Slopefessor Fr1day – Page 33 – High School Geometry

The depressed cubic as area

We investigated the area property of our very own triangle, and guess what! The depressed cubic showed up.

Ww!!! How???

Consider $\triangle ABC$ with vertices at $A(1,r^2),B(0,0),C(1+r,r+r^2)$ , where $0,\pm 1\neq r$ . PROVE that its area $\Delta$ can be given by $\Delta=\frac{r^3-r}{2}$ .

Easy-peasy.

We were given the coordinates $A(1,r^2),B(0,0),C(1+r,r+r^2)$ . We simply use the well-known area formula

$\Delta=\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{2}$

and, by taking $(x_1,y_1)=(1,r^2)$ , $(x_2,y_2)=(0,0)$ , $(x_3,y_3)=(1+r,r+r^2)$ , we obtain

$\begin{equation*} \begin{split} \Delta&=\frac{r^2(1+r-0)+0(1-(1+r))+(r+r^2)(0-1)}{2}\\ &=\frac{r^2(1+r)+(r+r^2)(-1)}{2}\\ &=\frac{r^2(r+1)-r(r+1)}{2}\\ &=\frac{(r+1)(r^2-r)}{2}\\ &=\frac{(r-1).r.(r+1)}{2}\\ \textrm{Or,}~\Delta&=\frac{r^3-r}{2} \end{split} \end{equation*}$

Observe that the numerator contains three consecutive factors, namely $r-1,r$ , and $r+1$ . Thus, if $r$ is an integer, then at least one of the three will be even, and so the resulting product $(r-1).r.(r+1)$ will be even. This means that the area $\Big(=\frac{(r-1).r.(r+1)}{2}\Big)$ will be an integer — and a multiple of a triangular number.

The depressed cubic

A cubic equation of the form $\boxed{x^3+Ax=B}$ in which the second degree term $x^2$ is missing, is usually called a depressed cubic. For a discussion on how to solve such equations, see here.

Suppose we re-arrange our area formula $\Delta=\frac{r^3-r}{2}$ as

$r^3-r = 2\Delta,$

then we obtain not just a depressed cubic, but a restricted, simplified, specialized, depressed cubic. As such, its solution also takes a relatively simplified form.

PROVE that the area formula obtained previously, namely $\Delta=\frac{r(r^2-1)}{2}$ , is an odd function.

Recall that an odd function $f(x)$ satisfies $f(-x)=-f(x)$ . Let’s write $\Delta(r)=\frac{r(r^2-1)}{2}$ . Then

$\begin{equation*} \begin{split} \Delta(-r)&=\frac{(-r)\Big((-r)^2-1\Big)}{2}\\ &=\frac{(-r)\Big(r^2-1\Big)}{2}\\ &=-\frac{r\Big(r^2-1\Big)}{2}\\ &=-\Delta(r) \end{split} \end{equation*}$

One implication of this “oddness” is that two opposite values of $r$ will yield the “same” area. As such, when we solve the depressed cubic equation

$r^3-r=2\Delta$

for $r$ , we must remember that any solution we obtain has a corresponding opposite, but the associated opposite doesn’t arise as a solution of the depressed cubic. For example, if $r=2$ is a solution of the depressed cubic, then we must also consider $r=-2$ , despite the fact that $r=-2$ is not a solution of the depressed cubic (see the exercises at the end).

For the depressed cubic $r^3-r=2\Delta$ , PROVE that $r$ can be given in terms of $\Delta$ by: $\boxed{r=\sqrt[3]{\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}}-\sqrt[3]{\frac{-9\Delta+\sqrt{81\Delta^2-3}}{9}}}$

Let’s do this!

Following the standard method of solving a depressed cubic, we seek $s$ and $t$ such that $3st=-1$ and $s^3-t^3=2\Delta$ . Then, $r=s-t$ is a solution of $r^3-r=2\Delta$ . Indeed, we have:

$\begin{equation*} \begin{split} r^3-r&=(s-t)^3-(s-t)\\ &=(s^3-3s^2t+3st^2-t^3)-(s-t)\\ &=(s^3-t^3)-3st(s-t)-(s-t)\\ &=2\Delta-(-1)(s-t)-(s-t)\quad\textrm{by assumption}\\ &=2\Delta+(s-t)-(s-t)\\ &=2\Delta \end{split} \end{equation*}$

So it remains to find such $s$ and $t$ that satisfy $s^3-t^3=2\Delta$ and $3st=-1$ . Isolate $t$ from the second equation: $t=-\frac{1}{3s}$ . Substituting, the equation $s^3-t^3=2\Delta$ becomes:

$s^3-\Big(-\frac{1}{3s}\Big)^3=2\Delta \implies s^3+\frac{1}{27s^3}=2\Delta$

Clear fractions and bring all terms to the left side:

$27s^6-(54\Delta)s^3+1=0 \implies 27(s^3)^2-(54\Delta)s^3+1=0$

Since this is now a quadratic in $s^3$ , we can use the quadratic formula to isolate $s^3$ . Doing this gives:

$s^3=\frac{54\Delta\pm\sqrt{(54\Delta)^2-4\times 27\times 1}}{2\times 27}$

Simplifying,

$s^3=\frac{9\Delta\pm\sqrt{81\Delta^2-3}}{9},~\textrm{or}~s=\sqrt[3]{\frac{9\Delta\pm\sqrt{81\Delta^2-3}}{9}}.$

It suffices to take $s=\sqrt[3]{\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}}$ . Then, $s^3-t^3=2\Delta$ gives $t^3=s^3-2\Delta$ :

$\begin{equation*} \begin{split} t^3&=\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}-2\Delta\\ &=\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}-\frac{18\Delta}{9}\\ &=\frac{-9\Delta+\sqrt{81\Delta^2-3}}{9}\\ \implies t&=\sqrt[3]{\frac{-9\Delta+\sqrt{81\Delta^2-3}}{9}} \end{split} \end{equation*}$

Finally,

$r=s-t=\sqrt[3]{\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}}-\sqrt[3]{\frac{-9\Delta+\sqrt{81\Delta^2-3}}{9}}.$

Using the preceding example, PROVE that $\Delta=\frac{\sqrt{3}}{9}$ if, and only if, $r=\frac{2\sqrt{3}}{3}$ .

First suppose that $\Delta=\frac{\sqrt{3}}{9}$ . Then $81\Delta^2-3=0$ . And so

$r=\sqrt[3]{\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}}-\sqrt[3]{\frac{-9\Delta+\sqrt{81\Delta^2-3}}{9}}$

reduces to

$\begin{equation*} \begin{split} r&=\sqrt[3]{\Delta}-\sqrt[3]{-\Delta}\\ &=2\sqrt[3]{\Delta}\\ &=2\sqrt[3]{\frac{\sqrt{3}}{9}}\\ &=\frac{2\sqrt{3}}{3} \end{split} \end{equation*}$

Conversely, if $r=\frac{2\sqrt{3}}{3}$ , then from the depressed cubic equation $r^3-r=2\Delta$ , we obtain

$\begin{equation*} \begin{split} \Delta&=\frac{r^3-r^}{2}\\ &=\frac{r(r^2-1)}{2}\\ &=\frac{\frac{2\sqrt{3}}{3}\Big[\Big(\frac{2\sqrt{3}}{3}\Big)^2-1\Big]}{2}\\ &=\frac{2\sqrt{3}}{3}\times\frac{1}{3}\times\frac{1}{2}\\ &=\frac{\sqrt{3}}{9} \end{split} \end{equation*}$

Discuss the nature of the roots of the depressed cubic $r^3-r=2\Delta$ .

We use the cubic discriminant to this end. For a general cubic $ax^3+bx^2+cx+d$ , its discriminant is given by the quantity

$\boxed{b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd}$

Putting $a=1$ , $b=0$ , $c=-1$ , and $d=-2\Delta$ , we obtain, for the depressed cubic $r^3-r=2\Delta$ :

$0^2(-1)^2-4(1)(-1)^3-4(0^3)(-2\Delta)-27(1^2)(-2\Delta)^2+18(1)(0)(-1)(-2\Delta),$

which reduces to $4-108\Delta^2$ . Consequently:

if $4-108\Delta^2=0$ , then the depressed cubic has three real roots, one of which is repeated;
if $4-108\Delta^2> 0$ , then the depressed cubic has three distinct real roots;
if $4-108\Delta^2< 0$ , the depressed cubic has one real root and two complex roots.

Solve the depressed cubic equation $r^3-r=2\frac{\sqrt{3}}{9}$ .

Observe that this is the case in which $\Delta=\frac{\sqrt{3}}{9}$ . The corresponding cubic discriminant $4-108\Delta^2$ becomes $4-108\times\frac{1}{27}=0$ . As such, we expect a repeated root. From Example 4, one of the roots is $r=\frac{2\sqrt{3}}{3}$ . Accordingly, $\Big(r-\frac{2\sqrt{3}}{3}\Big)$ is a factor of $r^3-r-2\frac{\sqrt{3}}{9}$ . By long division, we get the quadratic factor $\Big(r^2+\frac{2\sqrt{3}}{3}r+\frac{1}{3}\Big)$ , which is a perfect square:

$\Big(r^2+\frac{2\sqrt{3}}{3}r+\frac{1}{3}\Big)=\Big(r+\frac{\sqrt{3}}{3}\Big)^2.$

Thus, the complete solution to $r^3-r=2\frac{\sqrt{3}}{9}$ is $r=-\frac{\sqrt{3}}{3},~\frac{2\sqrt{3}}{3}$ . Notice how this solution set becomes “increased” in the next example.

Example

$\triangle ABC$ has an area of $\Delta=\frac{\sqrt{3}}{9}$ . The slopes of its sides form a geometric sequence $1,r,r^2$ . Find possible coordinates for the vertices $A,B,C$ .

Since the slopes of the sides form the geometric sequence $1,r,r^2$ , we can place the vertices at $A(1,r^2)$ , $B(0,0)$ , and $C(1+r,r+r^2)$ . The area will then be $\frac{r^3-r}{2}$ , in view of Example 1. But we’re given that the area is $\Delta=\frac{\sqrt{3}}{9}$ :

$\frac{r^3-r}{2}=\frac{\sqrt{3}}{9}\implies r^3-r=2\frac{\sqrt{3}}{9},$

which is the depressed cubic solved in the preceding example. Recall that the cubic has $r=\frac{-\sqrt{3}}{3},~\frac{2\sqrt{3}}{3}$ as solutions.

Being that the area formula $\Delta=\frac{r^3-r}{2}$ is an odd function (Example 2), we should also consider $r=\frac{\sqrt{3}}{3},~-\frac{2\sqrt{3}}{3}$ as possible values for the common ratio $r$ , even though these are not solutions to the depressed cubic.

for $r=\frac{-\sqrt{3}}{3}$ , the coordinates will be $A\left(1,\frac{1}{3}\right)$ , $B(0,0)$ , $C\left(1-\frac{\sqrt{3}}{3},~-\frac{\sqrt{3}}{3}+\frac{1}{3}\right)=\left(\frac{3-\sqrt{3}}{3},\frac{1-\sqrt{3}}{3}\right)$ ;
for $r=\frac{\sqrt{3}}{3}$ , the coordinates will be $A\left(1,\frac{1}{3}\right)$ , $B(0,0)$ , $C\left(1+\frac{\sqrt{3}}{3},~\frac{\sqrt{3}}{3}+\frac{1}{3}\right)=\left(\frac{3+\sqrt{3}}{3},\frac{1+\sqrt{3}}{3}\right)$ ;
for $r=\frac{-2\sqrt{3}}{3}$ , the coordinates will be $A\left(1,\frac{4}{3}\right)$ , $B(0,0)$ , $C\left(1-\frac{2\sqrt{3}}{3},~-\frac{2\sqrt{3}}{3}+\frac{4}{3}\right)=\left(\frac{3-2\sqrt{3}}{3},\frac{4-2\sqrt{3}}{3}\right)$ ;
for $r=\frac{2\sqrt{3}}{3}$ , the coordinates will be $A\left(1,\frac{4}{3}\right)$ , $B(0,0)$ , $C\left(1+\frac{2\sqrt{3}}{3},~\frac{2\sqrt{3}}{3}+\frac{4}{3}\right)=\left(\frac{3+2\sqrt{3}}{3},\frac{4+2\sqrt{3}}{3}\right)$ .

Since the area $\Delta=\frac{\sqrt{3}}{9}$ is so small, each case produces a rare-listic diagram when drawn.

Example (“Fixed point”)

Find coordinates for the vertices of a $\triangle ABC$ whose slopes are $1,r,r^2$ and whose area is $\sqrt{3}$ sq. units.

We solve the cubic equation $r^3-r=2\Delta$ using $\Delta=\sqrt{3}$ :

$r^3-r=2\sqrt{3}\implies (r-\sqrt{3})(r^2+\sqrt{3}r+2)=0\implies r=\sqrt{3};$

the quadratic part has complex roots. Again, we should consider $r=-\sqrt{3}$ , despite the fact that it’s not a solution to the current cubic equation.

Next, let’s use the coordinates $A(1,r^2)$ , $B(0,0)$ , $C(1+r,r+r^2)$ :

$A(1,3),B(0,0),C(1+\sqrt{3},3+\sqrt{3}),\quad\textrm{OR}\quad A(1,3),B(0,0),C(1-\sqrt{3},3-\sqrt{3}).$

In this case, the area and the common ratio are the same.

Give an example of a scalene triangle with non-zero side slopes, irrational side lengths, and an integer area.

The only challenge with this question is the non-zero slopes requirement; otherwise we could just have taken a triangle with one side on the $x$ -axis and the rest becomes easy.

Luckily, we can always turn to triangles with slopes in geometric progressions. In this case we take $\triangle ABC$ with vertices at $A(1,4),B(0,0),C(3,6)$ . The side slopes are $1,2,4$ . For the side lengths we have $AB=\sqrt{17}$ , $BC=3\sqrt{5}$ , and $CA=2\sqrt{2}$ , all irrational and all different, so the triangle is scalene. Its area is $3$ sq. units.

Let $\alpha$ be a root of the depressed cubic $r^3-r-2\Delta=0$ . PROVE that $r^3-r-2\Delta=(r-\alpha)(r^2+\alpha r+\frac{2\Delta}{\alpha})$ .

Since $\Delta\neq 0$ (an area), it follows that $\alpha\neq 0$ as well. This ensures that the quantity $\frac{2\Delta}{\alpha}$ is defined.

Set

$r^3-r-2\Delta=(r-\alpha)(r^2+br+c)=r^3+(b-\alpha)r^2+(c-\alpha b)r-\alpha c.$

Comparing coefficients of like terms, $b-\alpha=0,~c-\alpha b=-1$ , and $\alpha c=2\Delta$ . So $b=\alpha$ , $c=\alpha b-1=\alpha^2-1$ , and again $c=\frac{2\Delta}{\alpha}$ . The two expressions for $c$ yield $\alpha(\alpha^2-1)=2\Delta$ , which is valid because $\alpha$ is a root of the original cubic $r^3-r-2\Delta=0$ .

$\therefore r^3-r-2\Delta=(r-\alpha)(r^2+\alpha r+\frac{2\Delta}{\alpha}).$

Takeaway

If $\triangle ABC$ with vertices at $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ is such that the slopes of sides $AB,BC,CA$ are $a,ar,ar^2$ , respectively, then the area $\Delta$ of the triangle can be given by

$\Delta=\frac{1}{2}\Big(\frac{r-1}{r+1}\Big)\Big(\frac{1}{ar}\Big)(y_2-y_3)^2.$

By restricting the coordinates in a suitable way (see the exercises below), we obtain the depressed cubic $\frac{1}{2}r(r^2-1)$ as the area.

Tasks

(Special square) For any right triangle with vertices $(\alpha_1,\beta_1)$ , $(\alpha_2,\beta_2)$ , and $(\alpha_3,\beta_3)$ , PROVE that: $\Big(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_3|+|\beta_3-\beta_1|}\Big)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.$
(Notice how squaring respects addition in this case. In addition, not all four absolute values on the left are absolutely necessary $\cdots\vdots\cdots$ .)
(Special square) For any triangle with centroid $(\alpha_1,\beta_1)$ , circumcenter $(\alpha_2,\beta_2)$ , and orthocenter $(\alpha_3,\beta_3)$ , PROVE that: $\Big(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_3)+(\beta_3-\beta_1)}\Big)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.$
(Notice the absolute absence of absolute values in this case — so long as $(\alpha_2-\alpha_1)+(\beta_2-\beta_1)\neq 0$ and $(\alpha_3-\alpha_1)+(\beta_3-\beta_1)\neq 0$ . Also, any right triangle satisfies this property “TWICE” — with its regular vertices, and also with its three principal centers.)
(Opposite roots) PROVE that if a depressed cubic $x^3+Ax=B$ has two opposite solutions, namely $x=\alpha$ and $x=-\alpha$ , then $B=0$ .
(In our own case, the right member $B$ represents the area $\Delta$ of a certain, obscure triangle. Since $\Delta\neq 0$ , this explains why the depressed cubic equation $r^3-r=2\Delta$ cannot have two opposite solutions.)
has vertices at . Assume that the slopes of sides are , in that order.
- PROVE that its area $\Delta$ can be given by $\Delta=\frac{1}{2}\Big(\frac{r-1}{r+1}\Big)\Big(\frac{1}{ar}\Big)(y_2-y_3)^2.$
- Find an appropriate choice of $(y_2-y_3)$ for which the area $\Delta$ reduces to the depressed cubic $\Delta=\frac{a}{2}r(r^2-1).$
has vertices at , , . PROVE that:
- the coordinates of the foot of the altitude from vertex $A$ is $\Big(x_1+\frac{a^2r^3+1}{a^2r^2+1},~y_1+\frac{a^3r^4+ar}{a^2r^2+1}\Big)$ ;
- the length of the altitude from vertex $A$ is $h=\frac{ar(r-1)}{a^2r^2+1}$ ;
- the area of $\triangle ABC$ is $\frac{a}{2}(r-1)(r)(r+1)=\frac{a}{2}r(r^2-1)$ .
Consider with vertices at , , . Its side slopes are , a geometric progression in which and .
- Use the area formula in exercise 4 above to compute its area;
- Use the area formula in exercise 5 above to compute its area;
- Compute its area in the regular way, using $\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{2}$ ;
- Explain why the formula in exercise 5 did not work correctly in this case.
For with vertices at , , , PROVE that:
- its centroid is located at $\Big(\frac{2+r}{3},\frac{r+2r^2}{3}\Big)$ ;
- its circumcenter is at $\Big(\frac{r^3+r^2+r-1}{2(r-1)},\frac{r^3-r^2-r-1}{2(r-1)}\Big)$ ;
- its orthocenter is at $\Big(\frac{1+r^3}{1-r},\frac{-1-r^3}{1-r}\Big)$ ;
- the slope of its Euler line is $m_{E}=-\Big(\frac{r^3+r^2+r+3}{3r^3+r^2+r+1}\Big)$ .
(Same ratio) Given $A(1,r^2)$ , $B(0,0)$ , $C(1+r,r+r^2)$ , PROVE that the centroid, the circumcenter, and the orthocenter of $\triangle ABC$ lie on parallel lines $x+y=\frac{2}{3}(1+r+r^2)$ , $x+y=1+r+r^2$ , and $x+y=0$ , respectively.
(Notice that the line through the centroid is in the middle of the other two lines. Moreover, these lines preserve the usual $1:2$ ratio in which the centroid divides the circumcenter-orthocenter distance.)
If $\sqrt[3]{a+\sqrt{b}}=c+\sqrt{d}$ , PROVE that $\sqrt[3]{-a+\sqrt{b}}=-c+\sqrt{d}$ .
(Unit area) Find coordinates for the vertices of $\triangle ABC$ with slopes $1,r,r^2$ and an area of $1$ sq. unit.

A note on geometric sequences

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 $3$ ).