From nothing to infinity – High School Geometry

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 $n\geq 1$ .