A note on geometric sequences

We’re s--o excited about this!!! As in, EXCITED!!! Indeed, this is expected, considering what is to be expounded: “any” geometric sequence a,ar,ar^2,\cdots,ar^{n-1} can be encapsulated as slopes of a simple triangle like the one below:

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And the procedure is easy: first find a triangle with slopes a,ar,ar^2, then the remaining terms ar^3,ar^4,ar^5,\cdots can be obtained as slopes of line segments drawn from a fixed vertex A to a fixed side BC.

Positive common ratio

In our first three examples, we’ll generate the entire geometric sequence

    \[1,2,4,8,16,32,\cdots, 2^{n-1},\cdots\]

that is, (positive) powers of 2.

Example

Find coordinates for the vertices of \triangle ABC with slopes 1,2,4.

There are different ways to obtain a triangle whose slopes are a,ar,ar^2. One can begin with an arbitrary point B(x_1,y_1). Then the two other points can be chosen as A(x_1+1,y_1+ar^2) and C(x_1+1+r,y_1+ar+ar^2). With these, the slopes of sides AB,BC,CA will be ar^2,ar,a, respectively.

In the present case, let’s first choose B(x_1,y_1) as (0,0), for simplicity. Since the sequence is 1,2,4, we have a=1 and r=2. So A(x_1+1,y_1+ar^2)=A(1,4) and C(x_1+1+r,y_1+ar+ar^2)=C(3,6).

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The entire geometric sequence 1,2,4,8,16,32,\cdots, 2^{n-1},\cdots can be obtained from the above triangle.

Example

For each n\geq 1, PROVE that X_{n}=\Big(\frac{2^{n+1}-2}{2^{n+1}-1},\frac{2^{n+2}-4}{2^{n+1}-1}\Big) lies within BC, where B is the point (0,0) and C is the point (3,6).

Note that X_{n} is a point on BC. Let’s use distances to show that it’s indeed internal. We prove that BC=BX_{n}+CX_{n}.

    \begin{equation*} \begin{split} BC^2&=(3-0)^2+(6-0)^2\\ \therefore BC&=3\sqrt{5}\\ &\cdots\vdots\cdots\\ BX_{n}^2&=\Big(\frac{2^{n+1}-2}{2^{n+1}-1}-0\Big)^2+\Big(\frac{2^{n+2}-4}{2^{n+1}-1}-0\Big)^2\\ &=\frac{20(2^n-1)^2}{(2^{n+1}-1)^2}\\ \therefore BX_{n}&=\frac{2\sqrt{5}(2^n-1)}{2^{n+1}-1}\\ &\cdots\vdots\cdots\\ CX_{n}^2&=\Big(\frac{2^{n+1}-2}{2^{n+1}-1}-3\Big)^2+\Big(\frac{2^{n+2}-4}{2^{n+1}-1}-6\Big)^2\\ &=\frac{5(2^{n+2}-1)^2}{(2^{n+1}-1)^2}\\ \therefore CX_{n}&=\frac{\sqrt{5}(2^{n+2}-1)}{2^{n+1}-1}\\ &\cdots\vdots\cdots\\ BX_{n}+CX_{n}&=\frac{2\sqrt{5}(2^n-1)}{2^{n+1}-1}+\frac{\sqrt{5}(2^{n+2}-1)}{2^{n+1}-1}\\ &=\frac{\sqrt{5}}{2^{n+1}-1}\Big(6(2^n)-3\Big)\\ &=\frac{3\sqrt{5}}{2^{n+1}-1}\Big(2(2^n)-1\Big)\\ &=\frac{3\sqrt{5}}{2^{n+1}-1}\Big(2^{n+1}-1\Big)\\ &=3\sqrt{5}\\ \therefore BX_{n}+CX_{n}&=BC \end{split} \end{equation*}

This proves that for each n\geq 1, the point X_{n}:=\Big(\frac{2^{n+1}-2}{2^{n+1}-1},\frac{2^{n+2}-4}{2^{n+1}-1}\Big) lies internally on BC.

Example

For each n\geq 1, let X_{n}=\Big(\frac{2^{n+1}-2}{2^{n+1}-1},\frac{2^{n+2}-4}{2^{n+1}-1}\Big). PROVE that the slope of AX_{n} is 4(2^n), where A is the point (1,4).

This follows by direct calculation:

    \begin{equation*} \begin{split} \textrm{slope of}~AX_{n}&=\frac{4-\Big(\frac{2^{n+2}-4}{2^{n+1}-1}\Big)}{1-\Big(\frac{2^{n+1}-2}{2^{n+1}-1}\Big)}\\ &=\frac{4(2^{n+1}-1)-(2^{n+2}-4)}{2^{n+1}-1-(2^{n+1}-2)}\\ &=\frac{4(2^{n+1})-2^{n+2}}{1}\\ &=2^n\times 2^3-2^n\times 2^2\\ &=2^n(8-4)\\ &=4(2^n) \end{split} \end{equation*}

Thus, when n=1,2,3\cdots, the corresponding slopes will be 4(2^1),4(2^2),4(2^3)\cdots; that is, 8,16,32,\cdots.

By defining X_{n}:=\Big(\frac{2^{n+1}-2}{2^{n+1}-1},\frac{2^{n+2}-4}{2^{n+1}-1}\Big) and using \triangle ABC with vertices at A(1,4),B(0,0),C(3,6), we’ve seen that the entire geometric progression 1,2,4,8,16,\cdots,2^{n-1} can be obtained as slopes of the line segments AX_{n}, together with BC with slope 2. Notice that when n=0, X_{n} becomes (0,0)=B and when n=-2, X_{n} becomes (3,6)=C. What happens to X_{n} when n=-1?

Put n=1. Then X_1 has coordinates \Big(\frac{2}{3},\frac{4}{3}\Big). Together with A(1,4), the slope of line segment AX_{1} is

    \[\boxed{\frac{4-\frac{4}{3}}{1-\frac{2}{3}}=\frac{\frac{8}{3}}{\frac{1}{3}}=8=4(2^1)}\]

as expected.

Put n=2, then X_2 has coordinates \Big(\frac{6}{7},\frac{12}{7}\Big). Together with A(1,4), the slope of line segment AX_{2} is

    \[\boxed{\frac{4-\frac{12}{7}}{1-\frac{6}{7}}=\frac{\frac{16}{7}}{\frac{1}{7}}=16=4(2^2)}\]

as expected.

Put n=3, then X_3 has coordinates \Big(\frac{14}{15},\frac{28}{15}\Big). Together with A(1,4), the slope of line segment AX_{3} is

    \[\boxed{\frac{4-\frac{28}{15}}{1-\frac{14}{15}}=\frac{\frac{32}{15}}{\frac{1}{15}}=32=4(2^3)}\]

as expected.

Three things to note from the above calculation:

  • the points X_{n} are defined relative to the chosen coordinates for \triangle ABC;
  • the line segments AX_{n} that describe the geometric progression do not get to the midpoint of side BC;
  • the roles of vertex A and side BC cannot be altered; doing so will distort the orderly arrangement of the line segments. See the next two examples.

Example

Given \triangle ABC with vertices A(1,4),B(0,0),C(3,6), find coordinates for a point Y on AB such that the slope of the line segment CY is 16.

Let’s find the equations of AB and CY and then solve the resulting linear system:

    \begin{equation*} \begin{split} \textrm{equation of}~AB:~y&=4x\\ \textrm{equation of}~CY:~y&=16x-42\\ &\cdots\vdots\cdots\\ 4x&=16x-42\\ x&=\frac{7}{2}\\ y&=14 \end{split} \end{equation*}

Thus, Y is the point (\frac{7}{2},14). It is external to side AB — we don’t want that.

Example

Given \triangle ABC with vertices A(1,4),B(0,0),C(3,6), find coordinates for a point Z on CA such that the slope of the line segment BZ is 16.

Let’s find the equations of CA and BZ and then solve the resulting linear system:

    \begin{equation*} \begin{split} \textrm{equation of}~CA:~y&=x+3\\ \textrm{equation of}~BZ:~y&=16x\\ &\cdots\vdots\cdots\\ 16x&=x+3\\ x&=\frac{1}{5}\\ y&=\frac{16}{5} \end{split} \end{equation*}

Thus, Z is the point (\frac{1}{5},\frac{16}{5}). It is external to side CA — we don’t want that.

In the next two examples, we generate the geometric progression

    \[3,6,12,24,48,\cdots,3(2^{n-1}),\cdots\]

Example

Find coordinates for the vertices of \triangle ABC whose side slopes are 3,6,12.

Take A(0,2),B(-1,-10),C(2,8). Then the slope of CA is 3, the slope of BC is 6, and the slope of AB is 12.

Example

Let X_{n}:=\Big(-\frac{1}{2^{n+1}-1},\frac{-2^{n+3}-2}{2^{n+1}-1}\Big). For each n\geq 1, PROVE that the slope of the line segment AX_{n} is 12(2^n), where A is the point A(0,2).

Observe that each X_{n} lies within BC (B:=(-1,-10),C:=(2,8)). By direct calculation:

    \begin{equation*} \begin{split} \textrm{slope of}~AX_{n}&=\frac{\textrm{rise}}{\textrm{run}}\\ &=\frac{2-\Big(\frac{-2^{n+3}-2}{2^{n+1}-1}\Big)}{0-\Big(\frac{-1}{2^{n+1}-1}\Big)}\\ &=\frac{2(2^{n+1}-1)+(2^{n+3}+2)}{2^{n+1}-1}\times\frac{2^{n+1}-1}{1}\\ &=2^{n+2}+2^{n+3}\\ &=12(2^{n}) \end{split} \end{equation*}

Thus, we obtain the geometric progression 24,48,96,\cdots by putting n=1,2,3,\cdots. Together with the first three terms that are the slopes of the original triangle, we obtain the entire geometric progression 3,6,12,24,48,96,\cdots.

Now to the positive powers of 3; that is, the geometric progression

    \[1,3,9,27,\cdots,3^{n-1},\cdots\]

Example

Explain how to obtain the entire geometric progression 1,3,9,27,\cdots,3^{n-1},\cdots as slopes.

Easy-peasy.

Begin with a \triangle ABC in which the slopes of the sides are 1,3,9. To this end, take A:=(1,9),B:=(0,0),C:=(4,12). Then the slope of AB is 9, the slope of BC is 3, and the slope of CA is 1.

Next, let X_{n}:\Big(\frac{3^{n+1}-3}{3^{n+1}-1},\frac{3^{n+2}-9}{3^{n+1}-1}\Big). Then, by direct calculation (as in and ), the slope of AX_{n} is 9(3^{n}), for n\geq 1. Therefore, letting n=1,2,3,\cdots successively, we obtain the terms 27,81,243,\cdots. If we now append the slopes of the sides of \triangle ABC (which are 1,3,9), we obtain the entire geometric progression 1,3,9,27,\cdots,3^{n-1},\cdots.

Reciprocal terms

In addition to obtaining a,ar,ar^2,\cdots, ar^{n-1}, we also get \frac{1}{a},\frac{1}{ar},\frac{1}{ar^2},\cdots,\frac{1}{ar^{n-1}}, once we go past a certain point; let’s call it a terminal point (see the exercises at the end \cdots).

Example

Given \triangle ABC with vertices A(1,4),B(0,0),C(3,6), find coordinates for a point R on BC such that the slope of AR is \frac{1}{2}.

In Example we had X_{n}=\Big(\frac{2^{n+1}-2}{2^{n+1}-1},\frac{2^{n+2}-4}{2^{n+1}-1}\Big). Put n=-3, then X_{-3}=(\frac{7}{3},\frac{14}{3}). Take R:=(\frac{7}{3},\frac{14}{3}). Then R is an internal point on BC; further, the slope of AR is

    \[\frac{4-\frac{14}{3}}{1-\frac{7}{3}}=\frac{\frac{-2}{3}}{\frac{-4}{3}}=\frac{1}{2},\]

as desired. Similarly, by putting n=-4,-5,-6\cdots, we obtain the terms \frac{1}{4},\frac{1}{8},\frac{1}{16},\cdots. Thus, inside a simple, single triangle lies the infinite geometric progression

    \[\cdots,\frac{1}{16},\frac{1}{8},\frac{1}{4},\frac{1}{2},1,2,4,8,16,\cdots\]

disguised as slopes.

Example

Given \triangle ABC with vertices A(1,4),B(0,0),C(3,6), let X_{n}=\Big(\frac{2^{n+1}-2}{2^{n+1}-1},\frac{2^{n+2}-4}{2^{n+1}-1}\Big). PROVE that BX_{1}=CX_{-3}.

Put n=1, then X_{1}=(\frac{2}{3},\frac{4}{3}). Put n=-3, then X_{-3}=(\frac{7}{3},\frac{14}{3}), as also found in a previous example. By the distance formula:

    \begin{equation*} \begin{split} BX_{1}&=\sqrt{\Big(\frac{2}{3}-0\Big)^2+\Big(\frac{4}{3}-0\Big)^2}\\ &=\frac{2}{3}\sqrt{5}\\ CX_{-3}&=\sqrt{\Big(3-\frac{7}{3}\Big)^2+\Big(6-\frac{14}{3}\Big)^2}\\ &=\sqrt{\Big(\frac{9}{3}-\frac{7}{3}\Big)^2+\Big(\frac{9}{3}-\frac{14}{3}\Big)^2}\\ &=\sqrt{\Big(-\frac{2}{3}\Big)^2+\Big(-\frac{4}{3}\Big)^2}\\ &=\frac{2}{3}\sqrt{5}\\ &\cdots\vdots\cdots\\ \therefore BX_{1}&=CX_{-3} \end{split} \end{equation*}

Observe that the slopes of the sides of \triangle AX_{1}X_{-3} are \frac{1}{2},2,8; they form a geometric progression with a common ratio of 4, which is the square of the common ratio of the geometric progression 1,2,4 formed by the slopes of the sides of the parent \triangle ABC.

Takeaway

Slopes are usually introduced sometime in grade 9 (or earlier), while geometric sequences are usually studied in grade 11 (or earlier). Here, we’ve managed to show a close connection between the two concepts, using the medium of a triangle.

Throughout we worked with positive integer common ratios. But the common ratios can also be negative, and in such cases things become much more exciting.

Tasks

  1. (Depressed cubic) Consider \triangle ABC with vertices A(1,r^2),B(0,0),C(1+r,r+r^2). PROVE that:
    • the foot of the altitude from vertex A is \Big(\frac{1+r^3}{1+r^2},\frac{r+r^4}{1+r^2}\Big);
    • the length of the altitude from vertex A is h=\frac{r(r-1)}{\sqrt{r^2+1}};
    • the area of \triangle ABC is \frac{1}{2}r(r^2-1) (this becomes a depressed cubic if written in the form \frac{1}{2}(r^3-r). Moreover, if r is an integer (\neq -1,0,1 of course), then this area is always a positive integer).
  2. Let X_{n}=\Big(\frac{2^{n+1}-2}{2^{n+1}-1},\frac{2^{n+2}-4}{2^{n+1}-1}\Big). PROVE that X_{n}\rightarrow(1,2) as n\rightarrow\infty.
  3. Let X_{n}=\Big(\frac{2^{n+1}-2}{2^{n+1}-1},\frac{2^{n+2}-4}{2^{n+1}-1}\Big). PROVE that X_{n}\rightarrow(2,4) as n\rightarrow -\infty.
  4. (Special points) Given \triangle ABC with vertices A(1,4),B(0,0),C(3,6), PROVE that there are (internal) points V and H on BC such that:
    • the slope of AV is undefined and the slope of AH is zero (V for vertical, H for horizontal);
    • V and H trisect side BC (BV=VH=HC);
    • the areas of \triangle ABV,\triangle AVH, \triangle AHC are 1 sq. unit each (making the overall area of \triangle ABC to be 3 sq. units).
  5. (Higher powers) Given \triangle ABC with vertices at A(1,16),B(0,0),C(5,20), the slopes of its sides form a geometric progression with common ratio r_{1}=4. Find two points R and S on BC such that the slopes of the sides of \triangle ARS form a geometric progression with common ratio r_{2}=256~(=4^4).
  6. (Negative region) Given \triangle ABC with vertices A(1,25),B(0,0),C(6,30), find two points V and H on BC such that the slope of any line segment AX, where X is a point between V and H, is negative.
  7. (Base five) Explain how to generate the geometric progression \cdots,\frac{1}{125},\frac{1}{25},\frac{1}{5},1,5,25,125,\cdots using slopes.
  8. (Base six) Explain how to generate the geometric progression \cdots,\frac{1}{216},\frac{1}{36},\frac{1}{6},1,6,36,216,\cdots using slopes.
  9. (Base seven) Explain how to generate the geometric progression -1,-7,-49,-343,\cdots using slopes.
  10. (Terminal point) Given \triangle ABC with vertices A(1,9),B(0,0),C(4,12), determine the coordinates of the terminal point T on BC (a point on BC such that the slope of the line segment AT is \frac{1}{3}; this is also where the slopes of the line segments from vertex A change from positive powers of 3 to negative powers of 3).