side
is the longest side.
- In
, let
be the slopes of sides
, respectively. If
, PROVE that side
is NOT always the longest side.
- PROVE that the two infinite geometric sequences
and
can be obtained from the same triangle, using slopes.
- (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
.)
- 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
.
- 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
.
- Let
- (Right inside) Let
(
) be the slopes of sides
in
. PROVE that there are points
and
on
for which
.
- 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:
.
- 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:
.
- 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?
- For a positive integer
- (Almost right) Let
be the slopes of sides
in
. PROVE that there are points
and
on
such that
(Compare this with exercise 10 above.)
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 C–A–N–`–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. Hopefully this poster’s gratitude to the Heavenly Father doesn’t ruffle your feathers. If it does, please (don’t bother to) read further. May you find that “spark” you need to start what will lead you to succeed.
Thanksgiving
To the reader:
Wish: