Slopefessor Fr1day – Page 28 – High School Geometry

An identity involving altitudes

We’re getting close to reaching (part of) our goal of having a clone of the right triangle, as this equation shows:

(1) $\begin{equation*} \frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{1}{h_A^2}\frac{r^2+1}{(r+1)^2}. \end{equation*}$

Ve $r$ y close. Ve $r(\rightarrow 0)$ y close. Ve $r(\rightarrow\infty)$ y close.

Succinct strategy

In $\triangle ABC$ , let $k,kr,kr^2$ be the slopes of sides $AB,BC,CA$ . Let $h_A,h_B,h_C$ be the lengths of the altitudes from each vertex. We’ll derive equation (1) in two ways: The first approach is straight-forward, utilizing a previous post and a fact from this site, namely

(2) $\begin{equation*} h_A=\frac{bc}{2R},~h_B=\frac{ac}{2R},~h_C=\frac{ab}{2R},\end{equation*}$

where $R$ is the radius of the circumcircle of $\triangle ABC$ and $a,b,c$ are the side-lengths. The second method is long-winded, painstaking, old-fashioned.

Example 1

In any $\triangle ABC$ , PROVE that $\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{1}{h_A^2}\left(\frac{b^2+c^2}{a^2}\right)$ .

Using equation (2), we have:

$\begin{equation*} \begin{split} \frac{1}{h_B^2}+\frac{1}{h_C^2}&=\frac{4R^2}{a^2c^2}+\frac{4R^2}{a^2b^2}\\ &=\frac{4R^2}{a^2}\left(\frac{b^2+c^2}{b^2c^2}\right)\\ &=\frac{4R^2}{b^2c^2}\left(\frac{b^2+c^2}{a^2}\right)\\ &=\left(\frac{2R}{bc}\right)^2\left(\frac{b^2+c^2}{a^2}\right)\\ &=\frac{1}{h_A^2}\left(\frac{b^2+c^2}{a^2}\right). \end{split} \end{equation*}$

It follows immediately that in a right triangle with $\angle A=90^{\circ}$ we have

$\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{1}{h_A^2}.$

Example 2

PROVE that the equation

$\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{1}{h_A^2}\frac{r^2+1}{(r+1)^2}$

holds in a $\triangle ABC$ with $k,kr,kr^2$ as the slopes of sides $AB,BC,CA$ .

We use an identity from one of our previous posts, namely

$AB^2+AC^2=\frac{r^2+1}{(r+1)^2}BC^2.$

In terms of the standard notation, this is the same as $c^2+b^2=\frac{r^2+1}{(r+1)^2}a^2$ . Or:

$\frac{b^2+c^2}{a^2}=\frac{r^2+1}{(r+1)^2}$

In view of example 1, we then have:

$\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{1}{h_A^2}\frac{r^2+1}{(r+1)^2}.$

Point-blank.

Verbose verification

In $\triangle ABC$ with slopes $a,ar,ar^2$ for sides $AB,BC,CA$ , find the length of the altitude from vertex $A$ . Take $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ .

The length of the altitude from vertex $A$ is

$h_A=\frac{r-1}{r+1}\frac{y_2-y_3}{\sqrt{a^2r^2+1}}.$

(Absolute values may be necessary. Also notice how we switched the slopes from $k,kr,kr^2$ to $a,ar,ar^2$ .)

In $\triangle ABC$ with slopes $a,ar,ar^2$ for sides $AB,BC,CA$ , find the length of the altitude from vertex $B$ . Take $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ .

It is:

$h_B=(r-1)\frac{y_2-y_3}{\sqrt{a^2r^4+1}}.$

(Again, absolute values may be necessary.)

In $\triangle ABC$ with slopes $a,ar,ar^2$ for sides $AB,BC,CA$ , find the length of the altitude from vertex $C$ . Take $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ .

It is:

$h_C=\frac{r-1}{r}\frac{y_2-y_3}{\sqrt{a^2+1}}.$

(Absolute values may be necessary.)

Using examples 3 to 5, derive equation (1) directly.

From examples 3,4,5 we had:

$h_A=\frac{r-1}{r+1}\frac{y_2-y_3}{\sqrt{a^2r^2+1}},~h_B=(r-1)\frac{y_2-y_3}{\sqrt{a^2r^4+1}},~h_C=\frac{r-1}{r}\frac{y_2-y_3}{\sqrt{a^2+1}}.$

Thus:

$\begin{equation*} \begin{split} \frac{1}{h_A^2}&=\frac{(a^2r^4+1)+(a^2r^2+r^2)+2r(a^2r^2+1)}{(r-1)^2(y_2-y_3)^2}\\ \frac{1}{h_A^2}&=\frac{1}{h_B^2}+\frac{1}{h_C^2}+\frac{2r(a^2r^2+1)}{(a^2r^2+1)(r+1)^2h_A^2}\\ \left(1-\frac{2r}{(r+1)^2}\right)\frac{1}{h_A^2}&=\frac{1}{h_B^2}+\frac{1}{h_C^2}\\ \therefore \frac{r^2+1}{(r+1)^2}\frac{1}{h_A^2}&=\frac{1}{h_B^2}+\frac{1}{h_C^2} \end{split} \end{equation*}$

Given $\triangle ABC$

with vertices at $A(1,4)$

, $B(-1,2)$

, and $C(0,0)$

, verify that the three altitudes satisfy equation (1).

The slopes of sides $AB,BC,CA$ are $1,-2,4$ , respectively. They form a geometric progression in which $r=-2$ . As per our arrangement, we’ll take $B(x_2,y_2)=(-1,2)$ and $C(x_3,y_3)=(0,0)$ . Therefore the three altitudes are:

$h_A^2=\frac{36}{5},~h_B^2=\frac{36}{17},~h_C^2=\frac{9}{2}.$

And so

$\begin{equation*} \begin{split} \frac{1}{h_B^2}+\frac{1}{h_C^2}&=\frac{17}{36}+\frac{2}{9}\\ &=\frac{25}{36}\\ \frac{r^2+1}{(r+1)^2}\frac{1}{h_A^2}&=\frac{(-2)^2+1}{(-2+1)^2}\times\frac{5}{36}\\ &=\frac{25}{36} \end{split} \end{equation*}$

Given $\triangle ABC$

with vertices at $A(1,4)$

, $B(3,6)$

, and $C(0,0)$

, verify that the three altitudes satisfy equation (1).

The slopes of sides $AB,BC,CA$ are $1,2,4$ , respectively. They form a geometric progression in which $r=2$ . As per our arrangement, we’ll take $B(x_2,y_2)=(3,6)$ and $C(x_3,y_3)=(0,0)$ . Therefore the three altitudes are:

$h_A^2=\frac{4}{5},~h_B^2=\frac{36}{17},~h_C^2=\frac{9}{2}.$

And so

$\begin{equation*} \begin{split} \frac{1}{h_B^2}+\frac{1}{h_C^2}&=\frac{17}{36}+\frac{2}{9}\\ &=\frac{25}{36}\\ \frac{r^2+1}{(r+1)^2}\frac{1}{h_A^2}&=\frac{2^2+1}{(2+1)^2}\times\frac{5}{4}\\ &=\frac{25}{36} \end{split} \end{equation*}$

Find possible coordinates for the vertices of a triangle $ABC$

whose three altitudes are related according to $\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{5}{h_A^2}$

Familiar, friendly, and trendy. Set

$\frac{r^2+1}{(r+1)^2}=5\implies r=-2,-\frac{1}{2}.$

Therefore, any triangle $ABC$ whose side-slopes form a geometric progression with common ratio $r=-2$ will have three altitudes satisfying $\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{5}{h_A^2}$ . One such triangle has vertices at $A(1,4)$ , $B(-1,2)$ , and $C(0,0)$ .

Find possible coordinates for the vertices of a triangle $ABC$

whose three altitudes are related according to $\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{5}{8h_A^2}$

Familiar, friendly, trendy, and easy. Set

$\frac{r^2+1}{(r+1)^2}=\frac{5}{8}\implies r=3,\frac{1}{3}.$

Therefore, any triangle $ABC$ whose side-slopes form a geometric progression with common ratio $r=3$ will have three altitudes satisfying $\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{5}{8h_A^2}$ . One such triangle has vertices at $A(1,9)$ , $B(4,12)$ , and $C(0,0)$ . And there are many — infinitely many — others.

Takeaway

In terms of side-lengths, median-lengths, and slopes, we’ve shown how our triangle closely resembles the right type. Today, the list has now grown to include altitudes.

If anything happens to the right triangle, no need to groan because we already got its clone.

Tasks

PROVE that $\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{1}{h_A^2}$ in a $\triangle ABC$ with $\angle A=90^{\circ}$ .
Find possible coordinates for the vertices of $\triangle ABC$ whose three altitudes are related via $\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{13}{18h_A^2}.$
Find possible coordinates for the vertices of $\triangle ABC$ whose three altitudes are related via $\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{5}{2h_A^2}.$
For with vertices at , , and , verify that:
- the slopes form a geometric progression with common ratio $r=3$ ;
- $h_A=\frac{24}{\sqrt{37}},~h_B=\frac{96}{\sqrt{325}},~h_C=\frac{32}{\sqrt{5}}$ ;
- $\frac{1}{h_B^2}+\frac{1}{h_C^2}=\frac{5}{8h_A^2}$ .
(Triple arithmetic) For any triangle in which the slopes of the sides form a geometric progression with common ratio satisfying , PROVE that:
- the squares of the side-lengths form an arithmetic progression;
- the squares of the median-lengths form an arithmetic progression;
- the reciprocals of the squares of the the altitude-lengths form an arithmetic progression.
  (An equilateral triangle belongs here. But there are scalene triangles that belong here too. A right isosceles triangle in which $r=-2$ satisfies the first and third conditions above.)

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

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 $A$ .