An identity involving altitudes – High School Geometry

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.)