Geometric sequence on a staircase – High School Geometry

In case you’ve used a staircase before, you’ve “experienced” a special case of a geometric sequence as a consequence.

For a case study, consider the diagram below:

Join $A$ to $D$ , $D$ to $G$ , and $G$ to $A$ to form $\triangle ADG$ :

We’ll prove that the slopes of the sides of $\triangle ADG$ form a geometric sequence, if $AC=EG$ .

Example

A portion of a staircase is shown below:

Verify that the slopes of the sides of $\triangle ADG$ form a geometric sequence.

Easy-peasy.

The slopes of sides $DG,GA,AD$ are $\frac{1}{2},1,2$ , respectively. They form a geometric sequence with a common ratio $2$ .

(The fact that the slope of side $GA$ is $1$ implies that it makes an angle of $45^{\circ}$ with the horizontal. In real life, however, the angle is usually around $30^{\circ}$ .)

Example

A portion of a staircase is shown below:

Verify that the slopes of the sides of $\triangle ADG$ form a geometric sequence.

Easy-peasy as before.

The slopes of sides $DG,GA,AD$ are $\frac{11}{40},\frac{11}{20},\frac{11}{10}$ , respectively. They form a geometric sequence with a common ratio of $2$ .

(It shouldn’t come as a surprise that the common ratio is $2$ this time too. Indeed, it is almost always the case in a staircase, because the “rise” and “run” stay constant throughout the entire span of the staircase. For this reason, the geometric sequence obtained on a staircase is a special case.)

Basic background

From Example 3 to Example 6 below, we lay the foundation for our staircase story. Don’t worry if it seems to “escalate” quite quickly; remember we’re on a staircase $\cdots$ .

PROVE that if the slopes of the sides of a triangle form a geometric sequence with a positive common ratio, then the triangle contains an obtuse angle.

Actually, any triangle in which the three sides have positive slopes contains an obtuse angle (for now we’re content with slopes in geometric sequence).

Let $\triangle ABC$ be such that sides $AB,BC,CA$ have slopes $a,ar,ar^2$ , respectively. If the common ratio $r$ is positive, then the angle at $A$ is obtuse. To see this, let the lengths of sides $AB,BC,CA$ be $l_1,l_2,l_3$ , respectively. Then, as we’ll show in our next post, the following pseudo-pythagorean property holds:

$l_1^2+l_3^2=\frac{1+r^2}{(1+r)^2}l_2^2.$

Thus, we can use the cosine law to compute the cosine of the angle at $A$ :

$\begin{equation*} \begin{split} \cos A&=\frac{l_1^2+l_3^2-l_2^2}{2\times l_1\times l_3}\\ &=\frac{\left(\frac{1+r^2}{(1+r)^2}l_2^2\right)-l_2^2}{2\times l_1\times l_3}\\ &=\frac{\left(1+r^2-(1+r)^2\right)l_2^2}{2\times(1+r)^2\times l_1\times l_3}\\ &=\frac{-2r\times l_2^2}{2\times (1+r)^2\times l_1\times l_3}\\ &=\frac{-r}{(1+r)^2}\frac{l_2^2}{l_1l_3}\\ \implies\cos A& < 0\\ \therefore\angle A&~~\textrm{is OBTUSE} \end{split} \end{equation*}$

The obtuse angle at $A$ is well-oriented in such a way that a special $90^{\circ}$ can be cut out from it.

In $\triangle ABC$ , let $a,ar,ar^2~(r> 0)$ be the slopes of sides $AB,BC,CA$ , respectively. Then a point $X$ divides $BC$ in the ratio $r:1$ if and only if it shares the same $x$ -coordinate as vertex $A$ .

Thus, $AX$ will be a vertical line. Take note of this.

The problem deals with coordinates, so let’s begin by specifying the coordinates of the vertices of the parent triangle $ABC$ . Assume $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ are the vertices. Since the slopes of sides $AB,BC,CA$ are $a,ar,ar^2$ , there are relationships among the coordinates:

(1) $\begin{equation*} \begin{split} x_1&=x_3+\frac{1}{ar(r+1)}(y_2-y_3)\\ y_1&=\frac{r}{r+1}y_2+\frac{1}{r+1}y_3\\ x_2&=x_3+\frac{1}{ar}(y_2-y_3) \end{split} \end{equation*}$

Now, suppose a point $X$ divides side $BC$ in the ratio $r:1$ , with $r> 0$ . Using the formula for internal division of a line segment, we can write the $x$ -coordinate of point $X$ in terms of the $x$ -coordinates of $B(x_2,y_2)$ and $C(x_3,y_3)$ as follows:

$\begin{equation*} \begin{split} x&=\frac{rx_3+1x_2}{r+1}\\ &=\frac{rx_3+1\left(x_3+\frac{1}{ar}(y_2-y_3)\right)}{r+1}\\ &=\frac{(r+1)x_3+\frac{1}{ar}(y_3-y_3)}{r+1}\\ &=x_3+\frac{1}{ar(r+1)}(y_2-y_3)\\ &=x_1 \end{split} \end{equation*}$

Therefore, point $X$ shares the same $x$ -coordinate as vertex $A(x_1,y_1)$ .

Conversely, suppose that point $X$ shares the same $x$ -coordinate as vertex $A(x_1,y_1)$ . We proceed to find the $y$ -coordinate of $X$ . Since the slope of $BC$ is $ar$ and it goes through $(x_3,y_3)$ , its equation is $y-y_3=ar(x-x_3)$ . Using the fact that $x=x_1=x_3+\frac{1}{ar(r+1)}(y_2-y_3)$ at point $X$ , we have:

$\begin{equation*} \begin{split} y-y_3&=ar\left(x_3+\frac{1}{ar(r+1)}(y_3-y_3)-x_3\right)\\ y&=ar\left(\frac{1}{ar(r+1)}(y_3-y_3)\right)+y_3\\ &=\frac{1}{r+1}(y_2-y_3)+\left(\frac{r+1}{r+1}\right)y_3\\ \therefore y&=\frac{y_2+ry_3}{r+1} \end{split} \end{equation*}$

Let’s manipulate $x_1=x_3+\frac{1}{ar(r+1)}(y_2-y_3)$ slightly:

$\begin{equation*} \begin{split} x_1&=\left(\frac{r+1}{r+1}\right)x_3+\frac{1}{ar(r+1)}(y_2-y_3)\\ &=\frac{rx_3+x_3+\frac{1}{ar}(y_2-y_3)}{r+1}\\ &=\frac{rx_3+x_2}{r+1} \end{split} \end{equation*}$

Thus, $X$ is the point $\left(\frac{x_2+rx_3}{r+1},\frac{y_2+ry_3}{r+1}\right)$ . By examining its coordinates, we see that it actually divides side $BC$ in the ratio $r:1$ .

In $\triangle ABC$ , let $a,ar,ar^2~(r> 0)$ be the slopes of sides $AB,BC,CA$ , respectively. Then a point $Y$ divides $BC$ in the ratio $1:r$ if and only if it shares the same $y$ -coordinate as vertex $A$ .

Thus, $AY$ will be a horizontal line. Together with the previous example, the parent $\triangle ABC$ contains a special right triangle $AXY$ in which both legs $AX$ and $AY$ are parallel to the coordinate axes. Further, $BX=CY$ . The presence of this right triangle is key to our staircase story.

Let’s find the $y$ coordinate of point $Y$ which divides $BC$ in the ratio $1:r$ . Using equation (1) and the formula for internal division of a line segment:

$\begin{equation*} \begin{split} y&=\frac{ry_2+y_3}{r+1}\\ &=y_1\\ \end{split} \end{equation*}$

Easier than thought. Conversely, suppose that $Y$ shares the same $y$ -coordinate as vertex $A(x_1,y_1)$ , so that $y=y_1=\frac{ry_2+y_3}{r+1}$ . We’ll find its $x$ -coordinate, using the fact that it lies on side $BC$ whose equation is $y-y_3=ar(x-x_3)$ :

$\begin{equation*} \begin{split} \left(\frac{ry_2+y_3}{r+1}\right)-y_3&=ar(x-x_3)\\ \left(\frac{ry_2+y_3}{r+1}\right)-\left(\frac{r+1}{r+1}\right)y_3&=ar(x-x_3)\\ \frac{r}{r+1}(y_2-y_3)&=ar(x-x_3)\\ \frac{1}{a(r+1)}(y_2-y_3)&=x-x_3\\ x&=x_3+\frac{1}{a(r+1)}(y_2-y_3)\\ &=\left(\frac{r+1}{r+1}\right)x_3+\frac{1}{a(r+1)}(y_2-y_3)\\ &=\frac{x_3+rx_3+\frac{1}{a}(y_2-y_3)}{r+1}\\ &=\frac{x_3+r\left(x_3+\frac{1}{ar}(y_2-y_3)\right)}{r+1}\\ \therefore x&=\frac{x_3+rx_2}{r+1} \end{split} \end{equation*}$

These calculations show that $Y$ is the point $\left(\frac{rx_2+x_3}{r+1},\frac{ry_2+y_3}{r+1}\right)$ , and it divides side $BC$ in the ratio $1:r$ if we inspect its coordinates.

In the above diagram, point $C$ divides $AG$ in the ratio $r:1$ while point $E$ divides $AG$ in the ratio $1:r$ .

(Main goal)

Let $\triangle ABC$ be a triangle. Then the following two statements are equivalent:

the slopes of the sides form a geometric sequence with positive common ratio;
there are internal points $X$ and $Y$ on $BC$ such that $AX$ is vertical, $AY$ is horizontal, and $BX=CY$ .

The implication $(1)\implies(2)$ follows from the three previous examples. For $(2)\implies (1)$ , consider the diagram below:

We’ll show that

(2) $\begin{equation*}\left(\frac{f-b}{e-a}\right)\times\left(\frac{b-d}{a-c}\right)=\left(\frac{f-d}{e-c}\right)^2. \end{equation*}$

Since $BX=CY$ , we have:

$(a-c)^2+(v-d)^2=(f-b)^2+(e-u)^2.$

Because $B,X,Y,C$ are co-linear, we have $\frac{f-b}{e-u}=\frac{v-d}{a-c}$ , and so $f-b=(e-u)\left(\frac{v-d}{a-c}\right)$ . Using this in the previous equation, we have:

$\begin{equation*} \begin{split} (a-c)^2+(v-d)^2&=\left[(e-u)\left(\frac{v-d}{a-c}\right)\right]^2+(e-u)^2\\ (a-c)^2+(v-d)^2&=(e-u)^2\left[\frac{(v-d)^2+(a-c)^2}{(a-c)^2}\right]\\ \therefore (a-c)^2&=(e-u)^2 \end{split} \end{equation*}$

The latter means that either $a-c=e-u$ or $a-c=-(e-u)$ . However, in our diagram we have $a> c$ and $e> u$ , thus ruling out the possibility of $a-c=-(e-u)$ . We therefore settle for $a-c=e-u$ .

Returning to (2) and using some equality relations among the slopes of the co-linear points $B,X,Y,C$ , we have:

$\begin{equation*} \begin{split} \left(\frac{f-d}{e-c}\right)^2&=\left(\frac{b-v}{u-a}\right)^2\\ \left(\frac{f-b}{e-a}\right)\left(\frac{b-d}{a-c}\right)&=\left[(e-u)\frac{b-v}{u-a}\right].\frac{1}{e-a}\left(\frac{b-d}{a-c}\right)\\ &=\left(\frac{b-v}{u-a}\right)\left(\frac{b-d}{e-a}\right)\\ &=\left(\frac{b-v}{u-a}\right)\left(\frac{b-d}{u-c}\right)\\ &=\left(\frac{b-v}{u-a}\right)\left(\frac{b-v}{u-a}\right) \end{split} \end{equation*}$

This proves that the slopes of the sides of $\triangle ABC$ form a geometric progression. The common ratio is positive in view of the diagram.

In $\triangle ABC$ , let $a,ar,ar^2~(r> 0)$ be the slopes of sides $AB,BC,CA$ , respectively. For any positive integer $n\neq 1,r,\frac{1}{r}$ , PROVE that there are points $X$ and $Y$ on $BC$ such that the slopes of the sides of $\triangle AXY$ form a geometric sequence $\left(\frac{n-r}{nr-1}\right)ar,ar,\left(\frac{nr-1}{n-r}\right)ar$ .

Note that both Example 4 and Example 5 result from the special case $n=r$ , but the associated slopes do not form a geometric sequence. Also, $n$ need not be a positive integer; any positive number works as well.

We need to construct points $X$ and $Y$ first.

Let $X$ be the point which divides side $BC$ in the ratio $n:1$ . (The restriction that $n\neq 1$ is essential to exclude the midpoint of $BC$ .) Using equation (1) and the formula for internal division of a line segment, we can determine the coordinates of $X$ as follows:

$\begin{equation*} \begin{split} X_x&=\frac{nx_3+x_2}{n+1}\\ &=\frac{nx_3+x_3+\frac{1}{ar}(y_2-y_3)}{n+1}\\ &=\frac{(n+1)x_3+\frac{1}{ar}(y_2-y_3)}{n+1}\\ &=x_3+\frac{1(y_2-y_3)}{ar(n+1)}\\ X_y&=\frac{ny_3+y_2}{n+1} \end{split} \end{equation*}$

So, $X$ is the point $\left(x_3+\frac{(y_2-y_3)}{ar(n+1)},\frac{ny_3+y_2}{n+1}\right)$ . Next, let’s construct point $Y$ by dividing side $BC$ in the ratio $1:n$ .

$\begin{equation*} \begin{split} Y_x&=\frac{x_3+nx_2}{n+1}\\ &=\frac{x_3+nx_3+\frac{n}{ar}(y_2-y_3)}{n+1}\\ &=x_3+\frac{n}{ar(n+1)}(y_2-y_3)\\ Y_y&=\frac{y_3+ny_2}{n+1} \end{split} \end{equation*}$

Thus, $Y$ is the point $\left(x_3+\frac{n(y_2-y_3)}{ar(n+1)},\frac{y_3+ny_2}{n+1}\right)$ . Together with equation (1), the vertices of $\triangle AXY$ are:

$\begin{equation*} \begin{split} A&\left(x_3+\frac{(y_2-y_3)}{ar(r+1)},\frac{ry_2+y_3}{r+1}\right)\\ X&\left(x_3+\frac{(y_2-y_3)}{ar(n+1)},\frac{ny_3+y_2}{n+1}\right)\\ Y&\left(x_3+\frac{n(y_2-y_3)}{ar(n+1)},\frac{y_3+ny_2}{n+1}\right) \end{split} \end{equation*}$

From these, we obtain $\left(\frac{nr-1}{n-r}\right)ar,~ar,~\left(\frac{n-r}{nr-1}\right)ar$ as the slopes of sides $AX,XY,YA$ , respectively. Note that $n=0$ yields $a,ar,ar^2$ as the slopes of $AX,XY,YA$ — same slopes as those of the parent triangle $ABC$ .

The next example shows that if we exclude three points from $BC$ , namely, the midpoint of $BC$ , the point on $BC$ that shares the same $x$ -coordinate as $A$ , and the point on $BC$ that shares the same $y$ -coordinate as $A$ , then we obtain what may be regarded as a stronger version of Example 6.

For any $\triangle ABC$ , the following statements are equivalent:

the slopes of sides $AB,BC,CA$ form a geometric sequence with positive common ratio;
for any internal point $X$ on $BC$ , there is an internal point $Y$ on $BC$ such that $BX=YC$ and the slopes of $\triangle AXY$ form a geometric sequence.

Bear in mind the three excluded points on $BC$ . Basically, these are the points where the slope of $AX$ will be undefined, the slope of $AY$ will be zero, and where $X=Y$ . Naturally, these were excluded to ensure that we have proper slopes.

To show $(1)\implies (2)$ , let $X$ be the point which divides $BC$ in the ratio $n:1$ , where $n\neq 1$ is a positive integer. Choose $Y$ as the point which divides $BC$ in the ratio $1:n$ . Then $BX=CY$ . By Example 7, the slopes of the sides of $\triangle AXY$ form a geometric progression.

To show $(2)\implies (1)$ , take $X=B$ . Then $Y$ has to be taken as $C$ , so that the equality $BX=YC$ is satisfied (both sides being zero).

PROVE that the two infinite geometric progressions

(3) $\begin{equation*} \cdots,-\frac{1}{4},-\frac{1}{2},-1,-2,-4,\cdots \end{equation*}$

(4) $\begin{equation*} \cdots,\frac{1}{4},\frac{1}{2},1,2,4,\cdots \end{equation*}$

can be obtained from the same triangle.

You already know how to generate (4); if not, see here. So we focus on obtaining (3), following a similar procedure and using the same triangle that yields (4).

Begin with a triangle whose slopes are $1,2,4$ — like the one shown below:

where $A(1,4)$ , $B(0,0)$ , $C(3,6)$ are the vertices.

For an integer $n$ , set $X_{n}:=\left(\frac{2^{n-1}+2}{2^{n-1}+1},\frac{2^{n}+4}{2^{n-1}+1}\right)$ . Then $X_n$ is an internal point on $BC$ ; further, $X_{n}\rightarrow (1,2)$ as $n\rightarrow\infty$ and $X_{n}\rightarrow (2,4)$ as $n\rightarrow -\infty$ . These limits ensure that $X_{n}$ always resides within side $BC$ .

Next, let’s compute the slope of cevian $AX_{n}$ for each $n$ :

$\begin{equation*} \begin{split} \textrm{slope of}~AX_{n}&=\frac{4-\left(\frac{2^{n}+4}{2^{n-1}+1}\right)}{1-\left(\frac{2^{n-1}+2}{2^{n-1}+1}\right)}\\ &=\frac{4\left(2^{n-1}+1\right)-\left(2^{n}+4\right)}{2^{n-1}+1-\left(2^{n-1}+2\right)}\\ &=\frac{2(2^n)-2^n}{-1}\\ &=-2^n \end{split} \end{equation*}$

Letting $n=\cdots, -2,-1,0,1,2,\cdots$ gives rise to the infinite geometric sequence in (3).

That special right triangle

We conclude by isolating a unique property of the special right triangle that appears in staircases.

PROVE that a triangle is a right triangle with legs parallel to the coordinate axes if and only if the slopes of its three medians form a geometric sequence with common ratio $-2$ .

First suppose that we have a right triangle with legs parallel to the coordinate axes. This direction of the proof was done in our previous post, but let’s go through it briefly. Fix the vertices at $A(0,a)$ , $B(0,0)$ , and $C(c,0)$ ; a little modification will be needed for the general case.

Now calculate the slopes of medians $CN,BM,AL$ :

$\begin{equation*} \begin{split} \textrm{slope of median}~ CN&=-\frac{a}{2c}\\ \textrm{slope of median}~ BM&=\frac{a}{c}\\ \textrm{slope of median}~ AL&=-\frac{2a}{c} \end{split} \end{equation*}$

These form a geometric progression with common ratio of $-2$ .

Conversely, suppose that we have $\triangle ABC$ with vertices $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_3)$ in such a way that the slopes of the three medians form a geometric sequence with common ratio $-2$ .

If we denote the slopes of the medians from vertices $A,B,C$ by $m_A,m_B,m_C$ respectively, then:

$\begin{equation*} \begin{split} m_A&=\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}\\ m_B&=\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\\ m_C&=\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3} \end{split} \end{equation*}$

Suppose that the geometric sequence of median slopes is $m_C,m_B,m_A$ , then $m_B=-2m_C$ , and $m_A=-2m_B$ :

$\begin{equation*} \begin{split} \frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}&=-2\left(\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}\right)\\ \frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}&=-2\left(\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\right)\\ \end{split} \end{equation*}$

Clear fractions from the first equation above:

(5) $\begin{equation*} \begin{split} (x_1+x_2-2x_3)(y_1+y_3-2y_2)+2(x_1+x_3-2x_2)(y_1+y_2-2y_3)&=0\\ x_1(y_1-y_3)+x_2(-y_1-2y_2+3y_3)-2x_3(y_3-y_2)&=0 \end{split} \end{equation*}$

Now clear fractions from the second equation:

(6) $\begin{equation*} \begin{split} (x_1+x_3-2x_2)(y_2+y_3-2y_1)+2(x_2+x_3-2x_1)(y_1+y_3-2y_2)&=0\\ x_1(2y_1-3y_2+y_3)+2x_2(-y_1+y_2)+x_3(y_2-y_3)&=0 \end{split} \end{equation*}$

It turns out that $y_1$ can be eliminated if we perform the operation $2\times$ (5)-(6):

$\begin{equation*} \begin{split} x_1(3y_2-3y_3)+x_2(-6y_2+6y_3)+x_3(-3y_3+3y_2)&=0\\ \implies (x_1+x_3-2x_2)(y_2-y_3)&=0\\ &\vdots\cdots\vdots\\ x_1+x_3-2x_2&=0~\textrm{or}~y_2-y_3=0 \end{split} \end{equation*}$

Since the slopes of the three medians form a geometric sequence, we can’t have $x_1+x_3-2x_2=0$ . Otherwise this would result in an undefined slope for the median from vertex $B$ , namely $m_B=\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}$ , and a geometric sequence cannot contain such a term. So the only option is to take $y_2-y_3=0$ , whence $y_2=y_3$ . Our triangle now has one side parallel to the $x$ -axis.

Put $y_2=y_3$ in equation (5):

$\begin{equation*} \begin{split} x_1(y_1-y_3)+x_2(-y_1-2y_2+3y_3)-2x_3(y_3-y_2)&=0\\ \implies x_1(y_1-y_2)+x_2(-y_1-2y_2+3y_2)-2x_3(y_2-y_2)&=0\\ \implies (x_1-x_2)(y_1-y_2)&=0\\ \implies x_1-x_2&=0\quad\textrm{or}~y_1-y_2=0 \end{split} \end{equation*}$

Since we already had $y_2=y_3$ , it’s impossible to have $y_1=y_2$ in addition, otherwise this would lead to $y_1=y_2=y_3$ , which is not allowed for a triangle. So we take $x_1-x_2=0$ , whence $x_1=x_2$ . Our triangle now has a side that is parallel to the $y$ -axis.

Therefore, $\triangle ABC$ has side $AB$ parallel to the $y$ -axis, and side $BC$ parallel to the $x$ -axis.

Takeaway

We’ve used staircases in various cases and in various places — homes, airports, train stations, offices, etc. As we ascend our staircase, we “run” forward and “rise” higher, so our slopes are positive. In turn, the geometric sequence we describe has a positive common ratio — in most cases, $2$ .

Embellished with nice properties, triangles with slopes in geometric progression are established as objects to cherish and relish. And, they not only appeal in theory, they appear in real life too.

Tasks

(Three out) Given $\triangle ABC$ with vertices $A(1,9), B(0,0), C(4,12)$ , the slopes of the sides form a geometric sequence with common ratio $3$ . Find two points $X$ and $Y$ on $BC$ for which the slopes of the sides of $\triangle AXY$ do not form a geometric sequence.
(The third point is the midpoint of $BC$ , namely $(2,6)$ . Apart from these three points, given any other point $X'$ within $BC$ , there is a corresponding point $Y'$ within $BC$ for which the slopes of the sides of $\triangle AX'Y'$ form a geometric sequence.)
Consider the diagram below:

in which $C$ and $D$ have the same $x$ -coordinates, $D$ and $E$ share the same $y$ -coordinates, and $AC=EG$ .
- If also $CE=EG$ , PROVE that the slopes of the sides of $\triangle ADG$ form a geometric sequence with a common ratio of $2$ ;
- Give an example to show that if $C$ and $E$ do not trisect $AG$ , then the common ratio is no longer $2$ .
PROVE that if the slopes of the sides of a triangle are all positive (and non-zero), then the triangle contains an obtuse angle.
(Similarly, if the slopes are all negative, there’ll be an obtuse angle.)
In $\triangle ABC$ , let $a,ar,ar^2$ be the slopes of sides $AB,BC,CA$ , respectively. If $r> 0$ , PROVE that side $BC$ is the longest side.
In $\triangle ABC$ , let $a,ar,ar^2$ be the slopes of sides $AB,BC,CA$ , respectively. If $r< 0$ , PROVE that side $BC$ is NOT always the longest side.
PROVE that the two infinite geometric sequences $\cdots,-\frac{1}{9},-\frac{1}{3},-1,-3,-9,\cdots$ and $\cdots,\frac{1}{9},\frac{1}{3},1,3,9,\cdots$ can be obtained from the same triangle, using slopes.
(Close enough) Let $ABC$ be a triangle in which sides $AB,BC,CA$ have slopes $a,ar,ar^2$ , with $a\neq 0$ and $r> 0$ . Let $X$ and $Y$ be points on $BC$ which divide $BC$ in the ratios $r:1$ and $1:r$ . PROVE that the area of $\triangle AXY$ is $\left(\frac{r-1}{r+1}\right)$ times the area of $\triangle ABC$ .
(In the limit $r\rightarrow\infty$ , we have area of $\triangle AXY$ $\approx$ area of $\triangle ABC$ .)
Find coordinates for the vertices $A,B,C$ of a non-right triangle $ABC$ which contains a right triangle $AXY$ such that the area of $\triangle AXY$ is $\frac{1}{3}$ times the area of $\triangle ABC$ .
Let be such that the slopes of sides form the geometric sequence , with .
- Let $X$ be the point which divides $BC$ in the ratio $r:1$ and let $X'$ be the midpoint of $X$ and $BC$ . PROVE that the slope of cevian $AX'$ is $-3$ times the slope of $BC$ .
- Let $Y$ be the point which divides $BC$ in the ratio $1:r$ and let $Y'$ be the midpoint of $BC$ and $Y$ . PROVE that the slope of cevian $AY'$ is $-\frac{1}{3}$ times the slope of $BC$ .
- Conclude that every triangle with slopes $a,ar,ar^2$ (where $r> 0$ ) contains a sub-triangle with slopes $-\frac{1}{3}ar,ar,-3ar$ , a geometric sequence with common ratio $-3$ .
(Right inside) Let $a,ar,ar^2$ ( $r> 0$ ) be the slopes of sides $AB,BC,CA$ in $\triangle ABC$ . PROVE that there are points $X$ and $Y$ on $BC$ for which $XY^2=AX^2+AY^2$ .
Let $a,ar,ar^2$ ( $r> 0$ ) be the slopes of sides $AB,BC,CA$ in $\triangle ABC$ . PROVE that there two points $R$ and $S$ on $BC$ for which the slopes of $\triangle ARS$ form an arithmetic sequence with a common difference of $2ar$ , namely: $-3ar,-ar,ar$ .
Let $a,ar,ar^2$ ( $r> 0$ ) be the slopes of sides $AB,BC,CA$ in $\triangle ABC$ . PROVE that there two points $U$ and $T$ on $BC$ for which the slopes of $\triangle ATU$ form an arithmetic sequence with a common difference of $ar$ , namely: $-ar,0,ar$ .
Let , , be a right triangle in the first quadrant. Choose any vertex, say , and draw two cevians and to side .
- For a positive integer $n$ , let $P$ and $Q$ divide side $BC$ in the ratios $n:1$ and $1:n$ , respectively. PROVE that $\triangle APQ$ and $\triangle ABC$ have the same centroid
- Can this be extended to all triangles?
(Almost right) Let $a,ar,ar^2,~r> 0,$ be the slopes of sides $AB,BC,CA$ in $\triangle ABC$ . PROVE that there are points $X$ and $Y$ on $BC$ such that $AB^2+AC^2=BC.XY.$
(Compare this with exercise 10 above.)

Thanksgiving

Today is a special day in this poster’s life. As such, he gives praise to the Owner of life.

Mind always goes back to that marked Thursday, June 14, 2018. In turn, it C–A–N–`–T hold back from giving thanks.

What happened was very basic (as simple as HTML), but its effect became “drastic” (actually, dramatic, dynamic), and so one’s gratitude is made public.

To the reader:

Hopefully this poster’s gratitude to the Heavenly Father doesn’t ruffle your feathers. If it does, please (don’t bother to) read further.

Wish:

May you find that “spark” you need to start what will lead you to succeed.