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 1:1, we’ll modify this slightly for our triangle and rather divide the sides in the ratio 1:r, where ~1\neq r> 0.

In \triangle ABC, let X and Y be points on BA and CA which divide the sides in the ratio 1:n, where n is a positive integer. PROVE that the line segment XY is parallel to BC and that XY=\frac{n}{n+1}BC.

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In the diagram above, BX:XA=1:n and CY:YA=1:n. If we place the vertices of \triangle ABC at A(x_1,y_1), B(x_2,y_2), and C(x_3,y_3), then we can obtain the coordinates of points X and Y using the formula for internal division of a line segment:

    \[X\left(\frac{x_1+nx_2}{n+1},\frac{y_1+ny_2}{n+1}\right),~Y\left(\frac{x_1+nx_3}{n+1},\frac{y_1+ny_3}{n+1}\right).\]

The slope of line segment XY is:

    \begin{equation*} \begin{split} &=\left[\frac{y_1+ny_2}{n+1}-\left(\frac{y_1+ny_3}{n+1}\right)\right]\div\left[\frac{x_1+nx_2}{n+1}-\left(\frac{x_1+nx_3}{n+1}\right)\right]\\ &=\frac{n}{n+1}(y_2-y_3)\div\left(\frac{n}{n+1}(x_2-x_3)\right)\\ &=\frac{y_2-y_3}{x_2-x_3} \end{split} \end{equation*}

So XY is parallel to BC. In addition we obtain XY=\frac{n}{n+1}BC.

Easy-peasy.

Derailing development

If we apply the above idea to our own triangle, look what we get.

Suppose that \triangle ABC has vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3), and slopes a,ar,ar^2 for sides AB,BC,CA, respectively. For 1\neq r> 0, find coordinates for the point which divides AB in the ratio 1:r.

Note the “direction” of the division. The desired point will be \left(\frac{x_2+rx_1}{r+1},\frac{y_2+ry_1}{r+1}\right), using the formula for internal division of a line segment. In view of the relationship among the coordinates, this becomes

    \[\left(x_3+\frac{(2r+1)(y_2-y_3)}{ar(r+1)^2},\frac{(r^2+r+1)y_2+ry_3}{(r+1)^2}\right).\]

You see that quadratic rearing its beautiful head again?

Suppose that \triangle ABC has vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3), and slopes a,ar,ar^2 for sides AB,BC,CA, respectively. For 1\neq r> 0, find coordinates for the point which divides CB in the ratio 1:r.

Ordinarily, this would be the point \left(\frac{x_2+rx_3}{r+1},\frac{y_2+ry_3}{r+1}\right). Using the relationship among the coordinates, we get

    \[\left(x_3+\frac{y_2-y_3}{ar(r+1)},\frac{ry_2+y_3}{r+1}\right)\]

Suppose that \triangle ABC has vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3), and slopes a,ar,ar^2 for sides AB,BC,CA, respectively. For 1\neq r> 0, find coordinates for the point which divides CA in the ratio 1:r.

This would be the point \left(\frac{x_1+rx_3}{r+1},\frac{y_1+ry_3}{r+1}\right). The relationship among the coordinates then gives

    \[\left(x_3+\frac{y_2-y_3}{ar(r+1)^2},\frac{ry_2+(r^2+r+1)y_3}{(r+1)^2}\right)\]

(Main goal)

Suppose that \triangle ABC has vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3), and slopes a,ar,ar^2 for sides AB,BC,CA, respectively. For 1\neq r> 0, PROVE that there is a point X on AB, a point Y on BC, and a point Z on CA such that the slopes of the sides of \triangle XYZ form an arithmetic progression.

Basically, the side-slopes form a geometric progression, while the sub-triangle’s side-slopes form an arithmetic progression. For. Any. 1\neq r> 0.

Let X be the point which divides AB in the ratio 1:r, let Y be the point which divides CB in the ratio 1:r, and let Z be the point which divides CA in the ratio 1:r. As in the three preceding examples, these points are:

    \begin{equation*} \begin{split} X&\left(x_3+\frac{(2r+1)(y_2-y_3)}{ar(r+1)^2},\frac{(r^2+r+1)y_2+ry_3}{(r+1)^2}\right)\\ Y&\left(x_3+\frac{y_2-y_3}{ar(r+1)},\frac{ry_2+y_3}{r+1}\right)\\ Z&\left(x_3+\frac{y_2-y_3}{ar(r+1)^2},\frac{ry_2+(r^2+r+1)y_3}{(r+1)^2}\right) \end{split} \end{equation*}

By the modified midpoint theorem, XY would be parallel to CA, so its slope is ar^2 (alternatively can be confirmed by direct calculation). Similarly, YZ would be parallel to AB, so its slope is a.

By direct calculation, the slope of ZX is \frac{a+ar^2}{2}.

Thus, the slopes of the sides of \triangle XYZ are a,\frac{a+ar^2}{2},ar^2, an arithmetic progression.

Given \triangle ABC with vertices at A(1,4), B(3,6), and C(0,0), find coordinates for points X,Y,Z on AB,BC,CA such that the slopes of \triangle XYZ are 1,\frac{5}{2},4.

Observe that the slopes of sides AB,BC,CA are 1,2,4, respectively.

Thus, by Example 5, let X be the point which divides AB in the ratio 1:2, let Y be the point which divides CB in the ratio 1:2, and let Z be the point which divides CA in the ratio 1:2. Lots of repetitions there \cdots.

Using the formula for internal division of a line segment, we find these points to be X\left(\frac{5}{3},\frac{14}{3}\right), Y(1,2), and Z\left(\frac{1}{3},\frac{4}{3}\right). Then:

    \begin{equation*} \begin{split} \textrm{slope of}~XY&=4\\ \textrm{slope of}~YZ&=1\\ \textrm{slope of}~ZX&=\frac{5}{2} \end{split} \end{equation*}

Given \triangle ABC with vertices at A(1,9), B(4,12), and C(0,0), find coordinates for points X,Y,Z on AB,BC,CA such that the slopes of \triangle XYZ are 1,5,9.

In this case, the slopes of sides AB,BC,CA are 1,3,9, respectively. A geometric progression in which r=3.

By Example 5, let X,Y,Z divide each of the three sides AB,CB,CA in the ratio 1:3.

Using the formula for internal division of a line segment, we find these points to be X\left(\frac{7}{4},\frac{39}{4}\right), Y(1,3), and Z\left(\frac{1}{4},\frac{9}{4}\right).

    \begin{equation*} \begin{split} \textrm{slope of}~XY&=9\\ \textrm{slope of}~YZ&=1\\ \textrm{slope of}~ZX&=5 \end{split} \end{equation*}

Given \triangle ABC with vertices at A(1,4), B(-1,2), and C(0,0), find coordinates for (external) points X,Y,Z on AB,BC,CA such that the slopes of \triangle XYZ are 1,\frac{5}{2},4.

Observe that the slopes of sides AB,BC,CA are 1,-2,4, respectively. They form a geometric progression in which r=-2.

Let X,Y,Z be points which divide AB,CB,CA all in the ratio 1:-2. We obtain the external points X\left(3,6\right), Y(1,-2), and Z\left(-1,-4\right). Then:

    \begin{equation*} \begin{split} \textrm{slope of}~XY&=4\\ \textrm{slope of}~YZ&=1\\ \textrm{slope of}~ZX&=\frac{5}{2} \end{split} \end{equation*}

The scene here may be worth seeing \cdots

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Notice that C is the midpoint of both AZ and BY, while A is the midpoint of BX. It follows that the length of XY is twice that of AC.

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 ABC be a right isosceles triangle with slopes a,ar,ar^2. PROVE that there are two points X and Y on sides AB,AC such that the slope of the line segment XY is -\frac{3}{2} times the slope of BC.

By Example 5, we can find points X,Y,Z such that the slopes of the sides of \triangle XYZ are a,\frac{a+ar^2}{2},ar^2.

Using the fact that r^2+3r+1=0 for a right isosceles triangle, we have:

    \begin{equation*} \begin{split} \frac{a+ar^2}{2}&=\frac{a(1+r^2)}{2}\\ &=\frac{a(-3r)}{2}\\ &=-\frac{3}{2}(ar)\\ \end{split} \end{equation*}

Plus or minus two

Let ABC be an equilateral triangle with slopes a,ar,ar^2. PROVE that there are two points X and Y on two sides such that the slope of the line segment XY is \pm 2.

By Example 5, we can find points X,Y,Z on AB,BC,CA such that the slopes of the sides of \triangle XYZ are a,\frac{a+ar^2}{2},ar^2.

Using the fact that ar=\pm 1 and r^2+4r+1=0 for an equilateral triangle, we have:

    \begin{equation*} \begin{split} \frac{a+ar^2}{2}&=\frac{a(1+r^2)}{2}\\ &=\frac{a(-4r)}{2}\\ &=-2(ar)\\ &=\pm 2 \end{split} \end{equation*}

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 transformation.

Tasks

  1. (Two out) Consider \triangle ABC with vertices at A(1,4), B(3,6), and C(0,0). PROVE that one can obtain the infinite arithmetic sequence \cdots,-3,-2,-1,0,1,\boxed{2},3,\cdots as slopes of line segments drawn from vertex A to side BC.
    (Notice that 2 is out because it’s already the slope of side BC. Other than that, every other integer can be obtained this way. What’s somewhat interesting here is that the line segments from A to BC that generate the sequence terms all stay within the triangle; none of them is external.)
  2. (Fibonacci sequence) Explain how the terms of a Fibonacci sequence 0,1,1,2,3,5,8,13,\cdots can be generated as slopes.
  3. (Always odd) Let r be a positive integer, and consider \triangle ABC with vertices at A(1,r^2), B(0,0), and C(1+r,r+r^2). PROVE that there is a point X on AC and a point Y on CB such that the slope of the line segment XY is 2r-1.
    (If r is an integer, then any number of the form 2r-1 is odd.)
  4. Let \triangle ABC be such that sides AB,BC,CA have slopes a,ar,ar^2. Let X,Y,Z be points on AB,AC,CB which divide these sides all in the ratio 1:r. PROVE that:
    • the x-coordinates of \triangle XYZ form an arithmetic progression
    • the line connecting the centroid of \triangle XYZ to the centroid of \triangle ABC is parallel to side BC.
  5. Find an appropriate choice of coordinates for the vertices of a right isosceles triangle whose side-slopes form a geometric progression.