Take vertex of triangle
, together with the orthocenter
of the triangle, and our new point
with coordinates
(1)
then the resulting triangle has the following property: the slopes of its sides form a geometric progression. Same with vertices
and
as well as the corresponding triangles
and
(examples 6 and 10). This construction works for all triangles, except those triangles with one side parallel to the
or
axis.
Rigid transversals
If we draw three line segments from to each of the three vertices
of any triangle
, then the line segments
either all make acute angles with the
-axis or all make obtuse angles with the
-axis. See example 2 below.
In , let
be the slopes of sides
, respectively. PROVE that the slopes of the line segments
are
,
,
, in that order.
Let the coordinates of the vertices be ,
, and
. With
as in equation (1), we can calculate the slope of
(those of
and
are similar).
The placement of the negative sign beside was deliberate: to make it easy to remember the formula for the slope of
as well as those of
and
. Here’s how it works: the slope of
is the product of the slopes of the two sides (
and
) that originate from vertex
divided by the negative of the slope of the side opposite vertex
(side
).
In any , PROVE that the line segments
either all make acute angles with the positive
-axis or all make obtuse angles with the positive
-axis.
The slopes from to
are given by:
(2)
Re-write the terms in (2) as
(3)
and consider four cases:
- Case I:
are all positive. In this case, the product
is also positive and so each term in equation (3) is negative.
- Case II:
are all negative. In this case, the product
is also negative and so each term in equation (3) is positive.
- Case III: two of
are positive while the third is negative; specifically, let
be positive while
is negative. In this case, the product
is negative, and so the terms in equation (3) are all positive.
- Case IV: one of
is positive while the other two are negative; specifically, let
be positive while
and
are negative. In this case, the product
will be positive and so the terms in equation (3) will be negative.
In , let
be the slopes of sides
, respectively. PROVE that the line segment
is perpendicular to side
if and only if
.
can also be perpendicular to
or
, but requiring it to be perpendicular to
is more convenient.
First suppose that is perpendicular to
. Since the slope of
is
and the slope of
is
, we have:
Conversely, suppose that . Let’s simplify the slope of
:
This shows that is perpendicular to
.
(The case where and
is a special case. The case where
and
is an extremely pleasant case.)
In , let
be the slopes of sides
, respectively. PROVE that
is parallel to
if and only if
(in other words, the slopes of sides
and
are negatives of each other).
First suppose that is parallel to
. Set their slopes equal:
However, since we can’t have in a triangle, we take
.
Conversely, suppose that and re-calculate the slopes of
and
:
What this means is that and
are parallel to side
(in fact,
is the midpoint of
under this condition).
Let be the orthocenter of
, and let
be the point with coordinates in equation (1). PROVE that the slope of
is the negative reciprocal of the product of the slopes of sides
.
If has vertices at
,
,
, then the coordinates of its orthocenter are given by:
(4)
These are very much the same as the coordinates of point , except for the extra term
in the numerator of the
-coordinate and the extra term
in the
-coordinate. It follows that the slope of the line segment
is:
(5)
where are the slopes of sides
.
(Main goal)
For any triangle with non-zero side-slopes, PROVE that the three triangles
,
, and
have their side-slopes in geometric progressions.
We do this for triangle . The same argument can be adapted for the other two triangles.
Let the slopes of sides be
. Assume all non-zero.
The slope from vertex to the orthocenter
is
(by the definition of the altitude from vertex
). The slope of the line segment
is
(by example 1), and the slope of the line segment
is
(by example 5). Since
the slopes of sides form a geometric progression.
Right triangles
In , let
. PROVE that the slope of
is the reciprocal of the slope of the hypotenuse
.
Let be the slopes of sides
. By example 1, the slope of the line segment
is
. Since the given triangle is right-angled at
, sides
and
are perpendicular, and so
, which in turn simplifies the slope of the line segment
:
Alternatively, one can also deduce this from example 5.
(The three triangles reduces to only two for a right triangle, since its orthocenter coincides with a vertex. But we do get a special quadrilateral.)
Related trapezium
Let be a right triangle in which the hypotenuse has slope
and the legs are not parallel to the coordinate axes. PROVE that the quadrilateral
is a trapezium.
Excluding the case where the legs are parallel to the coordinate axes is essential; if not, coincides with the orthocenter, which is one of the vertices of a right triangle.
Suppose we have and
. According to example 7, the slope of
will be
as well, meaning that
is parallel to
. This gives the required trapezium
. Similarly, if the slope of the hypotenuse
is
(that is
), then the slope of
is again
, and this gives a trapezium
.
(Note that the slopes of and
are reciprocals of each other under the above condition. Together with the slope of
as
, we get that the slopes of the line segments
form a geometric progression, even though the slopes of the sides of the parent triangle
do not form a geometric progression.)
Calculate the coordinates of point for
with vertices at
,
,
. Deduce that
is a trapezium.
Observe that is right-angled at
. Using equation (1) with
,
, and
, we get:
Thus, is located at
in the Cartesian plane. The slope of
is
, and the slope of
is
, so the two line segments are parallel. The slope of
is
, while the slope of
is
. Therefore, we obtain a trapezium.
(Notice that the slopes of form a geometric progression, despite the fact the the slopes of the parent triangle
do not form a geometric progression.)
Calculate the coordinates of for
with vertices at
,
, and
. Deduce that
is a trapezium. Verify also that the slopes of the sides of triangles
,
, and
form geometric progressions.
Unlike in the preceding example, the given triangle is not right-angled this time. So this example is saying that we can still obtain this special trapezium even when the starting triangle is not right-angled. Using equation (1) with
,
, and
, we get:
Thus, is the point
. With this, the slope of
is
. Since the slope of
is also
, it follows that
is parallel to
. The slope of
is
and the slope of
is
. So we obtain a trapezium
.
Behind the scences, we calculated the orthocenter of triangle ; it is the point
. Together with
and the given vertices
,
,
, we have the following slopes:
The slopes of the sides of are then
; the slopes of the sides of
are
; the slopes of the sides of
are
. These form geometric progressions with common ratios
, respectively.
(Notice that the slopes of form a geometric progression, despite the fact the the slopes of the parent triangle
do not form a geometric progression.)
Takeaway
Let be the point on the nine-point circle of
whose coordinates were given in equation (1). Let
be the orthocenter of
. Then the following statements are equivalent:
- the slopes of sides
and
are reciprocals of each other
- the line segment
is perpendicular to side
is the midpoint of the line segment
.
You’ll be getting used to equivalent statements by now.
Tasks
- For
with vertices at
,
,
, calculate the slopes of line segments
,
, and
. Verify that
is perpendicular to
.
- Let
be such that side
is parallel to the
-axis. PROVE that:
coincides with the foot of the altitude from vertex
- none of the triangles
has slopes in geometric progression.
(This is the only exception to the main point in today’s discussion.)
- In
, let
be as given in equation (1). PROVE that the slopes of
and
are reciprocals of each other if and only if the slope of side
is
- PROVE if the slopes of the sides of a triangle form a geometric progression with common ratio
, then the slopes of
form a geometric progression with common ratio
.
- In
, let
be the slopes of sides
. PROVE that the following statements are equivalent:
is parallel to
(Under this condition, the slopes of line segmentsform a geometric progression.)