A special rational function

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

  1. PROVE that if the coordinates of a triangle’s vertices are related according to (3), then its side-slopes form a geometric progression.
  2. 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).
  3. 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}.
  4. 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.
  5. (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.