Sloppy slopes swap sequences

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 , whee .

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

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 tansfomation.