A special rational function – High School Geometry

A depressed cubic, followed by two special quadratics, and now a special rational function:

(1) $\begin{equation*} f(x)=\frac{2x}{x^2+1} \end{equation*}$

Everything about our triangle is special. For anyone without clue as to how the above rational function can be construed from a geometric point of view, our triangle is here to the rescue.

Slipshod notation

Let $\triangle ABC$ be such that sides $AB,BC,CA$ have slopes $a,ar,ar^2$ , as usual.

Denote the length of the altitude from vertex $A$ by $h_A$ and the length of the median from vertex $A$ by $m_A$ . We prove the following height-median-ratio:

(2) $\begin{equation*} \frac{h_A}{m_A}= \frac{2(ar)}{(ar)^2+1} \end{equation*}$

Simple deductions

Example

PROVE that $h_A\leq m_A$ .

We expect this to be the case — $h_A$ is the altitude from vertex $A$ , so it is a shorter distance compared to $m_A$ .

Alternatively, we can use the most basic form of the AGM inequality:

$\frac{h_A}{m_A}=\frac{2(ar)}{(ar)^2+1}\leq \frac{(ar)^2+1}{(ar)^2+1}=1 .$

Suppose that the sides $AB,BC,CA$ of $\triangle ABC$ have slopes $a,ar,ar^2$ , respectively. PROVE that the length of the altitude from vertex $A$ coincides with the length of the median from vertex $A$ if and only if $ar=1$ .

Obviously, if $ar=1$ , then the geometric progression is essentially $\frac{1}{a},1,a$ . These slopes yield an isosceles triangle, from which the conclusion follows.

The advantage of having equation (2) is that it allows for “algebraic” argument. For example, recall that $\frac{h_A}{m_A}=\frac{2(ar)}{(ar)^2+1}$ . If $h_A=m_A$ , then $(ar)^2+1=2ar$ . That is, $(ar-1)^2=0$ . And so $ar=1$ . Conversely, if $ar=1$ , then $h_A=\frac{2(ar)m_A}{(ar)^2+1}=\frac{2(1)m_A}{1^2+1}=m_A$ .

Systematic derivation

Many of the nice properties of triangles with slopes in geometric progression ( $a,ar,ar^2$ ) hinge on the following relations among the coordinates ( $x_1\neq x_2\neq x_3,~y_1\neq y_2\neq y_3$ ):

(3) $\begin{equation*} x_1=x_3+\frac{y_2-y_3}{ar(r+1)},~y_1=\frac{ry_2+y_3}{r+1},~x_2=x_3+\frac{y_2-y_3}{ar} \end{equation*}$

In particular, what we establish below depend on these relations.

Let $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_2)$ be the vertices of $\triangle ABC$ in which sides $AB,BC,CA$ have slopes $a,ar,ar^2$ . PROVE that $BC=\frac{\sqrt{(ar)^2+1}}{ar}(y_2-y_3)$ .

Normally, $BC^2=(x_2-x_3)^2+(y_2-y_3)^2$ . Using (3), we get:

$\begin{equation*} \begin{split} BC^2&=\left(x_3+\frac{y_2-y_3}{ar}-x_3\right)^2+(y_2-y_3)^2\\ &=(y_2-y_3)^2\left(\frac{(ar)^2+1}{(ar)^2}\right)\\ \therefore BC&=\frac{(y_2-y_3)\sqrt{(ar)^2+1}}{ar} \end{split} \end{equation*}$

Let $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_2)$ be the vertices of $\triangle ABC$ in which sides $AB,BC,CA$ have slopes $a,ar,ar^2$ . Find the equation of the altitude from vertex $A$ .

The slope of the altitude from vertex $A$ is $-\frac{1}{ar}$ , since the slope of side $BC$ is $ar$ . Thus the equation of this altitude is

$y-y_1=-\frac{1}{ar}(x-x_1).$

Using the expressions for $x_1$ and $y_1$ in equation (3), we obtain:

(4) $\begin{equation*} y-\left(\frac{ry_2+y_3}{r+1}\right)=-\frac{1}{ar}\left(x-x_3-\frac{y_2-y_3}{ar(r+1)}\right) \end{equation*}$

Let $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_2)$ be the vertices of $\triangle ABC$ in which sides $AB,BC,CA$ have slopes $a,ar,ar^2$ . Find the coordinates of the foot of the altitude from vertex $A$ .

The equation of side $BC$ is $y-y_3=ar(x-x_3)$ . The altitude from vertex $A$ has equation given by (4), namely:

$y-\left(\frac{ry_2+y_3}{r+1}\right)=-\frac{1}{ar}\left(x-x_3-\frac{y_2-y_3}{ar(r+1)}\right)$

Solving, we obtain

$x=x_3+\frac{(a^2r^3+1)(y_2-y_3)}{ar(r+1)(a^2r^2+1)},~y=y_3+\frac{(a^2r^3+1)(y_2-y_3)}{(r+1)(a^2r^2+1)}$

Let $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_2)$ be the vertices of $\triangle ABC$ in which sides $AB,BC,CA$ have slopes $a,ar,ar^2$ . PROVE that length of the altitude from vertex $A$ is $h_A=\left(\frac{r-1}{r+1}\right)\frac{y_2-y_3}{\sqrt{a^2r^2+1}}$ .

We find the distance $h_A$ from $A(x_1,y_1)=\left(x_3+\frac{y_2-y_3}{ar(r+1)},~\frac{ry_2+y_3}{r+1}\right)$ to the foot of the altitude

$\left(x_3+\frac{(a^2r^3+1)(y_2-y_3)}{ar(r+1)(a^2r^2+1)},~y_3+\frac{(a^2r^3+1)(y_2-y_3)}{(r+1)(a^2r^2+1)}\right)$

The $x$ -difference is

$x_3+\frac{(a^2r^3+1)(y_2-y_3)}{ar(r+1)(a^2r^2+1)}-\left(x_3+\frac{y_2-y_3}{ar(r+1)}\right)=\frac{ar(r-1)(y_2-y_3)}{(r+1)(a^2r^2+1)}$

The $y$ -difference is

$y_3+\frac{(a^2r^3+1)(y_2-y_3)}{(r+1)(a^2r^2+1)}-\left(\frac{ry_2+y_3}{r+1}\right)=\frac{(1-r)(y_2-y_3)}{(r+1)(a^2r^2+1)}$

By the distance formula:

$\begin{equation*} \begin{split} h_A^2&=\left(\frac{ar(r-1)(y_2-y_3)}{(r+1)(a^2r^2+1)}\right)^2+\left(\frac{(1-r)(y_2-y_3)}{(r+1)(a^2r^2+1)}\right)^2\\ &=\left(\frac{(r-1)(y_2-y_3)}{(r+1)(a^2r^2+1)}\right)^2(a^2r^2+1)\\ \therefore h_A&=\left(\frac{r-1}{r+1}\right)\frac{y_2-y_3}{\sqrt{a^2r^2+1}} \end{split} \end{equation*}$

Let $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_2)$ be the vertices of $\triangle ABC$ in which sides $AB,BC,CA$ have slopes $a,ar,ar^2$ . PROVE the height-median-ratio: $\frac{h_A}{m_A}=\frac{2(ar)}{(ar)^2+1}$ .

We use example 3 and example 6 and part of our previous post:

$\begin{equation*} \begin{split} BC&=\frac{\sqrt{(ar)^2+1}}{ar}(y_2-y_3)\\ h_A&=\left(\frac{r-1}{r+1}\right)\frac{y_2-y_3}{\sqrt{a^2r^2+1}}\\ m_A&=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC \end{split} \end{equation*}$

to obtain

$h_A=\frac{2m_A}{BC}\times\frac{1}{\sqrt{a^2r^2+1}}\times \frac{BC\times ar}{\sqrt{a^2r^2+1}}$

and then

$\frac{h_A}{m_A}=\frac{2(ar)}{(ar)^2+1}.$

Sample applications

No doubt you already know the drill.

Find coordinates for the vertices of a $\triangle ABC$ in which the length of the altitude from vertex $A$ is $\frac{3}{5}$ the length of the median from vertex $A$ .

Without the organized procedure that equation (2) provides, solving this type of problem may be somewhat random. So as our custom is, we use what we’ve got. In equation (2), set

$\frac{2(ar)}{(ar)^2+1}=\frac{3}{5}$

and solve the resulting quadratic for $ar$ :

$\begin{equation*} \begin{split} \frac{2(ar)}{(ar)^2+1}&=\frac{3}{5}\\ 3(ar)^2-10(ar)+3&=0\\ (ar-3)(3ar-1)&=0\\ ar&=3,\frac{1}{3} \end{split} \end{equation*}$

Fix $ar=3$ . It now remains to find the first and third terms of a geometric progression with second term $3$ . For simplicity, let’s choose $a=1$ . Then the third term will be $9$ . The basic set of coordinates are:

$(0,0),~(1,r^2),~(r+1,r+r^2)\implies (0,0),~(1,9),~(4,12).$

Choose $A(1,9),~B(4,12),~C(0,0)$ so that the slopes of $AB,BC,CA$ are $1,3,9$ as per our set up. The length of the median from vertex $A$ is $\sqrt{10}$ , and the length of the altitude from vertex $A$ is $\frac{3}{5}\sqrt{10}$ .

Find coordinates for the vertices of a $\triangle ABC$ in which the length of the altitude from vertex $A$ is $\frac{12}{13}$ the length of the median from vertex $A$ .

In equation (2), set

$\frac{2(ar)}{(ar)^2+1}=\frac{12}{13}$

and solve the resulting quadratic for $ar$ :

$\begin{equation*} \begin{split} \frac{2(ar)}{(ar)^2+1}&=\frac{12}{13}\\ 12(ar)^2-26(ar)+12&=0\\ 2(2ar-3)(3ar-2)&=0\\ ar&=\frac{3}{2},\frac{2}{3} \end{split} \end{equation*}$

Fix $ar=\frac{3}{2}$ . We now want a three-term geometric progression with $\frac{3}{2}$ as the second term. For simplicity, choose $a=1$ . Then we have $1,\frac{3}{2},\frac{9}{4}$ as the progression. The corresponding coordinates, using the most basic form, will be $(0,0)$ , $\left(1,\frac{3}{2}\right)$ , and $\left(\frac{5}{2},\frac{15}{4}\right)$ .

$\triangle ABC$ has coordinates at $A(1,16)=(x_1,y_1)$ , $B(5,20)=(x_2,y_2)$ , $C(0,0)=(x_3,y_3)$ . Find the length of the altitude from vertex $A$ .

The side-slopes are $1,4,16$ for $AB,BC,CA$ . They form a geometric progression in which $a=1$ and $r=4$ . The length of the altitude from vertex $A$ in such a case is given (see example 6) by

$h_A=\left(\frac{r-1}{r+1}\right)\frac{y_2-y_3}{\sqrt{a^2r^2+1}}=\left(\frac{4-1}{4+1}\right)\frac{20-0}{\sqrt{1^2\times 4^2+1}}=\frac{12}{\sqrt{17}}$

Takeaway

Be careful with the sign of $ar$ in the rational expression:

$\frac{h_A}{m_A}=\frac{2(ar)}{(ar)^2+1}$

because $ar$ can be negative, whereas the ratio on the left is positive.

Note also that the “strange” formulas we state work only in the case of triangles with slopes in geometric progression. (By the way, it now seems that these triangles provide a platform for studying some basic functions.)

Tasks

PROVE that if the coordinates of a triangle’s vertices are related according to (3), then its side-slopes form a geometric progression.
Let $f(x)=\frac{2x}{x^2+1}$ as in equation (1). For $x\neq 0$ , PROVE that $f(x)=f\left(\frac{1}{x}\right)$ .
Consider $\triangle ABC$ in which $\angle B=90^{\circ},~AB=1$ unit and $BC=x$ units. If $\angle BCA=\theta$ , PROVE that $\sin(2\theta)=\frac{2x}{x^2+1}$ .
Find coordinates for the vertices of a triangle $ABC$ in which the length of the altitude from vertex $A$ is $\frac{24}{25}$ the length of the median from vertex $A$ .
(Explicit error) Find coordinates for the vertices of $\triangle ABC$ for which the length of the altitude from vertex $A$ is twice the length of the median from vertex $A$ .