As shown by our slipshod, tongue-twisting title, slopes can be very “brittle”.
We demonstrate how to convert the slopes of the sides of a triangle from a geometric sequence to an arithmetic sequence. Hopefully it makes sense.
Modified midpoint theorem
A line which connects the midpoints of two sides of a triangle is parallel to the third side and its length is half of the third side’s length. So says the midpoint theorem.
Since a midpoint amounts to dividing a line segment in the ratio , we’ll modify this slightly for our triangle and rather divide the sides in the
atio
, whe
e
.
In , let
and
be points on
and
which divide the sides in the ratio
, where
is a positive integer. PROVE that the line segment
is parallel to
and that
.
In the diagram above, and
. If we place the vertices of
at
,
, and
, then we can obtain the coordinates of points
and
using the formula for internal division of a line segment:
The slope of line segment is:
So is parallel to
. In addition we obtain
.
Easy-peasy.
Derailing development
If we apply the above idea to our own triangle, look what we get.
Suppose that has vertices
,
,
, and slopes
for sides
, respectively. For
, find coordinates for the point which divides
in the ratio
.
Note the “direction” of the division. The desired point will be , using the formula for internal division of a line segment. In view of the
elationship among the coordinates, this becomes
You see that quadratic rearing its beautiful head again?
Suppose that has vertices
,
,
, and slopes
for sides
, respectively. For
, find coordinates for the point which divides
in the ratio
.
Ordinarily, this would be the point . Using the
elationship among the coordinates, we get
Suppose that has vertices
,
,
, and slopes
for sides
, respectively. For
, find coordinates for the point which divides
in the ratio
.
This would be the point . The
elationship among the coordinates then gives
(Main goal)
Suppose that has vertices
,
,
, and slopes
for sides
, respectively. For
, PROVE that there is a point
on
, a point
on
, and a point
on
such that the slopes of the sides of
form an arithmetic progression.
Basically, the side-slopes form a geometric progression, while the sub-triangle’s side-slopes form an arithmetic progression. Fo. Any.
.
Let be the point which divides
in the ratio
, let
be the point which divides
in the ratio
, and let
be the point which divides
in the ratio
. As in the three preceding examples, these points are:
By the modified midpoint theorem, would be parallel to
, so its slope is
(alternatively can be confirmed by direct calculation). Similarly,
would be parallel to
, so its slope is
.
By direct calculation, the slope of is
.
Thus, the slopes of the sides of are
, an arithmetic progression.
Given with vertices at
,
, and
, find coordinates for points
on
such that the slopes of
are
.
Observe that the slopes of sides are
, respectively.
Thus, by Example 5, let be the point which divides
in the ratio
, let
be the point which divides
in the ratio
, and let
be the point which divides
in the ratio
. Lots of repetitions there
.
Using the formula for internal division of a line segment, we find these points to be ,
, and
. Then:
Given with vertices at
,
, and
, find coordinates for points
on
such that the slopes of
are
.
In this case, the slopes of sides are
, respectively. A geometric progression in which
.
By Example 5, let divide each of the three sides
in the ratio
.
Using the formula for internal division of a line segment, we find these points to be ,
, and
.
Given with vertices at
,
, and
, find coordinates for (external) points
on
such that the slopes of
are
.
Observe that the slopes of sides are
, respectively. They form a geometric progression in which
.
Let be points which divide
all in the ratio
. We obtain the external points
,
, and
. Then:
The scene here may be worth seeing
Notice that is the midpoint of both
and
, while
is the midpoint of
. It follows that the length of
is twice that of
.
Touching negative common ratios, a couple of strangely exciting things happen. This is why we’ve been silent about them in many of our posts, but we have them on our radar and will treat them later.
Let be a right isosceles triangle with slopes
. PROVE that there are two points
and
on sides
such that the slope of the line segment
is
times the slope of
.
By Example 5, we can find points such that the slopes of the sides of
are
.
Using the fact that for a right isosceles triangle, we have:
Plus or minus two
Let be an equilateral triangle with slopes
. PROVE that there are two points
and
on two sides such that the slope of the line segment
is
.
By Example 5, we can find points on
such that the slopes of the sides of
are
.
Using the fact that and
for an equilateral triangle, we have:
Point-blank.
Takeaway
The construction we’ve done is just a variant of the medial triangle. Unlike the medial triangle, ours is different in the sense that it does not retain the slopes of the parent triangle. While the slopes of the parent triangle form a geometric progression, its own slopes form an arithmetic progression. Talk about tansfo
mation.
Tasks
- (Two out) Consider
with vertices at
,
, and
. PROVE that one can obtain the infinite arithmetic sequence
as slopes of line segments drawn from vertex
to side
.
(Notice thatis out because it’s already the slope of side
. Other than that, every other integer can be obtained this way. What’s somewhat interesting here is that the line segments from
to
that generate the sequence terms all stay within the triangle; none of them is external.)
- (Fibonacci sequence) Explain how the terms of a Fibonacci sequence
can be generated as slopes.
- (Always odd) Let
be a positive integer, and consider
with vertices at
,
, and
. PROVE that there is a point
on
and a point
on
such that the slope of the line segment
is
.
(Ifis an integer, then any number of the form
is odd.)
- Let
be such that sides
have slopes
. Let
be points on
which divide these sides all in the ratio
. PROVE that:
- the
-coordinates of
form an arithmetic progression
- the line connecting the centroid of
to the centroid of
is parallel to side
.
- the
- Find an appropriate choice of coordinates for the vertices of a right isosceles triangle whose side-slopes form a geometric progression.