Arithmetic and geometric slopes

Two posts ago, we promised to discuss the general theory behind triangles whose sides have slopes of 1,2,3 — and here we go!!! We also considered, in the previous post, triangles whose sides have slopes of the form k,-2k,4k. If you loved those two, and you read this one through, you’ll love this one too. Or even “befriend” it as we do.

That’s true.

Slopes in arithmetic progression

Example 1

Find coordinates for the vertices of a triangle whose sides have slopes of a,~a+d,~a+2d.

Let \triangle ABC have vertices as shown below:

Rendered by QuickLaTeX.com

Since the slopes of sides AB,~BC,~CA are a,~a+d,~a+2d, respectively, we have the following linear equations:

    \begin{equation*} \begin{split} y_2-y_1&=a(x_2-x_1)\\ y_3-y_2&=(a+d)(x_3-x_2)\\ y_3-y_1&=(a+2d)(x_3-x_1) \end{split} \end{equation}

which we can recast into a “homogeneous” system:

(1)   \begin{equation*} -ax_1+y_1+ax_2-y_2=0 \end{equation*}

(2)   \begin{equation*} (-a-2d)x_1+y_1+(a+2d)x_3-y_3=0 \end{equation*}

(3)   \begin{equation*} (-a-d)x_2+y_2+(a+d)x_3-y_3=0 \end{equation*}

Notice that there are six unknowns in three equations, so the system is underdetermined. Being also homogeneous, it has an infinite number of solutions, as our friend from L.A. (a.k.a Linear Algebra, not Los Angeles), just informed us. By row reduction, our system is found to have a rank of 3 as well as three degrees of freedom — the latter meaning that the solution can be expressed in terms of three parameters (or “free” variables). In fact, we have:

    \begin{equation*} \begin{split} x_1&=\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3\\ y_1&=\frac{a+2d}{2(a+d)}y_2+\frac{a}{2(a+d)}y_3\\ x_2&=\frac{1}{a+d}y_2+x_3-\frac{1}{a+d}y_3 \end{split} \end{equation}

where y_2,x_3,y_3 are our three “free” variables. Observe that the coefficients of these variables add up to one in the above equations, so in a sense they’re weighted.

To sum up, the required coordinates are:

    \begin{equation*} \begin{split} (x_1,y_1)&\mapsto\Big(\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3,~\frac{a+2d}{2(a+d)}y_2+\frac{a}{2(a+d)}y_3\Big)\\ (x_2,y_2)&\mapsto \Big(\frac{1}{a+d}y_2+x_3-\frac{1}{a+d}y_3,\quad y_2\Big)\\ (x_3,y_3)&\mapsto(x_3,y_3) \end{split} \end{equation}

As you may have thought, the formula breaks down when a+d=0, but this leads to an important special case that deserves a separate treatment.

Example 2

In the notation of our first diagram, PROVE that a+d=0 if, and only if, y_3-y_2=0.

This is immediate; just another way of saying that the slope of side BC is zero. However, if we want to be fancy, we can use equation (3). First assume that a+d=0. Then (3) reduces to y_3-y_2=0. Conversely, if y_3-y_2=0, then we’re left with

    \[(-a-d)x_2+(a+d)x_3=0\implies (a+d)(x_3-x_2)=0,\]

which in turn leaves us with two choices: a+d=0 or x_3-x_2=0. However, x_3-x_2=0 is impossible because otherwise it would mean that x_3=x_2. Together with the fact that y_3=y_2 (by the assumption in the converse), the points (x_2,y_2) and (x_3,y_3) will coincide, meaning that we don’t have a triangle. So we choose a+d=0 and this completes the proof.

As a by-product, we also obtain that a+d\neq 0 is equivalent to y_3-y_2\neq 0. We’ll have more to say about this case later.

Example 3

Find the x-coordinate of the centroid of a triangle whose sides have slopes of a,~a+d,~a+2d.

The x-coordinate of the centroid is given by \frac{x_1+x_2+x_3}{3}. Using the coordinates we found in Example 1, we have:

    \begin{equation*} \begin{split} &=\frac{\Big(\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3\Big)+\Big(\frac{1}{a+d}y_2+x_3-\frac{1}{a+d}y_3\Big)+x_3}{3}\\ &=\frac{\Big(\frac{3}{2(a+d)}y_2+3x_3-\frac{3}{2(a+d)}y_3\Big)}{3}\\ &=\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3\\ &=x_1 \end{split} \end{equation}

Example 4

PROVE that if the slopes of the sides of a triangle follow an arithmetic progression, then the triangle contains one vertical median.

Let \triangle ABC have vertices A(x_1,y_1),B(x_2,y_2),C(x_3,y_3) and let sides AB,BC,CA have slopes of a,~a+d,~a+2d, respectively. In Example 3, we saw that the x-coordinate of the centroid is precisely x_1. Since the centroid shares an x-coordinate with a triangle’s vertex (in this case A(x_1,y_1)), we know — from our previous post — that the median through that vertex is vertical.

Example 5

PROVE that if the slopes of the sides of a triangle follow an arithmetic progression a,~a+d,~a+2d, then so do the x-coordinates.

Note that the converse is not true.

To prove the above statement, we can use a result from our previous post, but let’s calculate it directly here. Enumerate the x-coordinates as follows:

    \begin{equation*} \begin{split} x_3&=x_3\\ &\cdots\vdots\cdots\\ x_1&=\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3\\ &=x_3+\frac{1}{2(a+d)}\Big(y_2-y_3\Big)\\ &\cdots\vdots\cdots\\ x_2&=\frac{1}{a+d}y_2+x_3-\frac{1}{a+d}y_3\\ &=x_3+\frac{1}{a+d}\Big(y_2-y_3\Big)\\ &=x_3+\frac{1}{2(a+d)}\Big(y_2-y_3\Big)+\frac{1}{2(a+d)}\Big(y_2-y_3\Big)\\ &=x_1+\frac{1}{2(a+d)}\Big(y_2-y_3\Big). \end{split} \end{equation}

So the x-coordinates differ consecutively by \frac{1}{2(a+d)}\Big(y_2-y_3\Big).

The fact that the x-coordinates follow an arithmetic progression whereas the y-coordinates do not (necessarily) follow an arithmetic progression makes the resulting triangle diagram quite weird, which was what we called “ugly-looking” two posts ago.

Slopes in geometric progression

Example 6

Find coordinates for the vertices of a triangle whose sides have slopes of k,~kr,~kr^2.

Of course, we want k\neq 0 and r\neq\pm 1. If there be any other restriction, it will become obvious when we’re done with the computations.

To find the coordinates, suppose we work with \triangle ABC whose vertices are as shown below:

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As before, it boils down to solving a linear system:

    \begin{equation*} \begin{split} y_2-y_1&=k(x_2-x_1)\\ y_3-y_2&=kr(x_3-x_2)\\ y_3-y_1&=kr^2(x_3-x_1) \end{split} \end{equation}

where, after row reduction and back-substitution, the solution is:

    \begin{equation*} \begin{split} x_1&=\frac{1}{kr(1+r)}y_2+x_3-\frac{1}{kr(1+r)}y_3\\ y_1&=\frac{r}{1+r}y_2+\frac{1}{1+r}y_3\\ x_2&=\frac{1}{kr}y_2+x_3-\frac{1}{kr}y_3 \end{split} \end{equation}

In each of the above expressions, the coefficients of the “free variables” y_2,x_3,y_3 add up to one, so in a sense they’re weighted.

Example 7

Find the centroid of the triangle with vertices given in Example 6 above.

The x-coordinate of the centroid is \frac{x_1+x_2+x_3}{3}:

    \begin{equation*} \begin{split} &=\frac{\Big(\frac{1}{kr(1+r)}y_2+x_3-\frac{1}{kr(1+r)}y_3\Big)+\Big(\frac{1}{kr}y_2+x_3-\frac{1}{kr}y_3\Big)+x_3}{3}\\ &=\frac{\frac{2+r}{kr(1+r)}y_2-\frac{2+r}{kr(1+r)}y_3+3x_3}{3}\\ &=\frac{(2+r)(y_2-y_3)}{3kr(1+r)}+x_3 \end{split} \end{equation}

If r=-2, then the x-coordinate coincides with x_3. (Note that we can’t have y_2=y_3 since otherwise one of the slopes will be zero, and we’re working with slopes in geometric progression.)

For the y-coordinate of the centroid, we use \frac{y_1+y_2+y_3}{3}:

    \begin{equation*} \begin{split} &=\frac{\Big(\frac{r}{1+r}y_2+\frac{1}{1+r}y_3\Big)+y_2+y_3}{3}\\ &=\frac{\frac{1+2r}{1+r}y_2+\frac{2+r}{1+r}y_3}{3}\\ &=\frac{1+2r}{3(1+r)}y_2+\frac{2+r}{3(1+r)}y_3 \end{split} \end{equation}

If r=-2, the y-coordinate of the centroid coincides with y_2.

COOL.

Example 8

Let \triangle ABC be such that sides AB,BC,CA have slopes k,kr,kr^2, respectively. PROVE that the slope of the median from vertex A is -kr.

The fact that the slope of the median from vertex A to side BC and the slope of side BC are negatives of each other has an important implication and application (see Example 9 below).

To prove the present statement, let’s refer to an earlier diagram:

Rendered by QuickLaTeX.com

We found that

    \begin{equation*} \begin{split} x_1&=\frac{1}{kr(1+r)}y_2+x_3-\frac{1}{kr(1+r)}y_3\\ y_1&=\frac{r}{1+r}y_2+\frac{1}{1+r}y_3\\ x_2&=\frac{1}{kr}y_2+x_3-\frac{1}{kr}y_3 \end{split} \end{equation}

The midpoint of BC is \Big(x_3+\frac{1}{2kr}(y_2-y_3),\frac{y_2+y_3}{2}\Big) and so the slope of the median from vertex A is:

    \begin{equation*} \begin{split} &=\frac{y_1-\Big(\frac{y_2+y_3}{2}\Big)}{x_1-\Big(x_3+\frac{1}{2kr}(y_2-y_3)\Big)}\\ &=\frac{\frac{r}{1+r}y_2+\frac{1}{1+r}y_3-\Big(\frac{y_2+y_3}{2}\Big)}{\frac{1}{kr(1+r)}y_2+x_3-\frac{1}{kr(1+r)}y_3-\Big(x_3+\frac{1}{2kr}(y_2-y_3)\Big)}\\ &=\frac{\frac{r-1}{2(1+r)}(y_2-y_3)}{\frac{1-r}{2kr(1+r)}(y_2-y_3)}\\ &=\frac{r-1}{2(1+r)}(y_2-y_3)\times \frac{2kr(1+r)}{(1-r)(y_2-y_3)}\\ &=-kr. \end{split} \end{equation}

Example 9

Let \triangle ABC be such that sides AB,BC,CA have slopes k,kr,kr^2, respectively. PROVE that AB=CA if, and only if, kr=\pm 1.

Suppose that kr=1 (a similar argument holds for kr=-1). By Example 8 above, the slope of the median from vertex A is -1. But then this means that the median from vertex A is perpendicular to side BC. Therefore, AB=CA and \triangle ABC is isosceles.

Conversely, suppose that AB=CA. Then the median from vertex A is perpendicular to side BC, which in turn means that the product of their slopes is -1. Since their slopes are kr and -kr, it follows that

    \begin{equation*} \begin{split} kr\times(-kr)&=-1\\ kr\times kr&=1\\ (kr)^2&=1\\ kr&=\pm 1. \end{split} \end{equation}

So we obtain a characterization of isosceles triangles, using slopes in geometric progression.

Example 10

Let \triangle ABC be a right-triangle in which sides AB,BC,CA have slopes k,kr,kr^2, respectively. PROVE that r is necessarily negative, and that side BC cannot be the hypotenuse.

Since the location of the right-angle was not specified, we have to consider all possibilities. We use the fact that the product of the slopes of two perpendicular lines is -1, and take the product of two slopes at a time:

    \begin{equation*} \begin{split} k\times kr&=-1\\ k\times kr^2&=-1\\ kr\times kr^2&=-1 \end{split} \end{equation}

From the first equation, r=-\frac{1}{k^2}, which is always negative because the denominator k^2 is always positive whatever (the real number) k is. From the third equation, we have r^3=-\frac{1}{k^2}, and again r is negative since it’s only the cube of a negative (real) number that can be negative.

Now from the second equation, we have (kr)^2=-1. Since the left side of this equation is always positive whereas the right side is always negative, the equation doesn’t have real solution. In turn, this means that side AB (with slope k) and side CA (with slope kr^2) do not form a right-angle. Therefore, side BC cannot be the hypotenuse.

Numerical examples

We now move from the abstract angle to actual examples.

  1. Let a=1 and d=1. Then the arithmetic sequence a,~a+d,~a+2d is 1,~2,~3 and the corresponding coordinates for the vertices of a triangle with slopes 1,2,3 are:

        \[x_1=x_3+\frac{1}{4}(y_2-y_3),~y_1=\frac{3}{4}y_2+\frac{1}{4}y_3,~x_2=x_3+\frac{1}{2}(y_2-y_3).\]

    x_3,y_2,y_3 are “free” variables. We obtain an assemblage of triangles by specifying values for x_3,y_2,y_3, so long as y_2\neq y_3.

    • Put x_3=0,~y_3=0 and y_2=4. Then x_1=1,~y_1=3,~x_2=2. The resulting triangle has coordinates at (1,3),(2,4),(0,0):

      Rendered by QuickLaTeX.com

      Notice the presence of a vertical median PB.

    • Put x_3=-1,~y_3=-2 and y_2=6. Then x_1=1,~y_1=4,~x_2=3. We obtain a triangle with coordinates at (1,4),(3,6),(-1,-2):

      Rendered by QuickLaTeX.com

      Notice the presence of a vertical median PB.

  2. Let a=1 and d=-2. Then the corresponding arithmetic sequence of slopes is 1,~-1,~-3; moreover:

        \[x_1=x_3-\frac{1}{2}(y_2-y_3),~y_1=\frac{3}{2}y_2-\frac{1}{2}y_3,~x_2=x_3-(y_2-y_3).\]

    Depending on what we choose for the “free variables”, we obtain different sets of triangles. Let x_3=1,~y_3=2,~y_2=3. Then x_1=\frac{1}{2},~y_1=\frac{7}{2},~x_2=0. This results in a triangle with coordinates at (\frac{1}{2},\frac{7}{2}),~(1,2),~(0,3):

    Rendered by QuickLaTeX.com

    Notice the presence of a vertical median PB. Also this is a right triangle.

  3. An example of a triangle with slopes in geometric progression. Let k=\frac{1}{2} and r=-4; the corresponding geometric sequence of slopes is \frac{1}{2},~-2,~8. If we put k=\frac{1}{2} and r=-4 in our formula, we obtain

        \[x_1=x_3+\frac{1}{6}(y_2-y_3),~y_1=\frac{4}{3}y_2-\frac{1}{3}y_3,~x_2=x_3-\frac{1}{2}(y_2-y_3).\]

    Next stop: choose values for the “free variables” x_3,y_3,y_2, bearing in mind that y_2\neq y_3. Let y_2=6 and x_3=y_3=0. We obtain x_1=1,~y_1=8,~x_2=-3 and the resulting triangle has vertices located at (1,8),(-3,6),(0,0):

    Rendered by QuickLaTeX.com

    Notice that this is a right triangle.

  4. Let k=1 and r=2; we obtain the geometric sequence of slopes 1,~2,~4. In turn:

        \[x_1=x_3+\frac{1}{6}(y_2-y_3),~y_1=\frac{2}{3}y_2+\frac{1}{3}y_3,~x_2=x_3+\frac{1}{2}(y_2-y_3).\]

    For convenience, let’s choose our “free variables” x_3=0=y_3 and y_2=6. We then obtain x_1=1,~y_1=4,~x_2=3 and a triangle with coordinates at (1,4),(3,6),(0,0) results:

    Rendered by QuickLaTeX.com

  5. Here’s an example in which some of the coordinates are irrational numbers. Let k=\sqrt{5} and r=\frac{1}{2}. Let’s also choose our “free variables” right away: x_3=0=y_3 and y_2=3\sqrt{5}. We obtain x_1=4,~y_1=\sqrt{5},~x_2=6 and a triangle with coordinates at (4,\sqrt{5}),(6,3\sqrt{5}),(0,0) results:

    Rendered by QuickLaTeX.com

    There’s something fascinating about this example that is worth mentioning. Observe that the given triangle doesn’t contain 90^{\circ}. Now the midpoints of sides PQ,~QR,~RP are

        \[\Big(3,\frac{3}{2}\sqrt{5}\Big),~\Big(2,\frac{\sqrt{5}}{2}\Big),~\Big(5,2\sqrt{5}\Big).\]

    Consequently, the lengths of the medians from R,~P,~Q are \frac{3}{2},~\frac{3\sqrt{21}}{2},~3\sqrt{5}, respectively. How are these related?

        \begin{equation*} \begin{split} \Big(\frac{3\sqrt{21}}{2}\Big)^2&=\frac{189}{4}\\ \Big(\frac{3}{2}\Big)^2+(3\sqrt{5})^2&=\frac{9}{4}+45\\ &=\frac{9}{4}+\frac{45\times 4}{1\times 4}\\ &=\frac{9}{4}+\frac{180}{4}\\ &=\frac{189}{4}\\ &=\Big(\frac{3\sqrt{21}}{2}\Big)^2. \end{split} \end{equation}

    COOL.

Takeaway

If the slopes of the sides of a triangle are k,kr,kr^2 (a geometric progression), and the slopes of the sides of another triangle are a,a+d,a+2d (an arithmetic progression), then properties that correspond to kr=\pm 1 have analogues in a+d=0. These properties are mainly lateral properties — isosceles or equilateral.

We encourage you to investigate different cases of slopes in arithmetic progression and slopes in geometric progression. You’ll encounter exciting surprises.

Tasks

  1. PROVE that if the slopes of the sides of a triangle form a geometric progression k, kr, kr^2, then the slopes of its three medians also form a geometric progression (with a different common ratio), provided r\neq -2,-\frac{1}{2}.
  2. PROVE that if the slopes of the sides of a triangle form an arithmetic progression, then the slopes of its three medians do not form an arithmetic progression.
  3. Let k,kr,kr^2 (r\neq -2,-\frac{1}{2}) be the slopes of the sides of a \triangle ABC. PROVE that the product of the slopes of the three medians is -(kr)^3.
  4. PROVE that if the slopes of the sides of a triangle are 1,r,r^2, then the triangle can never be a right triangle.
    \Big( Of course r\neq \pm 1. \Big)
  5. PROVE that if the slopes of the sides of a triangle are k,kr,kr^2, where k and r are rational numbers, then the triangle can never be equilateral.
    \Big( Actually if either k or r is a rational number, same conclusion. \Big)
  6. PROVE that if the slopes of the sides of a triangle are a,a+d,a+2d, where a and d are rational numbers, then the triangle can never be equilateral.
    \Big( It appears that equilateral triangles don’t like rational numbers? \Big)
  7. Let \triangle ABC be such that sides AB,BC,CA have slopes k,kr,kr^2, respectively. PROVE that the median from B and the median from C can never intersect at 90^{\circ}, except when r=-2.
    \Big( Any triangle whose side slopes form a geometric progression with common ratio r=-2 is a peach. \Big)
  8. PROVE that if the x-coordinates of the vertices of a triangle form an arithmetic progression, then the slopes of the sides do not necessarily form an arithmetic progression.
    \Big( A counter-example will do. \Big)
  9. If the slopes of the sides of a triangle form a geometric sequence with a common ratio of -2, PROVE that both the x-coordinates and the y-coordinates form arithmetic progressions.
    \Big( Cool. \Big)
  10. Let \triangle ABC be such that sides AB,BC,CA have slopes of k,kr,kr^2, respectively. PROVE that AB^2+BC^2=CA^2 if, and only if k^2r=-1.
    \Big( Pretend that you’re unaware of the Pythagorean theorem and perpendicular slopes. Solve the problem without recourse to these two. \Big)

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