General – Page 29 – High School Geometry

Half ratio times length

The derivation of the formula

(1) $\begin{equation*} m_A=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC \end{equation*}$

will be outlined in example 9, and will span (about) nine lines. Meanwhile, the formula and the title are not quite aligned $\cdots$

Copious verification

In equation (1) above, $m_A$ is the length of the median from vertex $A$ , and $BC$ is the length of side $BC$ . Our first four examples confirm that the relation (1) should hold for all values of $r$ , except the unwanted ones: $-1,0,1.$

Example

Consider $\triangle ABC$ with vertices at $A\left(1,\frac{9}{4}\left)$ , $B\left(\frac{5}{2},\frac{15}{4}\right)$ , and $C(0,0)$ . Find the length of the median from vertex $A$ , and verify that equation (1) is satisfied.

The slopes of sides $AB,BC,CA$ are $1,\frac{3}{2},\frac{9}{4}$ , respectively. They form a geometric progression in which $r=\frac{3}{2}$ .

Let’s first find the length of the median from vertex $A$ . The midpoint of side $BC$ is at $\left(\frac{5}{4},\frac{15}{8}\right)$ . Together with $A\left(1,\frac{9}{4}\left)$ , the distance formula gives:

$\begin{equation*} \begin{split} m_A^2&=\left(1-\frac{5}{4}\right)^2+\left(\frac{9}{4}-\frac{15}{8}\right)^2\\ &=\left(-\frac{1}{4}\right)^2+\left(\frac{3}{8}\right)^2\\ &=\frac{13}{64}\\ \therefore m_A&=\frac{1}{8}\sqrt{13} \end{split} \end{equation*}$

Thus, the length of the median from vertex $A$ is $m_A=\frac{1}{8}\sqrt{13}$ .

Next, the length of side $BC$ is, by the distance formula:

$\begin{equation*} \begin{split} BC^2&=\left(\frac{5}{2}-0\right)^2+\left(\frac{15}{4}-0\right)^2\\ &=\frac{25}{4}+\frac{225}{16}\\ &=\frac{325}{16}\\ \therefore BC&=\frac{5}{4}\sqrt{13} \end{split} \end{equation*}$

Finally, we check that both $m_A=\frac{1}{8}\sqrt{13}$ and $BC=\frac{5}{4}\sqrt{13}$ are related via equation (1):

$\begin{equation*} \begin{split} m_A&=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC\\ &=\frac{1}{2}\left(\frac{3/2-1}{3/2+1}\right)\frac{5}{4}\sqrt{13}\\ &=\frac{1}{2}\left(\frac{1/2}{5/2}\right)\frac{5}{4}\sqrt{13}\\ &=\frac{1}{8}\sqrt{13} \end{split} \end{equation*}$

Expected.

Example

Consider $\triangle ABC$ with vertices at $A\left(-\frac{7}{4},-\frac{15}{2}\right)$ , $B\left(\frac{1}{2},-3\right)$ , and $C(-1,6)$ . Find the length of the median from vertex $A$ and verify that equation (1) is satisfied.

The slopes of sides $AB,BC,CA$ are $2,-6,18$ , respectively. They form a geometric progression in which $r=-3$ . Special, special case.

The midpoint of side $BC$ is at $\left(-\frac{1}{4},\frac{3}{2}\right)$ . Together with $A\left(-\frac{7}{4},-\frac{15}{2}\right)$ , we can find the length of the median from vertex $A$ :

$\begin{equation*} \begin{split} m_A^2&=\left(-\frac{7}{4}+\frac{1}{4}\right)^2+\left(-\frac{15}{2}-\frac{3}{2}\right)^2\\ &=\left(-\frac{3}{2}\right)^2+\left(-\frac{18}{2}\right)^2\\ &=\frac{333}{4}\\ \therefore m_A&=\frac{3}{2}\sqrt{37} \end{split} \end{equation*}$

With $B$ at $\left(\frac{1}{2},-3\right)$ and $C$ at $(-1,6)$ :

$\begin{equation*} \begin{split} BC^2&=\left(-1-\frac{1}{2}\right)^2+\left(6+3\right)^2\\ &=\frac{9}{4}+81\\ &=\frac{333}{4}\\ \therefore BC&=\frac{3}{2}\sqrt{37} \end{split} \end{equation*}$

Surprised — that $m_A=BC$ .

If we put $r=-3$ in equation (1) we see that

$m_A=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC=\frac{1}{2}\left(\frac{-3-1}{-3+1}\right)BC=\frac{1}{2}\left(\frac{-4}{-2}\right)BC=BC.$

Surprise no longer applies. Subtract surprise from your dictionary.

If the slopes of the sides of a triangle form a geometric progression with common ration $r=-3$ , then one median and the side opposite it are equal in length.

Example

Consider $\triangle ABC$ with vertices at $A(1,2\sqrt{2})$ , $B(-1,\sqrt{2})$ , and $C(0,0)$ . Find the length of the median from vertex $A$ , and verify that equation (1) is satisfied.

The slopes of sides $AB,BC,CA$ are $\frac{\sqrt{2}}{2},-\sqrt{2},2\sqrt{2}$ , respectively. They form a geometric progression in which $r=-2$ .

The midpoint of side $BC$ is at $\left(-\frac{1}{2},\frac{\sqrt{2}}{2}\right)$ . Together with $A\left(1,2\sqrt{2}\right)$ , we find:

$\begin{equation*} \begin{split} m_A^2&=\left(1+\frac{1}{2}\right)^2+\left(2\sqrt{2}-\frac{\sqrt{2}}{2}\right)^2\\ &=\frac{9}{4}+\frac{18}{4}\\ &=\frac{27}{4}\\ \therefore m_A&=\frac{3}{2}\sqrt{3} \end{split} \end{equation*}$

With $B$ at $\left(-1,\sqrt{2}\right)$ and $C$ at $(0,0)$ :

$\begin{equation*} \begin{split} BC^2&=\left(-1-0\right)^2+\left(\sqrt{2}-0\right)^2\\ &=3\\ \therefore BC&=\sqrt{3} \end{split} \end{equation*}$

Put $r=-2$ in equation (1):

$\begin{equation*} \begin{split} m_A&=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC\\ &=\frac{1}{2}\left(\frac{-2-1}{-2+1}\right)\sqrt{3}\\ &=\frac{1}{2}\times 3\times \sqrt{3}\\ &=\frac{3}{2}\sqrt{3} \end{split} \end{equation*}$

YESpected.

Example

Consider $\triangle ABC$ with vertices at $A(1,4)$ , $B(3,6)$ , and $C(0,0)$ . Find the length of the median from vertex $A$ , and verify that equation (1) is satisfied.

Our examples will be incomplete without this ugly, lovely triangle.

The midpoint of side $BC$ is at $\left(\frac{3}{2},3\right)$ . Together with $A(1,4)$ :

$\begin{equation*} \begin{split} m_A^2&=\left(1-\frac{3}{2}\right)^2+\left(4-3\right)^2\\ &=\frac{1}{4}+1\\ &=\frac{5}{4}\\ \therefore m_A&=\frac{1}{2}\sqrt{5} \end{split} \end{equation*}$

With $B$ at $(3,6)$ and $C$ at $(0,0)$ , the length of $BC$ is seen to be

$BC=\sqrt{3^2+6^2}=\sqrt{45}=3\sqrt{5}.$

Since the slopes of the sides of $\triangle ABC$ form a geometric progression with common ratio $r=2$ , let’s put $r=2$ in equation (1):

$\begin{equation*} \begin{split} m_A&=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC\\ &=\frac{1}{2}\left(\frac{2-1}{2+1}\right)3\sqrt{5}\\ &=\frac{1}{2}\times\frac{1}{3}\times 3\sqrt{5}\\ &=\frac{1}{2}\sqrt{5} \end{split} \end{equation*}$

Passed.

Obvious observations

Maybe you’ve wondered why we omitted absolute value in equation (1):

$m_A=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC.$

Wonder no more.

Example

If $r> 1$ , PROVE that $\frac{r-1}{r+1}> 0$ .

This is because $r> 1\implies r-1> 0$ , and $r+1> 2> 0$ , so the quotient $\frac{r-1}{r+1}$ is also greater than $0$ .

Example

If $r< -1$ , PROVE that $\frac{r-1}{r+1}> 0$ .

If $r< -1$ , then we have $r-1< -2< 0$ and also $r+1< 0$ . So the quotient $\frac{r-1}{r+1}> 0$ .

Example

If $0< r< 1$ , PROVE that $\frac{r-1}{r+1}< 0$

If $r< 1$ , then $r-1< 0$ . The fact that $0< r$ ensures that $r+1> 0$ . So the numerator and denominator of $\frac{r-1}{r+1}$ have negative and positive signs, respectively. This makes the quotient negative.

However, if the common ratio of our three-term geometric progression happens to be sandwiched between $0$ and $1$ , we can simply use its reciprocal — which will then satisfy $\frac{1}{r}> 1$ — in equation (1). This is the case of Example 5.

Example

If $-1< r< 0$ , PROVE that $\frac{r-1}{r+1}< 0$ .

If $r< 0$ , then $r-1< -1< 0$ . If in addition $-1< r$ , then adding $1$ to both sides of the inequality gives $0< r+1$ . So the numerator and denominator of $\frac{r-1}{r+1}$ are negative and positive, respectively. This makes the quotient to be negative.

However, if the common ratio of our three-term geometric progression happens to be sandwiched between $-1$ and $0$ , then we can simply use its reciprocal — which will then satisfy $\frac{1}{r}< -1$ — in equation (1). This is the case of Example 6.

Tedious derivation

As we derive equation (1) in what follows, are we really being serious by hinting that the procedure is tedious? Follow us to see that the answer is obvious.

PROVE that equation (1) holds for any triangle $ABC$ with slopes $a,ar,ar^2$ for the sides $AB,BC,CA$ .

We’ll use two identities from our previous posts:

(2) $\begin{equation*} \begin{split} m_B^2+m_C^2&=\left(\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}\right)m_A^2\\ AB^2+AC^2&=\frac{r^2+1}{(r+1)^2}BC^2\\ \end{split} \end{equation*}$

and a well-known identity, namely

$m_a^2+m_b^2+m_c^2=\frac{3}{4}(a^2+b^2+c^2),$

which holds for any triangle. We’ll write the latter in the form

(3) $\begin{equation*}m_A^2+m_B^2+m_C^2&=\frac{3}{4}(AB^2+BC^2+CA^2).\end{equation*}$

Eliminate $m_B,m_C,AB,AC$ from equations (2) and (3):

$\begin{equation*} \left(\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}+1\right)m_A^2&=\frac{3}{4}\left(\frac{r^2+1}{(r+1)^2}+1\right)BC^2\\ \end{equation*}$

Isolate $m_A^2$ and continue the algebra:

$\begin{equation*} \begin{split} m_A^2&=\frac{3}{4}\left(\frac{r^2+1+(r+1)^2}{(r+1)^2}\right)\times\left(\frac{(r-1)^2}{(r+2)^2+(2r+1)^2+(r-1)^2}\right)BC^2\\ &=\frac{3}{4}\left(\frac{2(r^2+r+1)}{(r+1)^2}\right)\times \left(\frac{(r-1)^2}{6(r^2+r+1)}\right)BC^2\\ &=\frac{1}{4}\left(\frac{r-1}{r+1}\right)^2BC^2\\ \therefore m_A&=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC \end{split} \end{equation*}$

Just as we said.

Find coordinates for the vertices $A,B,C$ of a $\triangle ABC$ in which the length of one median is one-third of the length of the opposite side.

We want a triangle $ABC$ in which $m_A=\frac{1}{3}BC$ , for example. Without recourse to equation (1), solving this problem maybe somewhat random.

So we’ll use what we’ve got. In (1), set

$\frac{1}{2}\left(\frac{r-1}{r+1}\right)=\frac{1}{3}$

and solve for $r$ : $\quad 3r-3=2r+2\implies r=5$ .

It remains to find coordinates for the vertices of a triangle in which the slopes of the sides form a geometric progression with common ratio $r=5$ .

Use the most basic set of coordinates for the vertices:

$\begin{equation*} \begin{split} &(0,0)\\ &(1,r^2)\\ &(1+r,r+r^2)\\ &\vdots\vdots\vdots\\ &C(0,0)\\ &A(1,25)\\ &B(6,30) \end{split} \end{equation*}$

That’s it. $\triangle ABC$ with vertices above is what we want:

$\begin{equation*} \begin{split} BC^2&=6^2+30^2\\ \therefore BC&=6\sqrt{26}\\ &\vdots\vdots\\ m_A^2&=\left(1-3\right)^2+\left(25-15\right)^2\\ &=104\\ \therefore m_A&=2\sqrt{26}\\ &=\frac{1}{3}\times 6\sqrt{26}\\ &=\frac{1}{3}BC\\ \vdots\vdots\ddots&\ddots\vdots\vdots\\ \end{split} \end{equation*}$

Takeaway

Our main equation (1)

$m_A=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC$

is similar to what happens in a right triangle, where the length of the median to the hypotenuse is half of the length of the hypotenuse.

Tasks

Find coordinates for the vertices of $\triangle ABC$ in which one median has length equal to $\frac{3}{5}$ of the length of the opposite side.
Find coordinates for the vertices of a right triangle in which the side-slopes form a geometric progression with common ratio $r=-3$ .
(Polynomial equations) Let be such that sides have slopes , respectively. PROVE that:
- $a^2r^4+2a^2r^3+2r+1=0$ , if $l_1=l_2$ ;
- $a^2r^4-a^2r^2-r^2+1=0$ , if $l_1=l_3$ ;
- $2a^2r^3+a^2r^2+r^2+2r=0$ , if $l_2=l_3$ .
(Confusing notation) In any with vertices , , and , let the slopes of the medians be denoted by , and let the slopes of the sides be . PROVE that:
- $x_1(2m_A-m_B-m_C)+x_2(2m_B-m_A-m_C)+x_3(2m_C-m_A-m_B)=0$ ;
- $x_1(m_{AB}-m_{CA})+x_2(m_{BC}-m_{AB})+x_3(m_{CA}-m_{BC})=0$ .
  (In our slipshod style, we sometimes use $m_A$ to denote the length of a median and sometimes the slope of the same median. Are we short of notations?)
(Alternative derivation) Let be equilateral such that sides have slopes , respectively. PROVE that:
- $(ar)^2=1$ ;
- $r^2+4r+1=0$ .
  (You can apply exercise 3 above, which uses a different approach compared to what we did in a previous post.)
(Linear combinations) Let $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_3)$ be the vertices of $\triangle ABC$ in which sides $AB,BC,CA$ have slopes $a,-3a,9a$ . PROVE that $2x_1+x_2-3x_3=0$ , and $2y_1-3y_2+y_3=0$ .
Let $-1\neq r\in\mathbb{Z}$ . PROVE that $\frac{1}{2}\left(\frac{r-1}{r+1}\right)$ is also an integer if, and only if, $r=-3,1$ .
Let $0,1,-1\neq r\in\mathbb{Z}$ and consider $\triangle ABC$ with slopes $a,ar,ar^2$ for sides $AB,BC,CA$ . PROVE that the length $m_A$ of the median from vertex $A$ is an integer multiple of the length of side $BC$ if, and only if, $r=-3$ .
Find appropriate coordinates for the vertices of $\triangle ABC$ in which the side-slopes form a geometric progression and the length of the median from vertex $A$ is twice the length of side $BC$ .
Give an example of a triangle $ABC$ in which the length of the median from vertex $A$ is equal to the length of side $BC$ , but the slopes of the sides do not form a geometric progression. (This shows that the converse of example 2 is not true.)

Sloppy slopes swap sequences

As shown by our slipshod, tongue-twisting title, slopes can be very “brittle”.

We demonstrate how to convert the slopes of the sides of a triangle from a geometric sequence to an arithmetic sequence. Hopefully it makes sense.

Modified midpoint theorem

A line which connects the midpoints of two sides of a triangle is parallel to the third side and its length is half of the third side’s length. So says the midpoint theorem.

Since a midpoint amounts to dividing a line segment in the ratio $1:1$ , we’ll modify this slightly for our triangle and rather divide the sides in the $r$ atio $1:r$ , whe $r$ e $~1\neq r> 0$ .

In $\triangle ABC$ , let $X$ and $Y$ be points on $BA$ and $CA$ which divide the sides in the ratio $1:n$ , where $n$ is a positive integer. PROVE that the line segment $XY$ is parallel to $BC$ and that $XY=\frac{n}{n+1}BC$ .

In the diagram above, $BX:XA=1:n$ and $CY:YA=1:n$ . If we place the vertices of $\triangle ABC$ at $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_3)$ , then we can obtain the coordinates of points $X$ and $Y$ using the formula for internal division of a line segment:

$X\left(\frac{x_1+nx_2}{n+1},\frac{y_1+ny_2}{n+1}\right),~Y\left(\frac{x_1+nx_3}{n+1},\frac{y_1+ny_3}{n+1}\right).$

The slope of line segment $XY$ is:

$\begin{equation*} \begin{split} &=\left[\frac{y_1+ny_2}{n+1}-\left(\frac{y_1+ny_3}{n+1}\right)\right]\div\left[\frac{x_1+nx_2}{n+1}-\left(\frac{x_1+nx_3}{n+1}\right)\right]\\ &=\frac{n}{n+1}(y_2-y_3)\div\left(\frac{n}{n+1}(x_2-x_3)\right)\\ &=\frac{y_2-y_3}{x_2-x_3} \end{split} \end{equation*}$

So $XY$ is parallel to $BC$ . In addition we obtain $XY=\frac{n}{n+1}BC$ .

Easy-peasy.

Derailing development

If we apply the above idea to our own triangle, look what we get.

Suppose that $\triangle ABC$ has vertices $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ , and slopes $a,ar,ar^2$ for sides $AB,BC,CA$ , respectively. For $1\neq r> 0$ , find coordinates for the point which divides $AB$ in the ratio $1:r$ .

Note the “direction” of the division. The desired point will be $\left(\frac{x_2+rx_1}{r+1},\frac{y_2+ry_1}{r+1}\right)$ , using the formula for internal division of a line segment. In view of the $r$ elationship among the coordinates, this becomes

$\left(x_3+\frac{(2r+1)(y_2-y_3)}{ar(r+1)^2},\frac{(r^2+r+1)y_2+ry_3}{(r+1)^2}\right).$

You see that quadratic rearing its beautiful head again?

Ordinarily, this would be the point $\left(\frac{x_2+rx_3}{r+1},\frac{y_2+ry_3}{r+1}\right)$ . Using the $r$ elationship among the coordinates, we get

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

This would be the point $\left(\frac{x_1+rx_3}{r+1},\frac{y_1+ry_3}{r+1}\right)$ . The $r$ elationship among the coordinates then gives

$\left(x_3+\frac{y_2-y_3}{ar(r+1)^2},\frac{ry_2+(r^2+r+1)y_3}{(r+1)^2}\right)$

(Main goal)

Suppose that $\triangle ABC$ has vertices $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ , and slopes $a,ar,ar^2$ for sides $AB,BC,CA$ , respectively. For $1\neq r> 0$ , PROVE that there is a point $X$ on $AB$ , a point $Y$ on $BC$ , and a point $Z$ on $CA$ such that the slopes of the sides of $\triangle XYZ$ form an arithmetic progression.

Basically, the side-slopes form a geometric progression, while the sub-triangle’s side-slopes form an arithmetic progression. Fo $r$ . Any. $1\neq r> 0$ .

Let $X$ be the point which divides $AB$ in the ratio $1:r$ , let $Y$ be the point which divides $CB$ in the ratio $1:r$ , and let $Z$ be the point which divides $CA$ in the ratio $1:r$ . As in the three preceding examples, these points are:

$\begin{equation*} \begin{split} X&\left(x_3+\frac{(2r+1)(y_2-y_3)}{ar(r+1)^2},\frac{(r^2+r+1)y_2+ry_3}{(r+1)^2}\right)\\ Y&\left(x_3+\frac{y_2-y_3}{ar(r+1)},\frac{ry_2+y_3}{r+1}\right)\\ Z&\left(x_3+\frac{y_2-y_3}{ar(r+1)^2},\frac{ry_2+(r^2+r+1)y_3}{(r+1)^2}\right) \end{split} \end{equation*}$

By the modified midpoint theorem, $XY$ would be parallel to $CA$ , so its slope is $ar^2$ (alternatively can be confirmed by direct calculation). Similarly, $YZ$ would be parallel to $AB$ , so its slope is $a$ .

By direct calculation, the slope of $ZX$ is $\frac{a+ar^2}{2}$ .

Thus, the slopes of the sides of $\triangle XYZ$ are $a,\frac{a+ar^2}{2},ar^2$ , an arithmetic progression.

Given $\triangle ABC$ with vertices at $A(1,4)$ , $B(3,6)$ , and $C(0,0)$ , find coordinates for points $X,Y,Z$ on $AB,BC,CA$ such that the slopes of $\triangle XYZ$ are $1,\frac{5}{2},4$ .

Observe that the slopes of sides $AB,BC,CA$ are $1,2,4$ , respectively.

Thus, by Example 5, let $X$ be the point which divides $AB$ in the ratio $1:2$ , let $Y$ be the point which divides $CB$ in the ratio $1:2$ , and let $Z$ be the point which divides $CA$ in the ratio $1:2$ . Lots of repetitions there $\cdots$ .

Using the formula for internal division of a line segment, we find these points to be $X\left(\frac{5}{3},\frac{14}{3}\right)$ , $Y(1,2)$ , and $Z\left(\frac{1}{3},\frac{4}{3}\right)$ . Then:

$\begin{equation*} \begin{split} \textrm{slope of}~XY&=4\\ \textrm{slope of}~YZ&=1\\ \textrm{slope of}~ZX&=\frac{5}{2} \end{split} \end{equation*}$

Given $\triangle ABC$ with vertices at $A(1,9)$ , $B(4,12)$ , and $C(0,0)$ , find coordinates for points $X,Y,Z$ on $AB,BC,CA$ such that the slopes of $\triangle XYZ$ are $1,5,9$ .

In this case, the slopes of sides $AB,BC,CA$ are $1,3,9$ , respectively. A geometric progression in which $r=3$ .

By Example 5, let $X,Y,Z$ divide each of the three sides $AB,CB,CA$ in the ratio $1:3$ .

Using the formula for internal division of a line segment, we find these points to be $X\left(\frac{7}{4},\frac{39}{4}\right)$ , $Y(1,3)$ , and $Z\left(\frac{1}{4},\frac{9}{4}\right)$ .

$\begin{equation*} \begin{split} \textrm{slope of}~XY&=9\\ \textrm{slope of}~YZ&=1\\ \textrm{slope of}~ZX&=5 \end{split} \end{equation*}$

Given $\triangle ABC$ with vertices at $A(1,4)$ , $B(-1,2)$ , and $C(0,0)$ , find coordinates for (external) points $X,Y,Z$ on $AB,BC,CA$ such that the slopes of $\triangle XYZ$ are $1,\frac{5}{2},4$ .

Observe that the slopes of sides $AB,BC,CA$ are $1,-2,4$ , respectively. They form a geometric progression in which $r=-2$ .

Let $X,Y,Z$ be points which divide $AB,CB,CA$ all in the ratio $1:-2$ . We obtain the external points $X\left(3,6\right)$ , $Y(1,-2)$ , and $Z\left(-1,-4\right)$ . Then:

$\begin{equation*} \begin{split} \textrm{slope of}~XY&=4\\ \textrm{slope of}~YZ&=1\\ \textrm{slope of}~ZX&=\frac{5}{2} \end{split} \end{equation*}$

The scene here may be worth seeing $\cdots$

Notice that $C$ is the midpoint of both $AZ$ and $BY$ , while $A$ is the midpoint of $BX$ . It follows that the length of $XY$ is twice that of $AC$ .

Touching negative common ratios, a couple of strangely exciting things happen. This is why we’ve been silent about them in many of our posts, but we have them on our radar and will treat them later.

Let $ABC$ be a right isosceles triangle with slopes $a,ar,ar^2$ . PROVE that there are two points $X$ and $Y$ on sides $AB,AC$ such that the slope of the line segment $XY$ is $-\frac{3}{2}$ times the slope of $BC$ .

By Example 5, we can find points $X,Y,Z$ such that the slopes of the sides of $\triangle XYZ$ are $a,\frac{a+ar^2}{2},ar^2$ .

Using the fact that $r^2+3r+1=0$ for a right isosceles triangle, we have:

$\begin{equation*} \begin{split} \frac{a+ar^2}{2}&=\frac{a(1+r^2)}{2}\\ &=\frac{a(-3r)}{2}\\ &=-\frac{3}{2}(ar)\\ \end{split} \end{equation*}$

Plus or minus two

Let $ABC$ be an equilateral triangle with slopes $a,ar,ar^2$ . PROVE that there are two points $X$ and $Y$ on two sides such that the slope of the line segment $XY$ is $\pm 2$ .

By Example 5, we can find points $X,Y,Z$ on $AB,BC,CA$ such that the slopes of the sides of $\triangle XYZ$ are $a,\frac{a+ar^2}{2},ar^2$ .

Using the fact that $ar=\pm 1$ and $r^2+4r+1=0$ for an equilateral triangle, we have:

$\begin{equation*} \begin{split} \frac{a+ar^2}{2}&=\frac{a(1+r^2)}{2}\\ &=\frac{a(-4r)}{2}\\ &=-2(ar)\\ &=\pm 2 \end{split} \end{equation*}$

Point-blank.

Takeaway

The construction we’ve done is just a variant of the medial triangle. Unlike the medial triangle, ours is different in the sense that it does not retain the slopes of the parent triangle. While the slopes of the parent triangle form a geometric progression, its own slopes form an arithmetic progression. Talk about t $r$ ansfo $r$ mation.

Tasks

(Two out) Consider $\triangle ABC$ with vertices at $A(1,4)$ , $B(3,6)$ , and $C(0,0)$ . PROVE that one can obtain the infinite arithmetic sequence $\cdots,-3,-2,-1,0,1,\boxed{2},3,\cdots$ as slopes of line segments drawn from vertex $A$ to side $BC$ .
(Notice that $2$ is out because it’s already the slope of side $BC$ . Other than that, every other integer can be obtained this way. What’s somewhat interesting here is that the line segments from $A$ to $BC$ that generate the sequence terms all stay within the triangle; none of them is external.)
(Fibonacci sequence) Explain how the terms of a Fibonacci sequence $0,1,1,2,3,5,8,13,\cdots$ can be generated as slopes.
(Always odd) Let $r$ be a positive integer, and consider $\triangle ABC$ with vertices at $A(1,r^2)$ , $B(0,0)$ , and $C(1+r,r+r^2)$ . PROVE that there is a point $X$ on $AC$ and a point $Y$ on $CB$ such that the slope of the line segment $XY$ is $2r-1$ .
(If $r$ is an integer, then any number of the form $2r-1$ is odd.)
Let be such that sides have slopes . Let be points on which divide these sides all in the ratio . PROVE that:
- the $x$ -coordinates of $\triangle XYZ$ form an arithmetic progression
- the line connecting the centroid of $\triangle XYZ$ to the centroid of $\triangle ABC$ is parallel to side $BC$ .
Find an appropriate choice of coordinates for the vertices of a right isosceles triangle whose side-slopes form a geometric progression.