General – Page 31 – High School Geometry

Approximate Pythagorean Identity

Let $l_1,l_2,l_3$ be lengths of the sides of a triangle, and let the slopes of these sides be $a,ar,ar^2$ , respectively. Then, as we show in example 10:

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

This is analogous to the Pythagorean identity for right triangles. In theory, if we let $r\rightarrow\infty$ (or $r\rightarrow 0$ ), then $l_1^2+l_3^2\approx l_2^2$ . Thus, in a sense, a right triangle is a limiting case of a triangle with slopes in geometric progression.

Whatever a right triangle can do: $\ddots\ddots\ddots$

Whatever a right triangle can “do”, there is a triangle with slopes $a,ar,ar^2$ that can “do” the same thing. Did we just exaggerate?

Compare the hypotenuse of a right triangle with side $l_2$ of a triangle whose slopes are $a,ar,ar^2~~(a\neq 0, r\neq 0,\pm 1)$ .

By side $l_2$ , we mean the side whose slope is $ar$ , the geometric mean of the progression $a,ar,ar^2$ .

In terms of length: If $r> 0$ , then (in view of (1)) the side with length $l_2$ is the longest side of the triangle, just like the hypotenuse of a right triangle.
(Winner: Right triangle — because it always satisfies this property; our own triangle doesn’t always satisfy it if $r< 0$ .)
The median to the hypotenuse: In a right triangle, the length of the median to the hypotenuse is always half of the length of the hypotenuse. In a triangle with slopes $a,ar,ar^2$ , the length of the median to side $l_2$ has length $\frac{1}{2}\left(\frac{r-1}{r+1}\right)l_2$ (See the exercises). And so as $r\rightarrow 0,\infty$ , this becomes $\frac{1}{2}l_2$ .
(Winner: Right triangle again — because it always satisfies this property, while our own triangle satisfies it only when $r\rightarrow 0,\infty$ .)
Slope of the median to the hypotenuse: If the legs of a right triangle are parallel to the coordinate axes, then the slope of the median to the hypotenuse is the negative of the slope of the hypotenuse. On the other hand, in a triangle with slopes $a,ar,ar^2$ , the slope of the median to side $l_2$ is always the negative of the slope of side $l_2$ .
(Winner: The triangle with slopes $a,ar,ar^2$ wins here — because it always satisfies this property, whereas a right triangle satisfies it only when its legs are parallel to the coordinate axes.)

Final Score: Right triangle won $2$ to $1$ — courtesy of our generosity.

Consider $\triangle ABC$ with vertices $A(1,9)$ , $B(0,0)$ , and $C(-2,6)$ . Verify that the side lengths satisfy equation (1).

Observe that the slopes are $1,-3,9$ for sides $AC,CB,BA$ , respectively. They form a geometric progression in which $r=-3$ .

In (1), the side with length $l_2$ is the side whose slope is the geometric mean of the progression — in the present case, it is side $CB$ . This is important. $l_1$ and $l_3$ can then be chosen as the lengths of the other two sides.

Using the given vertices $A(1,9),~B(0,0),~C(-2,6)$ , let’s compute the side lengths:

$\begin{equation*} \begin{split} BA^2&=(0-1)^2+(0-9)^2\\ &=82\\ AC^2&=(-2-1)^2+(6-9)^2\\ &=18\\ CB^2&=(0+2)^2+(0-6)^2\\ &=40 \end{split} \end{equation*}$

Since $CB^2=40$ , we have to take $l_2^2=40$ . Let $l_1^2=AC^2=18$ and let $l_3^2=BA^2=82$ . Then:

$\begin{equation*} \begin{split} l_1^2+l_3^2&=18+82\\ &=100\\ \frac{1+r^2}{(1+r)^2}l_2^2&=\frac{1+(-3)^2}{(1+-3)^2}\times 40\\ &=\frac{10}{4}\times 40\\ &=100\\ &\vdots\ddots\ddots\vdots\\ \therefore l_1^2+l_3^2&=\frac{1+r^2}{(1+r)^2}l_2^2 \end{split} \end{equation*}$

Consider $\triangle ABC$ with vertices $A(-1,2)$ , $B(-0.5,-2)$ , and $C(1.5,-3)$ . Verify that the side lengths satisfy equation (1).

The slopes of sides $BC,CA,AB$ are $-\frac{1}{2},-2,-8$ , respectively. They form a geometric progression with common ratio $r=4$ . Further, the slope of side $CA$ , namely $-2$ , is the geometric mean of the progression. Thus the length of side $CA$ will be taken as $l_2$ in (1).

$\begin{equation*} \begin{split} CA^2&=(-1-1.5)^2+(2+3)^2\\ &=31.25~\textrm{or}~\frac{125}{4}\\ \therefore l_2^2&=\frac{125}{4}\\ AB^2&=(-1+0.5)^2+(2+2)^2\\ &=16.25~\textrm{or}~\frac{65}{4}\\ BC^2&=(-0.5-1.5)^2+(-2+3)^2\\ &=5 \end{split} \end{equation*}$

Since $l_2^2$ has been chosen as $\frac{125}{4}$ , we can take $l_1^2=\frac{65}{4}$ and $l_3^2=5$ . Then, in (1):

$\begin{equation*} \begin{split} l_1^2+l_3^2&=\frac{65}{4}+5\\ &=\frac{85}{4}\\ \frac{1+r^2}{(1+r)^3}l_2^2&=\frac{1+4^2}{(1+4)^2}\times \frac{125}{4}\\ &=\frac{17}{25}\times \frac{125}{4}\\ &=\frac{85}{4}\\ &\ddag\ddots\ddag\\ \therefore l_1^2+l_3^2&=\frac{1+r^2}{(1+r)^3}l_2^2 \end{split} \end{equation*}$

So the formula works for both negative and positive common ratios.

The slopes of the sides of $\triangle ABC$ are all non-zero and the side lengths satisfy $AB^2+CA^2=25BC^2$ . Find possible coordinates for the vertices $A,B,C$ .

The only challenge here is the non-zero slope specification; otherwise, we could just place two vertices along the $x$ -axis, making the third vertex easy to find.

Fortunately, we can always appeal to triangles whose slopes form geometric progressions under this situation. They’re cut out for these.

In (1), set $\frac{1+r^2}{(1+r)^2}=25$ and solve for $r$ :

$\begin{equation*} \begin{split} 25(1+2r+r^2)&=1+r^2\\ 24r^2+50r+24&=0\\ 12r^2+25r+12&=0\\ (4r+3)(3r+4)&=0\\ r&=-\frac{3}{4},-\frac{4}{3} \end{split} \end{equation*}$

The workload now reduces to finding coordinates for a triangle whose slopes form a geometric progression in which $r=-\frac{3}{4}$ .

Easy-peasy. We can use $B(0,0)$ , $A(1,r^2)$ , $C(1+r,r+r^2)$ and just put $r=-\frac{3}{4}$ .

We obtain $B(0,0)$ , $A\left(1,\frac{9}{16}\right)$ , and $C\left(\frac{1}{4},-\frac{3}{16}\right)$ . The side slopes are all non-zero: $m_{AB}=\frac{9}{16}$ , $m_{BC}=-\frac{3}{4}$ , and $m_{CA}=1$ .

For the side lengths, we have:

$\begin{equation*} \begin{split} AB^2&=\left(1-0\right)^2+\left(\frac{9}{16}-0\right)^2\\ &=\frac{337}{256}\\ CA^2&=\left(\frac{1}{4}-1\right)^2+\left(-\frac{3}{16}-\frac{9}{16}\right)^2\\ &=\frac{288}{256}\\ BC^2&=\left(\frac{1}{4}-0\right)^2+\left(-\frac{3}{16}-0\right)^2\\ &=\frac{25}{256}\\ &\ddots\ddots\ddots\\ AB^2+CA^2&=\frac{337}{256}+\frac{288}{256}\\ &=\frac{625}{256}\\ &=25\times\left(\frac{25}{256}\right)\\ &=25BC^2 \end{split} \end{equation*}$

DONE!!! $AB^2+CA^2=25BC^2$ , with $A\left(1,\frac{9}{16}\right)$ , $B(0,0)$ , and $C\left(\frac{1}{4},-\frac{3}{16}\right)$ .

Quantity digression

Let’s examine the quantity $\frac{1+r^2}{(1+r)^2}$ that appeared in formula (1).

For real $r\neq -1$ , PROVE that the minimum value of $\frac{1+r^2}{(1+r)^2}$ is $\frac{1}{2}$ .

Let’s do this for real. And, without using calculus. Set $\frac{1+r^2}{(1+r)^2}=k$ and clear fractions to obtain $k(1+r)^2=1+r^2$ .

(2) $\begin{equation*} \begin{split} k(r^2+2r+1)&=1+r^2\\ kr^2+2kr+k&=1+r^2\\ (k-1)r^2+(2k)r+(k-1)&=0 \end{split} \end{equation*}$

Now comes the main point. Since $k$ is a real number, the discriminant of the above quadratic must be greater than or equal to zero. So:

$\begin{equation*} \begin{split} (2k)^2-4\times(k-1)\times(k-1)&\geq 0\\ 4k^2-4(k^2-2k+1)&\geq 0\\ 8k-4&\geq 0\\ \implies k&\geq \frac{1}{2} \end{split} \end{equation*}$

Thus, $\frac{1+r^2}{(1+r)^2}=k\geq \frac{1}{2}$ .

Let $r$ be an integer. PROVE that $L(r)=\frac{1+r^2}{(1+r)^2}$ is also an integer if and only if $r=-2,0$ .

If you’ve been following our posts, you may have noticed our public, profuse admiration for $r=-2$ . It is a peach.

Put $r=-2$ to obtain $L(-2)=\frac{1+(-2)^2}{(1+-2)^2}=5$ ; put $r=0$ to get $L(0)=\frac{1+0^2}{(1+0)^2}=1$ .

Conversely, suppose that $L(r)=\frac{1+r^2}{(1+r)^2}$ is an integer, with $r$ an integer too. We’ll show that $r$ must be $-2$ or $0$ . Set $\frac{1+r^2}{(1+r)^2}=k$ , then we obtain what we had in (2):

$\begin{equation*} \begin{split} (k-1)r^2+(2k)r+(k-1)&=0\\ r&=\frac{-2k\pm\sqrt{(2k)^2-4\times (k-1)\times(k-1)}}{2(k-1)}\\ &=\frac{-2k\pm\sqrt{8k-4}}{2(k-1)}\\ r&=\frac{-k\pm\sqrt{2k-1}}{k-1} \end{split} \end{equation*}$

To obtain an integer solution, the quantity under the square root, namely $2k-1$ , must first of all be a perfect square. Since $2k-1$ is an odd number, it must be congruent to $1$ modulo $8$ if it’s a perfect square (a bit of number theory here $\cdots$ no worries $\ddots\ddots\ddots\vdots\vdots\vdots\ddots\ddots\vdots$ ). Basically, the difference $(2k-1)-1$ must be divisible by $8$ .

Continuing, let’s set $(2k-1)-1=8t$ , for some integer $t$ . As this leads to $2k-2=8t$ , we obtain $k=4t+1$ . Using this in the expression we obtained for $r$ :

$\begin{equation*} \begin{split} r&=\frac{-k\pm\sqrt{2k-1}}{k-1}\\ &=\frac{-(4t+1)\pm\sqrt{2(4t+1)-1}}{(4t+1)-1}\\ &=\frac{-4t-1\pm\sqrt{8t+1}}{4t} \end{split} \end{equation*}$

Now $8t+1$ is a perfect square if and only if $t=\frac{n(n+1)}{2}$ for some positive integer $n$ (see the exercises at the end). Thus we have

$\begin{equation*} \begin{split} r&=\frac{-4\times\left(\frac{n(n+1)}{2}\right)-1\pm\sqrt{8\times \left(\frac{n(n+1)}{2}\right)+1}}{4\times\left(\frac{n(n+1)}{2}\right)}\\ &=\frac{-2n(n+1)-1\pm(2n+1)}{2n(n+1)}\\ r&=-\frac{n}{n+1},~-\frac{n+1}{n} \end{split} \end{equation*}$

Consider $r=-\frac{n}{n+1}$ . It is an integer only when $n=-2,0$ — giving $r=-2,0$ , respectively.

Consider $r=-\frac{n+1}{n}$ . It is an integer only when $n=\pm 1$ . Put $n=1$ to obtain $r=-2$ . (Using $n=-1$ gives $r=0$ again. However, because of the discriminant of the quadratic equation we solved, we don’t want $n=-1$ . Similarly, we don’t want $n=-2$ above.)

Another way of proceeding is to use polynomial division to write $\frac{r^2+1}{(r+1)^2}$

as $1-\frac{2r}{(r+1)^2}$

. Since $r$

is an integer, the consecutive integers $r$

and $r+1$

have no common factor. Thus the denominator of $\frac{2r}{(r+1)^2}$

must be $1$

. In turn, $(r+1)=\pm 1$

, giving $r=-2,0$

, as before.

Our decision to follow a longer procedure is that it also yields the form of rational numbers — not just integers — that make the expression $L(r)=\frac{r^2+1}{(r+1)^2}$ an integer.

If a right triangle satisfies (1), PROVE that the side whose length is $l_2$ cannot be the hypotenuse.

We can use slopes to do this, but let’s use lengths instead.

Suppose the length of the hypotenuse is $l_2$ . Then we have $l_1^2+l_3^2=l_2^2$ by the Pythagorean theorem. Together with (1) we have the following quadratic-quadratic system:

$\begin{equation*} \begin{split} l_1^2+l_3^2&=l_2^2\\ l_1^2+l_3^2&=\frac{1+r^2}{(1+r)^2}l_2^2\\ &\ddots\vdots\ddots\\ \therefore 1&=\frac{1+r^2}{(1+r)^2}\\ (1+r)^2&=1+r^2\\ 1+2r+r^2&=1+r^2\\ \implies r&=0 \end{split} \end{equation*}$

Since $r=0$ is not allowed for a geometric progression, we conclude that $l_2$ cannot be the hypotenuse.

The peach of the bunch

Any triangle whose slopes form a geometric sequence with common ratio $r=-2$ is a peach. If it’s also a right triangle, then it becomes the peach of the pick of the bunch.

If a right triangle has $r=-2$ , PROVE that the $x$ -coordinates, the $y$ -coordinates, and the squares of the lengths form arithmetic progressions.

Our first post of this year showed that if $r=-2$ , then the $x$ -coordinates and the $y$ -coordinates form (different) arithmetic progressions — even when the triangle is not right.

Put $r=-2$ in (1) to obtain

$l_1^2+l_3^2=5l_2^2$

By the way, this is a well-known condition for perpendicular medians.

Since we’re dealing with a right triangle this time (and $l_2$ cannot be the hypotenuse, in view of the previous example), we can take $l_1$ as the hypotenuse, so that

$l_1^2=l_2^2+l_3^2$

Eliminate $l_1$ from the two equations: $(l_2^2+l_3^2)+l_3^2=5l_2^2\implies l_3^2=2l_2^2$ .

Then $l_1^2=l_2^2+l_3^2=l_2^2+2l_2^2=3l_2^2$ .

The squares of the lengths, in terms of $l_2$ , are: $l_2^2,2l_2^2,3l_2^2$ . They form an arithmetic sequence.

If a right triangle has $r=-2$ , PROVE that $a=\pm\frac{\sqrt{2}}{2}$ , or $a=\pm\frac{\sqrt{2}}{4}$ .

Assume the progression of the slopes is $a,ar,ar^2$ . If $r=-2$ , then this becomes $a,-2a,4a$ .

We must have $a\times (-2a)=-1\implies a=\pm\frac{1}{\sqrt{2}}=\pm\frac{\sqrt{2}}{2}$ .

Or we must have $4a\times (-2a)=-1\implies a=\pm\frac{\sqrt{2}}{4}$ .

Derivation of the formula

We now derive equation (1).

In $\triangle ABC$ , let $a,ar,ar^2$ be the slopes of sides $AB,BC,CA$ , respectively. If the lengths of these sides are also denoted by $l_1,l_2,l_3$ , PROVE that $l_1^2+l_3^2=\frac{r^2+1}{(r+1)^2}l_2^2$ .

Let the vertices be $A(x_1,y_1),~B(x_2,y_2),C(x_3,y_3)$ . Then the following relations exist among the coordinates:

(3) $\begin{equation*} \begin{split} x_1&=x_3+\frac{y_2-y_3}{ar(r+1)}\\ y_1&=\frac{ry_2+y_3}{r+1}\\ x_2&=x_3+\frac{y_2-y_3}{ar} \end{split} \end{equation*}$

Using the distance formula:

$\begin{equation*} \begin{split} l_1^2&=AB^2\\ &=\left[x_3+\frac{y_2-y_3}{ar(r+1)}-\left(x_3+\frac{y_2-y_3}{ar}\right)\right]^2+\left[\frac{ry_2+y_3}{r+1}-y_2\right]^2\\ &=\left(-\frac{y_2-y_3}{a(r+1)}\right)^2+\left(-\frac{y_2-y_3}{r+1}\right)^2\\ &=\left(\frac{a^2+1}{a^2(r+1)^2}\right)(y_2-y_3)^2\\ l_2^2&=BC^2\\ &=\left(\frac{a^2r^2+1}{a^2r^2}\right)(y_2-y_3)^2\\ l_3^2&=CA^2\\ &=\left(\frac{a^2r^4+1}{a^2r^2(r+1)^2}\right)(y_2-y_3)^2\\ &\ddots\ddots\ddots\\ &\ddots\ddots\ddots\\ l_1^2+l_3^2&=\left(\frac{a^2+1}{a^2(r+1)^2}\right)(y_2-y_3)^2+\left(\frac{a^2r^4+1}{a^2r^2(r+1)^2}\right)(y_2-y_3)^2\\ &=\frac{(r^2+1)(a^2r^2+1)}{a^2r^2(r+1)^2}(y_2-y_3)^2\\ \frac{r^2+1}{(r+1)^2}l_2^2&=\frac{r^2+1}{(r+1)^2}\times\left(\frac{a^2r^2+1}{a^2r^2}\right)(y_2-y_3)^2\\ &=\frac{(r^2+1)(a^2r^2+1)}{a^2r^2(r+1)^2}(y_2-y_3)^2\\ &=l_1^2+l_3^2\\ &\ddots\ddots\ddots\ddots\ddots\ddots\ddots\\ &\ddots\ddots\ddots\ddots\ddots\ddots\ddots\\ \therefore l_1^2+l_3^2&=\frac{r^2+1}{(r+1)^2}l_2^2. \end{split} \end{equation*}$

Takeaway

In the equation

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

the side whose length is $l_2$ plays a crucial role in the theory of triangles with slopes $a,ar,ar^2$ . In particular, it acts like the hypotenuse of a right triangle.

(For something totally irrelevant, we would like to point out that the title of this post — approximate pythagorean identity — reminds us of APIs, a truly endearing term in programming that abbreviates Application Programming Interface. We lve APIs!!!!!!!!!!)

Tasks

(Number theory) A triangle has non-zero side slopes, and its side lengths $l_1,l_2,l_3$ are related via $l_1^2+l_3^2=841 l_2^2$ . Find possible coordinates for the vertices of the triangle.
(Observe that $841=29^2=20^2+21^2$ . And, in our example, we had $25=5^2=3^2+4^2$ . The quantity $\frac{1+r^2}{(1+r)^2}$ that appeared in (1) is always a sum of two consecutive squares, if it is an integer. But it is not always a perfect square. Thus, this closely relates to a problem in number theory where integer solutions to equations of the form $n^2+(n+1)^2=u^2$ are sought.)
For $r\geq 0$ , PROVE that $\frac{1}{2}\leq \frac{1+r^2}{(1+r)^2}\leq 1$ .
Let $-1\neq r\in\mathbb{Q}$ . PROVE that $\frac{1+r^2}{(1+r)^2}$ is an integer if and only if $r$ is of the form $-\frac{n}{n+1}$ or $-\frac{n+1}{n}$ , for some (positive) integer $n$ .
Let $-1\neq r\in\mathbb{Q}$ . PROVE that the integer values of $\frac{1+r^2}{(1+r)^2}$ are always of the form $n^2+(n+1)^2$ .
PROVE that the roots of the quadratic equation $(k-1)r^2+(2k)r+k-1=0$ are reciprocals of each other.
(Alternatively, this means that if $r\neq 0$ , then both $r$ and $\frac{1}{r}$ yield the same values for the quantity $\frac{1+r^2}{(1+r)^2}$ .)
Let $t\geq 0$ be an integer. PROVE that $8t+1$ is a perfect square if and only if $t=\frac{n(n+1)}{2}$ for some (positive) integer $n$ .
(Quadratic proximity) Let be the slopes of the sides of a triangle.
- If the triangle is right isosceles, PROVE that $r^2+3r+1=0$ ;
- If the triangle is equilateral, PROVE that $r^2+4r+1=0$ .
In $\triangle ABC$ , let sides $AB,BC,CA$ have slopes $a,ar,ar^2$ , respectively. PROVE that $m_A=\frac{1}{2}\left(\frac{r-1}{r+1}\right)BC$ , where $m_A$ is the length of the median from vertex $A$ and $BC$ is the length of side $BC$ .
(In particular, when $r=0$ , we obtain $m_A=\frac{1}{2}BC$ , ignoring the negative sign. Thus, the length of the median to side $BC$ is half the length of $BC$ , similar to the median to the hypotenuse in a right triangle. But $r=0$ is not allowed here.)
Let .
- If $r\neq 0,1$ , PROVE that $M(r)=M\left(\frac{1}{r}\right)$ .
- If $r\in\mathbb{Z}$ , PROVE that $M(r)\in\mathbb{Z}$ if and only if $r\in\{-2,0,2,4\}$ .
- If $r\in\mathbb{Q}$ , PROVE that $M(r)\in\mathbb{Z}$ if and only if $r$ has the form $\frac{n+2}{n-1}$ or $\frac{n-1}{n+2}$ for some integer $n$ .
Find exact solutions to the equation $\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}=\frac{r^2+1}{(r+1)^2}$ .
(Later on we’ll show how $\frac{(r+2)^2+(2r+1)^2}{(r-1)^2}$ appears in median lengths of triangles with slopes in geometric progression.)

Geometric sequence on a staircase

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.