General – Page 34 – High School Geometry

From nothing to infinity

How can one start with nothing, then end up with infinitely many things? Well, when one starts with a triangle whose slopes are not necessarily in geometric progression and then “enters inside” it; one obtains an assemblage of “smaller” triangles whose slopes are in geometric progression.

Desirable properties from sub-triangles

A sub-triangle is simply a (smaller) triangle contained in another triangle. It can share a side with the original triangle, its vertices can lie on the sides of the original triangle, or its vertices can lie completely inside the parent triangle.

The medial triangle of any triangle is a good example of a sub-triangle in which its vertices lie on the midpoints of the sides of the original triangle (for instance, in the diagram below, $\triangle LMN$ is the medial triangle of $\triangle ABC$ ).

A sub-triangle can also lie completely inside the parent triangle, like below:

We show that any triangle with one side parallel to the $x$ -axis contains infinitely many sub-triangles with slopes in geometric progression; in particular, we obtain that “coveted” case in which the common ratio $r=-2$ .

Example (THEOREM)

Let $\triangle ABC$ be a right triangle in which one leg is on the $x$ -axis (or parallel to the $x$ -axis). For any real number $r$ , $1\neq r> 0$ , PROVE that there’s a sub-triangle of $\triangle ABC$ whose side slopes form a geometric progression with common ratio $r$ .

To make the proof simple, consider a right triangle $ABC$ with vertices at $A(0,0),~B(k,0),~C(0,l)$ . Every other right triangle that fits the description in our “theorem” is a translation of the named $\triangle ABC$ , so no loss in generality.

$\triangle CLB$ does the trick, if we define $L:=\Big(\frac{k}{r+1},\frac{l}{r+1}\Big)$ . The essence of requiring a positive $r$ is to ensure that $L$ lies inside the given triangle.

$\begin{equation*} \begin{split} m_{1}&=\frac{l-(\frac{l}{r+1})}{0-(\frac{k}{r+1})}\\ &=\frac{l(\frac{r}{r+1})}{-\frac{k}{r+1}}\\ &=r\Big(-\frac{l}{k}\Big)\\ &\cdots\vdots\cdots\\ m_{2}&=\frac{\frac{l}{r+1}-0}{\frac{k}{r+1}-k}\\ &=\frac{\frac{l}{r+1}}{-k(\frac{r}{r+1})}\\ &=\frac{1}{r}\Big(-\frac{l}{k}\Big)\\ &\cdots\vdots\cdots\\ m_{3}&=\frac{l-0}{0-k}\\ &=-\frac{l}{k} \end{split} \end{equation*}$

Thus, the slope sequence $m_{2},m_{3},m_{1}$ is geometric, with common ratio $r$ .

Considering how simple and basic the above proof is, does the result deserve the use of that reserved — and almost revered — word “theorem”? Or, should the “label” be reversed?

Example

PROVE that each of the sub-triangles constructed in Example 1 above contains an obtuse angle.

Recall the slopes: $m_{1}=r\Big(-\frac{l}{k}\Big),~m_{2}=\frac{1}{r}\Big(-\frac{l}{k}\Big),~m_{3}=-\frac{l}{k}$ . Since the common ratio $r$ is positive, it follows (depending on the sign of $l/k$ ) that the slopes of the sides are either all positive or all negative. Now, any triangle that contains all positive or all negative slopes must be obtuse-angled. (VERIFY this!)

Consequently, none of the sub-triangles constructed in Example 1 is equilateral or right-angled.

Example

For any right triangle with legs on the coordinate axes, PROVE that if the endpoints of the hypotenuse are joined to the centroid, then the slopes of the resulting sub-triangle form a geometric progression with common ratio $r=2$ .

COOL!

Let the coordinates be $A(0,0),~B(k,0),~C(0,l)$

. Then the centroid is $\textbf{G}(\frac{k}{3},\frac{l}{3})$

, as shown below:

Since the centroid has coordinates $G(\frac{k}{3},\frac{l}{3})=(\frac{k}{2+1},\frac{l}{2+1})$ , this is the case of $r=2$ in view of Example 1.

Note that the same result holds when the legs are parallel to the coordinate axes.

Two times infinity

Not only do we get one sub-triangle for a given $r$ ; we actually get two — in fact, more!!!

Example

PROVE that every triangle with slopes $k,kr,kr^2$ contains a sub-triangle with slopes $k,-kr,kr^2.$

We’ll apply the first proposition we established in our previous post; it states that the (non-zero) slopes of a triangle form a geometric progression if and only if there is a median whose slope is the negative of the slope of the side it meets.

Suppose that $\triangle ABC$ is such that sides $AB,BC,CA$ have slopes $k,kr,kr^2$ . Then the slope of the median from vertex $A$ is $-kr$ :

Let $\textbf{R}$ be the midpoint of $BC$ and let $\textbf{M}$ be the midpoint of $AC$ . Join $\textbf{RM}$ , the dashed red line segment shown above. Since $\textbf{RM}$ connects the midpoints of two sides, it is parallel to the third side, namely $AB$ . So, the slope of $\textbf{RM}$ is equal to the slope of $AB$ , which is $k$ . Furthermore, the slope of $MA$ is $kr^2$ , same as that of $AC$ .

Thus, $\triangle \textbf{ARM}$ has slopes $k,-kr,kr^2$ for sides $\textbf{RM},AR,MA$ , respectively.

Note that there are actually two sub-triangles with slopes $k,-kr,kr^2$ associated to a parent triangle with slopes $k,kr,kr^2$ .

Example (THEOREM)

Let $\triangle ABC$ be a non-right triangle in which one side is on the $x$ -axis (or parallel to the $x$ -axis). For any real number $r$ , $1\neq r> 0$ , PROVE that there’s a sub-triangle of $\triangle ABC$ whose side slopes form a geometric progression with common ratio $r$ .

This follows from Example 1. Since one side is parallel to the $x$ -axis, draw an altitude from the opposite vertex. The two resulting right triangles both conform to the description in Example 1.

Example (Main goal)

PROVE that every triangle with one side parallel to the $x$ -axis contains a sub-triangle in which one median is vertical and one median is horizontal.

The proof of Example 6 follows from Example 3 and Example 4.

EUREKA! Remember that “VHM” property? See here. Since the “VHM” property is extremely nice, we were curious as to whether it can be embedded into every triangle. Today’s post arose from that curiosity. We’ve now done it for triangles in which one side has zero slope; we’ll do something similar for general triangles.

Three times infinity

We improve on the number of sub-triangles we can have for any given $r$ .

Example (Theorem)

PROVE that any right triangle with legs parallel to the coordinate axes contains three different sub-triangles whose slopes form a geometric progression with common ratio $r$ , where $1\neq r> 0$ .

We prove this for the right triangle $ABC$ with coordinates $A(0,0),~B(k,0),~C(0,l)$ ; just a slight modification is needed for the general case when the legs are parallel to, but not on, the coordinate axes.

The desired sub-triangles, each with side slopes in geometric progression and common ratio $r$ , are: $\triangle CZY,~\triangle ZBX,~\triangle CZB$ . Using the given coordinates, we can confirm this by calculating slopes:

$\begin{equation*} \begin{split} \textrm{Slope of CZ}&=\frac{\frac{l}{r+1}-l}{\frac{k}{r+1}-0}\\ &=r\Big(-\frac{l}{k}\Big)\\ \textrm{Slope of ZY}&=\frac{\frac{l(r^2+r+1)}{(r+1)^2}-\frac{l}{r+1}}{\frac{kr}{(r+1)^2}-\frac{k}{r+1}}\\ &=r^2\Big(-\frac{l}{k}\Big)\\ \textrm{Slope of YC}&=-\frac{l}{k}\\ \end{split} \end{equation*}$

Thus, the slopes of the sides of $\triangle CZY$ differ by a constant multiple of $r$ . Since we’ve already shown that the slopes of $\triangle CZB$ differ by a constant multiple of $r$ , it just remains to do the same for $\triangle ZBX$ .

$\begin{equation*} \begin{split} \textrm{Slope of ZB}&=\frac{\frac{l}{r+1}-0}{\frac{k}{r+1}-k}\\ &=\frac{1}{r}\Big(-\frac{l}{k}\Big)\\ \textrm{Slope of ZX}&=\frac{\frac{lr}{(r+1)^2}-\frac{l}{r+1}}{\frac{k(r^2+r+1)}{(r+1)^2}-\frac{k}{r+1}}\\ &=\frac{1}{r^2}\Big(-\frac{l}{k}\Big)\\ \textrm{Slope of BX}&=-\frac{l}{k} \end{split} \end{equation*}$

Therefore, the slopes of the sides of $\triangle ZBX$ form a geometric progression with common ratio $r$ .

Observe that $\triangle GXY$ in the above diagram has side slopes also in geometric progression, but with common ratio $r^2$ .

Example (zig-zag theorem)

For any right triangle $RST$ with legs ( $RS,ST$ ) on the coordinate axes, PROVE that there are points $X$ and $Y$ on the hypotenuse $RT$ and an interior point $Z$ such that the slopes of $XZ,ZT,TR,RZ,ZY$ form a five-term geometric progression.

Let the vertices of the right triangle $RST$ be located at $R(0,l),~S(0,0),~T(k,0)$ . For $1\neq r> 0$ , define $Z:=(\frac{k}{r+1},\frac{l}{r+1}),~X:=\Big(\frac{k(r^2+r+1)}{(r+1)^2},\frac{lr}{(r+1)^2}\Big),Y:=\Big(\frac{kr}{(r+1)^2},\frac{l(r^2+r+1)}{(r+1)^2}\Big)$ . Then $Z$ is an interior point of $\triangle RST$ ; further, $X$ and $Y$ lie (internally) on the hypotenuse $RT$ . With the given coordinates, we have the following slopes:

$\begin{equation*} \begin{split} \textrm{Slope of XZ}&=\frac{1}{r^2}\Big(-\frac{l}{k}\Big)\\ \textrm{Slope of ZT}&=\frac{1}{r}\Big(-\frac{l}{k}\Big)\\ \textrm{Slope of TR}&=-\frac{l}{k}\\ \textrm{Slope of RZ}&=r\Big(-\frac{l}{k}\Big)\\ \textrm{Slope of ZY}&=r^2\Big(-\frac{l}{k}\Big) \end{split} \end{equation*}$

These slopes differ by a constant multiple of $r$ . Notice how the line segments $XZ,ZT,TR,RZ,ZY$ are traversed in a sort of “zig-zag” manner as shown below:

Numerical problems

Always a good idea to support theory with examples.

Example

Given $\triangle ABC$ with vertices at $A(0,9),~B(-6,0),~C(6,0)$ , find coordinates for its four sub-triangles that all satisfy the “VHM” property.

Note that the parent triangle doesn’t satisfy the “VHM” property; in particular, the $y$ -coordinates $(0,0,9)$ do not form an arithmetic progression, though the $x$ -coordinates $(-6,0,6)$ do.

A sample sub-triangle is $\triangle KMP$ with vertices $K(2,3),~M(1,6),~P(3,4.5)$ shown above. Notice that its $x$ -coordinates $(1,2,3)$ form an arithmetic progression, as well as its $y$ -coordinates $(3,4.5,6)$ .

How did we obtain $\triangle KMP$ ? Easy. We followed Example 3 and Example 4:

divide the parent triangle into two right triangles ( $\triangle ABO$ and $\triangle AOC$ )
find the centroids of each of the two resulting right triangles (for $\triangle AOC$ it is $K(2,3)$ )
connect the centroid to the endpoints of the hypotenuse (e.g., for $\triangle AOC$ , join $A$ to $K$ and $C$ to $K$ ; this way, the resulting triangle $AKC$ has common ratio $r=2$ )
connect the centroid to the midpoint of the hypotenuse ( $KP$ in the diagram)
join $P$ to the midpoint of $AK$ ( $PM$ in the diagram). DONE.

The centroid ( $K$ ), the midpoint of the hypotenuse ( $P$ ), and the midpoint of side $AK$ ( $M$ ) are all we need. Repeating this process for the lower sub-triangle $KCP$ as well as for the right triangle on the left, we obtain the following list of four sub-triangles satisfying that satisfying “VHM” property (you read that correctly):

$(1,6),~(2,3),~(3,4.5)$ ;
$(2,3),(3,4.5),(4,1.5)$ ;
$(-4,1.5),(-3,4.5),(-2,3)$ ;
$(-3,4.5),(-2,3),(-1,6)$ .

Of course, each is a peach.

Example

Given the right triangle $RST$ with coordinates $R(0,8),~S(0,0),~T(8,0)$ , find coordinates for two points $X$ and $Y$ on $RT$ and an interior point $Z$ such that there is a five-term geometric progression for the slopes of $XZ,ZT,TR,RZ,ZY$ .

Set $X:=(6.5,1.5),~Y(1.5,6.5),~Z(2,2)$ . Then we have:

$\begin{equation*} \begin{split} m_{XZ}&=\frac{1.5-2}{6.5-2}\\ &=-\frac{1}{9}\\ m_{ZT}&=\frac{2-0}{2-8}\\ &=-\frac{1}{3}\\ m_{TR}&=\frac{8-0}{0-8}\\ &=-1\\ m_{RZ}&=\frac{8-2}{0-2}\\ &=-3\\ m_{ZY}&=\frac{2-6.5}{2-1.5}\\ &=-9 \end{split} \end{equation}$

Thus the slopes of the segments $XZ,ZT,TR,RZ,ZY$ are $-\frac{1}{9},-\frac{1}{3},-1,-3,-9$ , respectively. They form a five-term geometric progression with a common ratio of $3$ .

Having read through this page, it’s now too late to hate these greats — triangles with slopes in geometric progression.

Takeaway

Sometimes, you already have what you seek — if you look “within”. Does that seem to click?

As we’ve shown, a parent triangle may not have a desirable property on the surface, until its interior is examined. To look down on people — or things — solely on their outward look, is not cool.

Tasks

(Embedded hexagon) For any triangle $ABC$ , PROVE that it is possible to embed a hexagon whose vertices lie completely inside $\triangle ABC$ , and such that three of its sides are $\frac{1}{3}$ the side lengths of $\triangle ABC$ and the remaining three are $\frac{1}{6}$ the side lengths of $\triangle ABC$ .

(In the diagram above, $RS=\frac{1}{2}VU=\frac{1}{6}BC;~TU=\frac{1}{2}RW=\frac{1}{6}AB;~VW=\frac{1}{2}TS=\frac{1}{6}CA$ . Simply find the coordinates of $R,S,T,U,V,W$ in terms of the coordinates of $A,B,C$ .)
(Power two) Given $\triangle ABC$ with vertices $A(1,4),~B(3,6),~C(0,0)$ , the slopes of sides $AB,BC,CA$ form a geometric progression with common ratio $r_1=2$ . Find coordinates for two points $X$ and $Y$ on $BC$ such that the slopes of the sides of $\triangle AXY$ form a geometric progression with common ratio $r_2=4$ .
(Notice that $r_2=r_{1}^2$ . This is always possible for any triangle whose slopes form a geometric progression with a positive common ratio.)
(Power two) Given $\triangle ABC$ with vertices $A(1,18),~B(4,24),~C(0,0)$ , the slopes of sides $AB,BC,CA$ form a geometric progression with common ratio $r_1=3$ . Find coordinates for two points $X$ and $Y$ on $BC$ such that the slopes of the sides of $\triangle AXY$ form a geometric progression with common ratio $r_2=9$ .
(Notice that $r_2=r_{1}^2$ . This is always possible for any triangle whose slopes form a geometric progression with a positive common ratio.)
(Five terms) For any triangle $ABC$ with slopes of sides $AB,BC,CA$ forming a geometric progression with positive common ratio, PROVE that there are points $X$ and $Y$ on $BC$ such that the slopes of the line segments $YA,AC,CB,BA,AX$ form a five-term geometric progression.
(This has very important implication and application. In particular, it helps us to associate every finite geometric sequence having a positive common ratio with a triangle.)
Given with vertices , its slopes () are in geometric progression. PROVE that it contains a sub-triangle satisfying the following two properties:
- its slopes are in arithmetic progression, namely $1,\frac{5}{2},4;$
- the midpoint of one of its sides is the centroid of the original triangle.
  (The above triangle has very nice properties; no wonder we’ve encountered it twice in these tasks.)
(Two sequences) Let $\triangle ABC$ be such that the slopes of sides $AB,BC,CA$ form a geometric progression $k,kr,kr^2$ with a positive common ratio $r$ . 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 $k,\frac{k+kr^2}{2},kr^2$ .
Let $\triangle ABC$ be such that the slopes of sides $AB,BC,CA$ form a geometric progression, with a positive common ratio. PROVE that there are points $X$ and $Y$ on $BC$ such that $XY=\Big(\frac{r}{r^2+1}\Big)BC$ .
Let $\triangle ABC$ be such that the slopes of sides $AB,BC,CA$ form a geometric progression, with a negative common ratio. PROVE that there is a point $X$ on $AB$ and a point $Y$ on $CA$ such that $XY=\Big(\frac{r}{r^2+1}\Big)BC$ .
(Apparently, negative common ratios change things slightly. Being biased, they are our favorite.)
(Internal division) Let $AB$ be a line segment, where $A$ has coordinates $(x_1,y_1)$ and $B$ has coordinates $(x_2,y_2)$ . For $n\geq 2$ , PROVE that the point $W\Big(\frac{x_1+(n-1)x_2}{n},\frac{y_1+(n-1)y_2}{n}\Big)$ lies within $AB$ .
(Smaller lengths) PROVE that every triangle contains a sub-triangle whose lengths are $\frac{1}{n}$ of the lengths of the original triangle, for any $n\geq 1$ .

Two propositions on geometric slopes

Basically, geometric slopes involve three numbers $m_{1},m_{2},m_{3}$ that are in geometric progression — like $1,2,4$ — but at the same time are slopes of the sides of a triangle. Note that this is not a standard terminology; we’re using it here to keep the title short.

We present two “propositions” that describe precisely when the slopes of the sides of a triangle form a geometric progression.

Notation and characterization

In $\triangle ABC$ , let’s denote the slopes of the three medians from vertices $A,B,C$ by $m_{A},~m_{B},~m_{C}$ , respectively. Similarly, the slopes of sides $AB,BC,CA$ will be denoted by $m_{AB},~m_{BC},~m_{CA}$ .

If $m_{A}+m_{BC}=0$ , PROVE that $x_{1}=\frac{x_{2}(y_1-y_2)+x_{3}(y_3-y_1)}{y_3-y_2}$ , where $\triangle ABC$ has vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ .

$m_{A}$ is the slope of the line segment joining $A(x_1,y_1)$ to the midpoint of $BC=\Big(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\Big)$ ; $m_{BC}$ is the slope of side $BC$ . So we have:

$m_{A}=\frac{\frac{y_2+y_3}{2}-y_1}{\frac{x_2+x_3}{2}-x_1}=\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1};\quad m_{BC}=\frac{y_3-y_2}{x_3-x_2}.$

Now, if $m_{A}+m_{BC}=0$ , write $m_{A}=-m_{BC}$ . Clear fractions:

$(x_3-x_2)(y_2+y_3-2y_1)=-(y_3-y_2)(x_2+x_3-2x_1).$

Expand and combine like terms:

$x_3y_2+x_3y_3-2x_3y_1-x_2y_2-x_2y_3+2x_2y_1=-(x_2y_3-x_2y_2+x_3y_3-x_3y_2-2x_1y_3+2x_1y_2).$

Obtain $2x_3y_3-2x_3y_1+2x_2y_1-2x_2y_2=2x_1y_3-2x_1y_2$ , from which $x_1$ can be isolated as:

$x_1=\frac{x_2(y_1-y_2)+x_3(y_3-y_1)}{y_3-y_2}.$

Example 2 (First proposition)

For $\triangle ABC$ with vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ and non-zero side slopes, PROVE that the slopes of sides $AB,BC,CA$ form a geometric progression if, and only if, $m_{A}+m_{BC}=0$ .

In other words, once one of the medians has a slope that is the negative of the slope of the side that contains its “foot”, then the slopes of the sides of the triangle form a geometric progression.

The non-zero requirement for the slopes of the sides is crucial. See Example 3 below. Also, click here for a pdf version of some of the equations in this post.

Suppose that $m_{A}+m_{BC}=0$ ; we’ll show that the slopes of sides $AB,BC,CA$ (namely $\frac{y_1-y_2}{x_1-x_2}, \frac{y_2-y_3}{x_2-x_3}, \frac{y_3-y_1}{x_3-x_1}$ ) form a geometric progression. It suffices to show that $\frac{y_1-y_2}{x_1-x_2}\times \frac{y_3-y_1}{x_3-x_1}=\Big(\frac{y_2-y_3}{x_2-x_3}\Big)^2$ .

Since $m_{A}+m_{BC}=0$ , we can use $x_1=\frac{x_2(y_1-y_2)+x_3(y_3-y_1)}{y_3-y_2}$ from Example 1 and obtain $\frac{y_1-y_2}{x_1-x_2}\times \frac{y_3-y_1}{x_3-x_1}$ :

$\begin{equation*} \boldmath\begin{split} &=\scriptstyle\frac{(y_1-y_2)(y_3-y_1)}{\Big[\frac{x_2(y_1-y_2)+x_3(y_3-y_1)}{y_3-y_2}-x_2\Big]\Big[x_3-\Big(\frac{x_2(y_1-y_2)+x_3(y_3-y_1)}{y_3-y_2}\Big)\Big]}\\ &=\scriptscriptstyle\frac{(y_1-y_2)(y_3-y_1)(y_3-y_2)^2}{[x_2(y_1-y_2)+x_3(y_3-y_1)-x_2(y_3-y_2)][x_3(y_3-y_2)-x_2(y_1-y_2)-x_3(y_3-y_2)]}\\ &=\scriptstyle\frac{(y_1-y_2)(y_3-y_1)(y_3-y_2)^2}{[x_2(y_1-y_2-y_3+y_2)+x_3(y_3-y_1)][x_3(y_3-y_2-y_3+y_1)-x_2(y_1-y_2)]}\\ &=\frac{(y_1-y_2)(y_3-y_1)(y_3-y_2)^2}{(x_3-x_2)(y_3-y_1)(x_3-x_2)(y_1-y_2)}\\ &=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2. \end{split} \end{equation*}$

$\therefore \frac{(y_1-y_2)(y_3-y_1)}{(x_1-x_2)(x_3-x_1)}&=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2$ , and so the slopes $\frac{y_1-y_2}{x_1-x_2}, \frac{y_2-y_3}{x_2-x_3}, \frac{y_3-y_1}{x_3-x_1}$ form a geometric progression. The converse was proved in Example 8 of our post on January 28, 2020.

Consider the right triangle $ABC$ with vertices at $A(0,0),~B(k,0),~C(0,l)$ . PROVE that the slope of the median from vertex $A$ is the negative of the slope of the hypotenuse $BC$ .

Using the given coordinates and the mid-point formula, the mid-point of the hypotenuse $BC$ is

$\Big(\frac{0+k}{2},\frac{l+0}{2}\Big)=\Big(\frac{k}{2},\frac{l}{2}\Big),$

and so the slope of the median from vertex $A$ is

$\frac{\frac{l}{2}-0}{\frac{k}{2}-0}=\frac{l}{k}.$

Now, the slope of the hypotenuse $BC$ is

$\frac{l-0}{0-k}=-\frac{l}{k},$

which is the negative of the slope of the median from vertex $A$ .

However, the slopes of the sides of $\triangle ABC$

, namely $(0,~-\frac{l}{k},~\infty)$

do not form a geometric progression. This doesn’t contradict the equivalent statements in our first proposition; it simply illustrates the fact that a hypothesis cannot be dismissed, else things can go amiss.

Example 4 (Second proposition)

In $\triangle ABC$ , PROVE that $m_{A}+m_{BC}=0$ if, and only if, $m_{a}=\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$ , where $m_{a}$ is the length of the median from vertex $A$ and the triangle’s vertices are located at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ .

Consequently, we obtain a second characterization of slopes in geometric progression.

First suppose that $m_{A}+m_{BC}=0$ . Since

$m_{A}=\frac{\frac{y_2+y_3}{2}-y_1}{\frac{x_2+x_3}{2}-x_1}=\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1};\quad m_{BC}=\frac{y_3-y_2}{x_3-x_2}$

$m_{A}+m_{BC}=0\implies y_2+y_3-2y_1=-\Big(\frac{y_3-y_2}{x_3-x_2}\Big)(x_2+x_3-2x_1)$ .

Now, the length $m_{a}$ of the median from vertex $A$ is the distance from $(x_1,y_1)$ to the midpoint of $BC=\Big(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\Big)$ :

$\begin{equation*} \begin{split} m_{a}^2&=\Big(\frac{x_2+x_3}{2}-x_1\Big)^2+\Big(\frac{y_2+y_3}{2}-y_1\Big)^2\\ &=\Big(\frac{x_2+x_3-2x_1}{2}\Big)^2+\Big(\frac{y_2+y_3-2y_1}{2}\Big)^2\\ &=\frac{1}{4}\Big((x_2+x_3-2x_1)^2+(y_2+y_3-2y_1)^2\Big)\\ &=\frac{1}{4}\Big[(x_2+x_3-2x_1)^2+\Big(-(\frac{y_3-y_2}{x_3-x_2})(x_2+x_3-2x_1)\Big)^2\Big]\\ &=\frac{1}{4}\frac{(x_2+x_3-2x_1)^2}{(x_3-x_2)^2}\Big((x_3-x_2)^2+(y_3-y_2)^2\Big)\\ \therefore m_{a}&=\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2} \end{split} \end{equation}$

A little care should be taken pertaining the sign of $\frac{x_2+x_3-2x_1}{x_3-x_2}$ , so its absolute value should be used instead.

Conversely, suppose that $m_{a}=\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$ . Usually, $m_{a}^2=\Big(\frac{x_2+x_3-2x_1}{2}\Big)^2+\Big(\frac{y_2+y_3-2y_1}{2}\Big)^2$ , in view of the given coordinates. Together with the assumption of the converse, this amounts to:

$\begin{equation*} \begin{split} \Big(\frac{x_2+x_3-2x_1}{2}\Big)^2+\Big(\frac{y_2+y_3-2y_1}{2}\Big)^2&=\Big(\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}\Big)^2\\ \Big(\frac{y_2+y_3-2y_1}{2}\Big)^2&=\Big(\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\Big)^2[(x_3-x_2)^2+(y_3-y_2)^2]-\Big(\frac{x_2+x_3-2x_1}{2}\Big)^2\\ \Big(\frac{y_2+y_3-2y_1}{2}\Big)^2&=(x_2+x_3-2x_1)^2\Big(\frac{(x_3-x_2)^2+(y_3-y_2)^2}{4(x_3-x_2)^2}-\frac{1}{4}\Big)\\ \therefore \frac{(y_2+y_3-2y_1)^2}{4}&=\frac{1}{4}(x_2+x_3-2x_1)^2\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2\\ \Big(\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}\Big)^2&=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2 \end{split} \end{equation}$

Extracting square roots, we get $\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}=\pm(\frac{y_3-y_2}{x_3-x_2})$ . Since the slope of a median cannot be equal to the slope of the side it meets, we must then take the negative square root, and so $\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}=-(\frac{y_3-y_2}{x_3-x_2})$ . In terms of our notation, this translates to $m_{A}=-m_{BC}$ , or $m_{A}+m_{BC}=0$ .

Combining our first and second propositions, we obtain the following three equivalent statements for $\triangle ABC$

with vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$

$m_{A}+m_{BC}=0$ (that is, the slope of the median from vertex $A$ is the negative of the slope of side $BC$ );

the (non-zero) slopes of sides $AB,BC,CA$ form a geometric progression;

the length of the median from vertex $A$ is $m_{a}=\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$ .

Applying the first proposition

Our first application goes to the right triangle considered in Example 3; it will be the basis of a future post — where we aim to unveil a hidden characteristic in every triangle.

PROVE that the right triangle $ABC$ with vertices at $A(0,0),~B(k,0),~C(0,l)$ contains a “sub-triangle” whose side slopes form a geometric progression with a common ratio of $3$ .

In view of Example 3, we know that the median from vertex $A$ has a slope that’s the negative of the slope of the hypotenuse $BC$ . Consider the midpoint of median from $A$ to $BC$ , located at $G\Big(\frac{k}{4},\frac{l}{4}\Big)$ , say.

The triangle with vertices $C(0,l),~G\Big(\frac{k}{4},\frac{l}{4}\Big),~B(k,0)$ is the desired “sub-triangle”. Indeed, the slopes of its sides are:

$\begin{equation*} \begin{split} m_{BC}&=-\frac{l}{k}\\ &\cdots\vdots\cdots\\ m_{CG}&=\frac{l-\frac{l}{4}}{0-\frac{k}{4}}\\ &=\frac{\frac{3l}{4}}{-\frac{k}{4}}\\ &=3\Big(-\frac{l}{k}\Big)\\ &\cdots\vdots\cdots\\ m_{GB}&=\frac{\frac{l}{4}-0}{\frac{k}{4}-k}\\ &=\frac{\frac{l}{4}}{-\frac{3k}{4}}\\ &=\frac{1}{3}\Big(-\frac{l}{k}\Big) \end{split} \end{equation}$

This verifies that the slopes of the “sub-triangle” $BCG$ form a geometric progression with a common ratio of $3$ .

PROVE that if the slopes of the sides of a triangle form a geometric progression (with common ratio $r$ ), then so do the slopes of the three medians, provided $r\neq -2$ (or $-\frac{1}{2}$ ).

In Example 7 and Example 8 below, we explain what underlies the restriction $r\neq -2$ .

If $\triangle ABC$ has sides $AB,BC,CA$ with slopes $m_{AB},m_{BC},m_{CA}$ in geometric progression, we’ll prove that the median slopes $m_{C},m_{A},m_{B}$ form a new geometric progression.

Let the coordinates of $\triangle ABC$ be $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ . We have

$m_{AB}=\frac{y_1-y_2}{x_1-x_2},~m_{BC}=\frac{y_2-y_3}{x_2-x_3},~m_{CA}=\frac{y_3-y_1}{x_3-x_1}$

and

$m_{A}=\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1},~m_{B}=\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2},~m_{C}=\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}.$

From our first proposition, $m_{AB},~m_{BC},~m_{CA}$ being in geometric progression means that $m_{A}+m_{BC}=0$ , and so $m_{A}^2=m_{BC}^2$ . That is, $\Big(\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}\Big)^2=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2$ . From Example 1, $x_1=\frac{x_3(y_3-y_1)+x_2(y_1-y_2)}{y_3-y_2}$ .

$\begin{equation*} \begin{split} m_{B}\times m_{C}&=\Big(\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\Big)\Big(\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}\Big)\\ &=\frac{y_1+y_2-2y_3}{\Big[\frac{x_3(y_3-y_1)+x_2(y_1-y_2)}{y_3-y_2}+x_2-2x_3\Big]}\times\frac{y_1+y_3-2y_2}{\Big[\frac{x_3(y_3-y_1)+x_2(y_1-y_2)}{y_3-y_2}+x_3-2x_2\Big]}\\ &=\frac{y_1+y_2-2y_3}{\frac{x_3(y_3-y_1)+x_2(y_1-y_2)+x_2(y_3-y_2)-2x_3(y_3-y_2)}{y_3-y_2}}\times\frac{y_1+y_3-2y_2}{\frac{x_3(y_3-y_1)+x_2(y_1-y_2)+x_3(y_3-y_2)-2x_2(y_3-y_2)}{y_3-y_2}}\\ &=\scriptscriptstyle\frac{(y_1+y_2-2y_3)(y_3-y_2)}{\Big[x_3(y_3-y_1-2y_3+2y_2)+x_2(y_1-y_2+y_3-y_2)\Big]}\times\frac{(y_1+y_3-2y_2)(y_3-y_2)}{\Big[x_3(y_3-y_1+y_3-y_2)+x_2(y_1-y_2-2y_3+2y_2)\Big]}\\ &=\scriptscriptstyle\frac{(y_1+y_2-2y_3)(y_3-y_2)}{[x_3(-y_1-y_3+2y_2)+x_2(y_1+y_3-2y_2)]}\times\frac{(y_1+y_3-2y_2)(y_3-y_2)}{[x_3(-y_1-y_2+2y_3)+x_2(y_1+y_2-2y_3)]}\\ &=\frac{(y_1+y_2-2y_3)(y_3-y_2)}{(x_2-x_3)(y_1+y_3-2y_2)}\times\frac{(y_1+y_3-2y_2)(y_3-y_2)}{(x_2-x_3)(y_1+y_2-2y_3)}\\ &=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2\\ &=m_{A}^2 \end{split} \end{equation*}$

So, the median slopes $m_{C},~m_{A},~m_{B}$ form a geometric progression, with $m_{A}$ as the geometric mean.

Suppose that the slopes of the sides of a triangle form a geometric progression with a common ratio of $-2$ . PROVE that the triangle contains a horizontal median.

We’ve given a proof of this before by explicitly determining the coordinates of the vertices in terms of the common ratio $r=-2$ ; here we present a direct approach that avoids row reduction (linear algebra).

Assume the geometric progression of the slopes is $m_{AB},~m_{BC},~m_{CA}$ , where $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ are the triangle’s vertices. Since the common ratio is $-2$ , we have:

$\frac{y_2-y_3}{x_2-x_3}=-2\Big(\frac{y_1-y_2}{x_1-x_2}\Big),~\frac{y_1-y_3}{x_1-x_3}=-2\Big(\frac{y_2-y_3}{x_2-x_3}\Big).$

Isolate $x_1$ from the first equation:

$x_1=x_2-\frac{2(y_1-y_2)(x_2-x_3)}{y_2-y_3}.$

Also, from the second equation, isolating $x_1$ gives

$x_1=x_3-\frac{(y_1-y_3)(x_2-x_3)}{2(y_2-y_3)}.$

Equate the two expressions obtained for $x_1$ :

$\begin{equation*} \begin{split} x_2-\frac{2(y_1-y_2)(x_2-x_3)}{y_2-y_3}&=x_3-\frac{(y_1-y_3)(x_2-x_3)}{2(y_2-y_3)}\\ x_2-x_3&=\frac{2(y_1-y_2)(x_2-x_3)}{y_2-y_3}-\frac{(y_1-y_3)(x_2-x_3)}{2(y_2-y_3)}\\ 2(x_2-x_3)(y_2-y_3)&=4(y_1-y_2)(x_2-x_3)-(y_1-y_3)(x_2-x_3)\\ \therefore 2(y_2-y_3)&=4(y_1-y_2)-(y_1-y_3),\quad\textrm{since}~x_2\neq x_3\\ 2y_2-2y_3&=4y_1-4y_2-y_1+y_3\\ 3y_1+3y_3-6y_2&=0\\ y_1+y_3-2y_2&=0 \end{split} \end{equation}$

Thus, the slope of the median from vertex $B$ is zero, that is, $m_{B}=\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}=\frac{0}{x_1+x_3-2x_2}=0$ , giving a horizontal median.

Suppose that the slopes of the sides of a triangle form a geometric progression with a common ratio of $-2$ . PROVE that the triangle contains a vertical median.

As in Example 7, the fact the geometric progression has a common ratio of $-2$ means we can write:

$\frac{y_2-y_3}{x_2-x_3}=-2\Big(\frac{y_1-y_2}{x_1-x_2}\Big),~\frac{y_1-y_3}{x_1-x_3}=-2\Big(\frac{y_2-y_3}{x_2-x_3}\Big).$

We obtain $\frac{y_1-y_3}{x_1-x_3}=4\Big(\frac{y_1-y_2}{x_1-x_2}\Big)$ , from which

$y_3=y_1-\frac{4(y_1-y_2)(x_1-x_3)}{x_1-x_2}.$

Also, from $\frac{y_2-y_3}{x_2-x_3}=-2\Big(\frac{y_1-y_2}{x_1-x_2}\Big)$ , isolating $y_3$ gives

$y_3=y_2+\frac{2(y_1-y_2)(x_2-x_3)}{x_1-x_2}.$

Equate the two expressions obtained for $y_3$ :

$\begin{equation*} \begin{split} y_1-\frac{4(y_1-y_2)(x_1-x_3)}{x_1-x_2}&=y_2+\frac{2(y_1-y_2)(x_2-x_3)}{x_1-x_2}\\ y_1-y_2&=\frac{2(y_1-y_2)(x_2-x_3)}{x_1-x_2}+\frac{4(y_1-y_2)(x_1-x_3)}{x_1-x_2}\\ y_1-y_2&=\frac{2(y_1-y_2)(x_2-x_3+2x_1-2x_2)}{x_1-x_2}\\ (y_1-y_2)(x_1-x_2)&=2(y_1-y_2)(2x_1+x_2-3x_3)\\ \therefore x_1-x_2&=2(2x_1+x_2-3x_3)\quad\textrm{since}~y_1\neq y_2\\ 0&=3x_1+3x_2-6x_3\\ 0&=x_1+x_2-2x_3 \end{split} \end{equation}$

Therefore, the slope of the median from vertex $C$ , given by $m_{C}=\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}=\frac{y_1+y_2-2y_3}{0}$ , is undefined, so this median is vertical.

Suppose that $\triangle ABC$ is isosceles and that the slope of the median from the “apex” $A$ is $m_{A}=\pm 1$ . PROVE that the slopes of its sides form a geometric progression.

This is based on the fact that the slope of the median from the “apex” of an isosceles triangle is the negative reciprocal of the slope of the base. In particular, if $m_{A}=1$ , then $m_{BC}=\frac{-1}{1}=-1$ , giving $m_{A}+m_{BC}=0$ . By our first proposition, this means the slopes of sides $AB,BC,CA$ form a geometric progression. Similarly, if $m_{A}=-1$ , then $m_{BC}=\frac{-1}{-1}=1$ , also giving $m_{A}+m_{BC}=0$ , and we again obtain a geometric progression for the slopes of sides $AB,BC,CA$ .

A numerical problem

$\triangle ABC$ has vertices at $A(r,s),~B(-2,1),~C(4,-3)$ . Determine suitable values of $r$ and $s$ for which the slopes of sides $AB,BC,CA$ form a geometric progression.

Using our first proposition, we can tackle this problem by somehow “forcing” a median from side $BC$ to have a slope that’s the negative of side $BC$ ‘s slope. (Note that we can also use Exercise 1 below, which may be easier.)

Let’s find the equation of this “forced” median. The midpoint of $BC$ is $(1,-1)$ , and its slope is $-\frac{2}{3}$ . So we want a median through $(1,-1)$ with slope $\frac{2}{3}$ . Its equation is:

$\begin{equation*} \begin{split} y-(-1)&=\frac{2}{3}\Big(x-1\Big)\\ y&=\frac{2}{3}x-\frac{5}{3} \end{split} \end{equation}$

Thus, any point on the line $y=\frac{2}{3}x-\frac{5}{3}$ — except the trio $(4,1),~(1,-1),~(-2,-3)$ — will do for the coordinates of vertex $A$ . Why exclude these three? First, the point $(1,-1)$ is the midpoint of $BC$ , so we can’t form a triangle with it. Next, the point $(4,1)$ shares the same $y$ -coordinate with vertex $B$ , which will make the slope of $AB$ zero if $A$ is chosen as the point $(4,1)$ . Similarly, we can’t take $(-2,-3)$ for $A$ because then $A$ will have the same $x$ -coordinate with vertex $C$ , resulting in an undefined slope.

With the exclusions out of the way, we can take any other point, say $(7,3)$ , as $A$ . Then $\triangle ABC$ has vertices at $A(7,3),~B(-2,1),~C(4,-3)$ , and the slopes of sides $AB,BC,CA$ become

$\frac{2}{9},-\frac{2}{3},2$

respectively. They form a geometric progression with a common ratio of $-3$ (or $-\frac{1}{3}$ if we reverse the numbers).

A few other points on the line $y=\frac{2}{3}x-\frac{5}{3}$ that can be chosen in place of $A(7,3)$ — and the corresponding geometric progressions for the slopes of the sides — are listed below:

$\begin{equation*} \begin{split} \textbf{(10,5)},(-2,1),(4,-3)&\mapsto\Big\{\frac{1}{3},-\frac{2}{3},\frac{4}{3}\Big\},~r=-2;\\ \textbf{(13,7)},(-2,1),(4,-3)&\mapsto\Big\{\frac{2}{5},-\frac{2}{3},\frac{10}{9}\Big\},~r=-\frac{5}{3};\\ \textbf{(16,9)},(-2,1),(4,-3)&\mapsto\Big\{\frac{4}{9},-\frac{2}{3},1\Big\},~r=-\frac{3}{2};\\ \textbf{(19,11)},(-2,1),(4,-3)&\mapsto\Big\{\frac{10}{21},-\frac{2}{3},\frac{14}{15}\Big\},~r=-\frac{7}{5};\\ \textbf{(-5,-5)},(-2,1),(4,-3)&\mapsto\Big\{\frac{2}{9},-\frac{2}{3},2\Big\},~r=-3 \end{split} \end{equation}$

Observe that the choice $(-5,-5)$ yields the same set of geometric progression as our initial choice $(7,3)$ .

As in, this whole thing is COOL!

Takeaway

In the domain of triangles, the GOAT (Greatest Of All Triangles) is the triangle whose slopes form a geometric progression. No jokes.

That told, our first proposition is the basis for our future application. In particular, what we did for the right triangle in Example 5 will be done for any triangle. As in, A-N-Y!

Tasks

(Generating formula) Let $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ be the coordinates of the vertices of $\triangle ABC$ . Require that $x_1\neq x_2\neq x_3$ and that $y_1\neq y_2\neq y_3$ . PROVE that the slopes of sides $AB,BC,CA$ form a geometric progression if, and only if, $x_1(y_3-y_2)+x_2(y_2-y_1)+x_3(y_1-y_3)=0.$
$\Big($ Thus, in order to obtain coordinates for the vertices of a triangle whose slopes form a geometric progression, begin by selecting “any” five numbers — ensuring that no two $x'$ s are equal and no two $y'$ s are equal — then use the above relation to determine the value of the remaining coordinate. $\Big)$
(Special ratio) Let be the coordinates of the vertices of . PROVE that:
- $m_{a}=\frac{3}{2}a$ if, and only if, $\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\times\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}=-1$ (that is, the medians through $B$ and $C$ are perpendicular)
- if the slopes $m_{AB},~m_{BC},~m_{CA}$ form a geometric progression, then $m_{a}=\frac{3}{2}a$ if, and only if, $x_1+x_3-2x_2=0$ (that is, the median through $B$ is vertical)
- if the slopes $m_{AB},~m_{BC},~m_{CA}$ form a geometric progression and $m_{a}=\frac{3}{2}a$ , then the median through $C$ is horizontal
- if the slopes $m_{AB},~m_{BC},~m_{CA}$ form a geometric progression and $m_{a}=\frac{3}{2}a$ , then the common ratio of the geometric progression is $r=-2.$
  (If the slopes of the sides of a triangle form a geometric progression, no other value for the common ratio yields the relation $m_{a}=\frac{3}{2}a$ ; only $r=-2$ does.)
(Half right) Let $ABC$ be a triangle in which the slopes $m_{AB},m_{BC},m_{CA}$ form a geometric progression. PROVE that the acute angle between the median from $A$ and side $BC$ is $45^{\circ}$ if, and only if, $m_{BC}=-1\pm\sqrt{2}$ or $m_{BC}=1\pm\sqrt{2}$ .
(Pseudo perpendicularity) Suppose that the slopes $m_{AB},m_{BC},m_{CA}$ of the sides $AB,BC,CA$ of $\triangle ABC$ form a geometric progression with common ratio $r$ . Let the lengths of sides $AB,BC,CA$ be $l_1,l_2,l_3$ , respectively. PROVE that the following pseudo-pythagorean property holds: $l_1^2+l_3^2=\frac{1+r^2}{(1+r)^2}l_2^2$ .
(If we let $r=0$ , then we obtain a full pythagorean relationship. However, $r=0$ is unacceptable for a geometric progression. Note that this doesn’t imply the impossibility of having right triangles with slopes in geometric progression; there are actually many of them — with an appropriately chosen first term and a negative common ratio. Now, if we let $r=-2$ , we get $l_1^2+l_3^2=5l_2^2$ , a peach.)
Suppose that the slopes of a right triangle form a geometric progression. PROVE that the slope of the hypotenuse cannot be the geometric mean of the progression.
(For a change, attempt this exercise without recourse to perpendicular slopes. Use our second proposition (Example 4) and the fact that the length of the median to the hypotenuse is half of the length of the hypotenuse.)
Let $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ be the coordinates of the vertices of an equilateral $\triangle ABC$ . If the slopes of sides $AB,BC,CA$ form a geometric progression, prove that $x_{1}=\frac{(1+\sqrt{3})x_2+(1-\sqrt{3})x_{3}}{2}$ , or $x_{1}=\frac{(1-\sqrt{3})x_{2}+(1+\sqrt{3})x_{3}}{2}.$
(Note the order of the slopes.)
$\triangle ABC$ has vertices at $A(-2,1),~B(4,-3),~C(r,s)$ . Find three different pairs $(r,s)$ for which the slopes of the sides form a geometric progression with a common ratio of $-2$ .
(Three possibilities) Let be the coordinates of the vertices of a triangle. If the slopes of the sides form a geometric progression, PROVE that at least one of these three is true:
- $r+s=5$ ;
- $rs-2r-2s+3=0$ ;
- $rs-3r-3s+8=0$ .
  (Use the generating formula.)
PROVE that every right triangle $ABC$ with legs $AB$ and $AC$ on — or parallel to — the coordinate axes contains a “sub-triangle” $GBC$ that has the following property: the length of the median from vertex $G$ is one-fourth the length of the hypotenuse $BC$ ; that is, $m_{g}=\frac{1}{4}\times BC$ .
(For instance, consider the “sub-triangle” we constructed in Example 5.)
Find suitable choice of coordinates for the vertices of $\triangle ABC$ that has the property that $m_{a}=\frac{1}{3}a$ , where $m_{a}$ is the length of the median from vertex $A$ and $a$ is the length of side $BC$ .
(Consider the case where the slopes of sides $AB,BC,CA$ form a geometric progression with a common ratio of $5$ .)