A note on geometric sequences – High School Geometry

We’re s--o excited about this!!! As in, E–X–C–I–T–E–D!!! 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:

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 $C(x_1+1+r,y_1+ar+ar^2)=C(3,6)$ .

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 $AX_{1}$

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

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

(Depressed cubic) Consider with vertices . 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).
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 $n\rightarrow\infty$ .
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 $n\rightarrow -\infty$ .
(Special points) Given with vertices , PROVE that there are (internal) points and on 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).
(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 $r_{2}=256~(=4^4)$ .
(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.
(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.
(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.
(Base seven) Explain how to generate the geometric progression $-1,-7,-49,-343,\cdots$ using slopes.
(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 $3$ ).