# Converging Triangle - 2

This note is about a variation in one of my problems. Have a look at that problem: Converging Triangle

In that problem, every point $\displaystyle A_k\,,\,k \geq 4$ was lying on segment $\displaystyle A_{k-3}A_{k-2}$ such that it divides that segment in a constant ratio $\large\frac{m}{n}$$=$$\large\frac{3}{7}$$\displaystyle \,\forall\; k \geq 4$

In general for any $m\,,\,n \in\mathbb{R^+}$(set of positive real numbers),

$\large\frac{A_1J}{JE}$ $\displaystyle =$ click to see the answer

Another variation in this question which comes to my mind is that instead of dividing the segment $\displaystyle A_{k-3}A_{k-2}$ in a constant ratio, we divide $\displaystyle\angle A_{k-3}A_{k-1}A_{k-2}$ in a constant ratio for all $\displaystyle k \geq 4$. Again $A_k$'s converges to some point $J$ and point $E$ is defined similarly. We have to find the ratio $\large\frac{A_1J}{JE}$.

A new problem what I want to ask is:

Problem - Let there be a triangle $\displaystyle A_1A_2A_3$ as shown in the figure below. Lines $\displaystyle A_iA_{i+1}$ are drawn with $\displaystyle A_{i+1}$ lying on segment $\displaystyle A_{i-2}A_{i-1}$ such that for some real number $\displaystyle r$ in between $\displaystyle 0$ and $\displaystyle 1$

$\displaystyle \frac{\angle A_{i-2}A_iA_{i+1}}{\angle A_{i-2}A_iA_{i-1}} = r\hspace{20pt} i \geq 3\,,\,i \in \mathbb{N}$

In this way $\displaystyle\triangle A_{i-2}A_{i-1}A_i$ becomes smaller as $\displaystyle i$ is getting larger and ultimately converges to a point, say $\displaystyle J$. Join $\displaystyle A_1$ and $\displaystyle J$ and extend it to meet $\displaystyle A_2A_3$ at $\displaystyle E$.

Find the ratio $\displaystyle\frac{A_1J}{JE}$

Unlike the previous problem, I think this problem is incomplete. Since this problem requires division of angles, angles of $\displaystyle \triangle A_1A_2A_3$ must be given in order to find the ratio $\displaystyle\frac{A_1J}{JE}$.

So for completeness of the problem suppose angles of $\displaystyle\triangle A_1A_2A_3$ are given as $\displaystyle A_1 = A\,,\,A_2 = B\,,\,A_3 = C$.

I have used the approach of solution given by Chris Lewis in my previous problem but not able to proceed after some steps.

Solution - First shift this figure into a cartesian plane with $\displaystyle A_1$ located at $\displaystyle (0,0)$, $\displaystyle A_2$ located at $\displaystyle (a,0)$ and $\displaystyle A_3$ located at $\displaystyle (b,c)$.

Let $\displaystyle a = a_{12} =$ length of side $\displaystyle A_1A_2$. So, $\displaystyle a_{12} > 0$.

Coordinates $\displaystyle (b,c)$ can be determined as:

$\displaystyle b = A_1A_3\cos A_1$

$\displaystyle c = A_1A_3\sin A_1$

Now, $\displaystyle \frac{A_1A_3}{\sin A_2} = \frac{A_1A_2}{\sin A_3}$ (sine rule of triangle)

$\displaystyle \Rightarrow A_1A_3 = \frac{A_1A_2\sin A_2}{\sin A_3} = \frac{a_{12}\sin A_2}{\sin A_3}$

So we get,

$\displaystyle b = \frac{a_{12}\sin A_2\cos A_1}{\sin A_3}$

$\displaystyle c = \frac{a_{12}\sin A_2\sin A_1}{\sin A_3}$

There is some nomenclature which I have introduced for easy writing and better understanding the terms.

• Length of segment $\displaystyle A_pA_q$ is denoted by $\displaystyle a_{pq}$.

$\hspace{20pt} \displaystyle\text{Length}(A_pA_q) = a_{pq}$ for all $\displaystyle p\,,\,q \in \mathbb{N}$

• For simplicity in writing we denote

$\hspace{20pt} \displaystyle\angle A_{i-2}A_iA_{i-1}$ with $\displaystyle A_i$ for all $\displaystyle i \geq 3$

• Ratio of length of one part of segment $\displaystyle A_iA_{i+1}$ to the segment itself is denoted by $\displaystyle s_i$.

$\hspace{20pt}\displaystyle s_i = \frac{A_iA_{i+3}}{A_iA_{i+1}} = \frac{a_{i\, i+3}}{a_{i\,i+1}}$ for all $\displaystyle i \in \mathbb{N}$

Let us denote the coordinates of $\displaystyle A$'s as

$\displaystyle A_n = (x_n, y_n)$ for all $\displaystyle n \in \mathbb{N}$

Then we are given

$\displaystyle x_1 = y_1 = 0$

$\displaystyle x_2 = a_{12}\;,\;y = 0$

$\displaystyle x_3 = \frac{a_{12}\sin A_2\cos A_1}{\sin A_3}\;,\; y_3 = \frac{a_{12}\sin A_2\sin A_1}{\sin A_3}$

Using section formula we get a recurrence relation as,

$\displaystyle (x_4,y_4) = (1 - s_1)(x_1,y_1) + s_1(x_2,y_2)$

and in general,

$\displaystyle (x_n,y_n) = (1 - s_{n-3})(x_{n-3},y_{n-3}) + s_{n-3}(x_{n-2},y_{n-2})$

That is

$\displaystyle x_n = (1 - s_{n-3})x_{n-3} + s_{n-3}x_{n-2}$ for all $\displaystyle n \geq 4$ and

$\displaystyle y_n = (1 - s_{n-3})y_{n-3} + s_{n-3}y_{n-2}$ for all $\displaystyle n \geq 4$

We see that both the relations are same and are homogenous linear recursion of degree 3 with variable coefficients.

In the sequence of $\displaystyle x_n$ there is another sequence $\displaystyle s_n$ that is need to be solved.

Taking $\displaystyle n = 1$,

$\displaystyle s_1 = \frac{A_1A_4}{A_4A_2} = \frac{a_{14}}{a_{12}} = \frac{area(\triangle A_1A_3A_4)}{area(\triangle A_1A_3A_2)} = \frac{\frac{1}{2}a_{13}a_{34}\sin (rA_3)}{\frac{1}{2}a_{13}a_{23}\sin A_3} = \frac{\sin A_2\sin(rA_3)}{\sin A_4\sin A_3}$

For $\displaystyle n = i > 1$,

$\displaystyle s_i = \frac{A_iA_{i+3}}{A_iA_{i+1}} = \frac{a_{i\,i+3}}{a_{i\,i+1}} = \frac{area(\triangle A_iA_{i+2}A_{i+3})}{area(\triangle A_iA_{i+2}A_{i+1})} = \frac{\frac{1}{2}a_{i\,i+2}a_{i+2\,i+3}\sin (rA_{i+2})}{\frac{1}{2}a_{i\,i+2}a_{i+1\,i+2}\sin A_{i+2}} = \frac{\sin[(1-r)A_{i+1}]\sin(rA_{i+2})}{\sin A_{i+3}\sin A_{i+2}}$

So we get the sequence $\displaystyle s_n$ defined as

$\hspace{20pt}\displaystyle s_1 = \frac{\sin A_2\sin(rA_3)}{\sin A_3\sin A_4}$

and $\displaystyle s_n = \frac{\sin[(1-r)A_{n+1}]\sin(rA_{n+2})}{\sin A_{n+2}\sin A_{n+3}}$ for all $\displaystyle n \geq 2$

We see that in the sequence $\displaystyle s_n$ there is a sequence of angles $\displaystyle A_n$. So we need to first solve for general value of $\displaystyle A_n$.

$\displaystyle A_1\,,\,A_2\,,\,A_3$ are the three initial angles given. It is easy to see that

$\displaystyle A_4 = A_1 + rA_3$

$\displaystyle A_5 = A_2 + rA_4$

seems like that next term will be

$\displaystyle A_6 = A_3 + rA_5$

but this is wrong, if you see in the diagram you will get that

$\displaystyle A_6 = (1-r)A_3 + rA_5$

and in general

$\displaystyle A_n = (1-r)A_{n-3} + rA_{n-1}$ for all $\displaystyle n \geq 6$

So we get the sequence $\displaystyle A_n$ defined as

$\displaystyle A_1 = A\,,\,A_2 = B\,,\,A_3 = C$ and

$\displaystyle A_n = A_{n-3} + rA_{n-1}$ for $\displaystyle n = 4, 5$

$\displaystyle A_n = (1-r)A_{n-3} + rA_{n-1}$ for all $\displaystyle n \geq 6$

So what we have done from starting is summarised below

Every point $\displaystyle A_n\,,n\geq 1$ has its coordinate as

$\displaystyle A_n = (x_n,y_n)$ for all $\displaystyle n\in\mathbb{N}$

Definition of sequence $\mathbf{x_n}$

$\displaystyle x_1 = 0\;\;,\;\;x_2 = a_{12}\;,\;x_3 = \frac{a_{12}\sin A_2\cos A_1}{\sin A_3}$

$\hspace{215pt}\displaystyle x_n = (1 - s_{n-3})x_{n-3} + s_{n-3}x_{n-2}\;\;$ for all $n \geq 4$

Definition of sequence $\mathbf{y_n}$

$\displaystyle y_1 = 0\;\;,\;\;y_2 = 0\;,\;y_3 = \frac{a_{12}\sin A_2\sin A_1}{\sin A_3}$

$\hspace{215pt}\displaystyle y_n = (1 - s_{n-3})y_{n-3} + s_{n-3}y_{n-2}\;\;$ for all $n \geq 4$

Definition of sequence $\mathbf{s_n}$

$\displaystyle s_1 = \frac{\sin A_2\sin(rA_3)}{\sin A_3\sin A_4}$

$\hspace{215pt}\displaystyle s_n = \frac{\sin[(1-r)A_{n+1}]\sin(rA_{n+2})}{\sin A_{n+2}\sin A_{n+3}}\;\;$ for all $n \geq 2$

Definition of sequence $\mathbf{A_n}$

$\displaystyle A_1 = A\;\;,\;\;A_2 = B\;,\;A_3 = C$

$\hspace{240pt}\displaystyle A_n = A_{n-3} + rA_{n-1}\;\;$ for $\displaystyle n = 4, 5$

$\hspace{225pt}\displaystyle A_n = (1-r)A_{n-3} + rA_{n-1}\;\;$ for all $\displaystyle n \geq 6$

We need to find the coordinates of point $\displaystyle J$ to get the required ratio $\displaystyle\frac{A_1J}{JE}$. Let coordinates of point $\displaystyle J$ be $\displaystyle(x_J, y_J)$. Then we have,

$\displaystyle x_J = \lim_{n \to \infty}x_n$ and $\displaystyle y_J = \lim_{n \to \infty}y_n$

In order to get the general term of $\displaystyle x_n\;,\;y_n$ we need the general term of $\displaystyle s_n$ for which we need the general term of $\displaystyle A_n$.

We see that sequence $\displaystyle A_n$ is the only independent sequence free of any other sequence. We need to find the general term of this sequence. We see that it is homogenous linear recurrence of degree $3$ with constant cofficients.

Let $\displaystyle A_n = \lambda^n\,,\, n \geq 3$ be the solution of the recurrence relation. Then putting this solution in the recurrence relation we get,

$\lambda^n = (1-r)\lambda^{n-3} + r\lambda^{n-1}$

$\Rightarrow \lambda^3 = (1-r) + r\lambda^2$

$\Rightarrow \lambda^3 - r\lambda^2 - (1-r) = 0$

There are the following three roots to this equation:

$\displaystyle\lambda_1 = 1\;,\;\;\;\lambda_2 = -\frac{1 - r}{2} + \frac{\sqrt{(3 + r)(1 - r)}}{2}i\;,\;\;\;\lambda_3 = -\frac{1 - r}{2} - \frac{\sqrt{(3 + r)(1 - r)}}{2}i\;\;\;$ where $i = \sqrt{-1}$

So we get,

$A_n = \alpha 1^n + \beta\lambda_2^n + \gamma\lambda_3^n$ for all $n \geq 3$

To get the values of $\alpha, \beta, \gamma$ we use the values of $A_3\,,\,A_4\,,\,A_5$ in terms of $A\,,\,B\,,\,C$.

We have

$A_3 = C\;,\;\;A_4 = A + rC\;,\;\;A_5 = B + r(A + rC) = B + rA + r^2C$

So,

$C = A_3 = \alpha + \beta\lambda_2^3 + \gamma\lambda_3^3\hspace{20pt}\cdots\text{Eq. 1}$

$A + rC = A_4 = \alpha + \beta\lambda_2^4 + \gamma\lambda_3^4\hspace{20pt}\cdots\text{Eq. 2}$

$B + rA + r^2C = A_5 = \alpha + \beta\lambda_2^5 + \gamma\lambda_3^5\hspace{20pt}\cdots\text{Eq. 3}$

In matrix form we get,

$\displaystyle\begin{bmatrix} 1 & \lambda_2^3 & \lambda_3^3 \\ 1 & \lambda_2^4 & \lambda_3^4 \\ 1 & \lambda_2^5 & \lambda_3^5 \end{bmatrix}\begin{bmatrix} \alpha \\ \beta \\ \gamma\end{bmatrix} = \begin{bmatrix} C \\ A + rC \\ B + rA + r^2C\end{bmatrix}$

Solving for $\alpha, \beta, \gamma$ from these equations in terms of $r$ by writing $\lambda_2$ and $\lambda_3$ in terms of $r$ we get, (I am not writing the intermediate steps, it is very tedious for writing here)

$\displaystyle\alpha = \frac{A + B + C}{3 - 2r} = \frac{180\degree}{3 - 2r}\hspace{20pt}[\because A, B, C$ are the angles of triangle$]$

$\displaystyle\beta = \Bigg[\frac{A}{2(2r-3)} + \frac{(1 + r - r^2)B}{2(1 - r)^2(2r - 3)} + \frac{(r^3 - r^2 + r - 2)C}{2(1 - r)^2(2r - 3)}\Bigg] + \Bigg[\frac{(1 + 2r - r^2)A}{2(1 - r)(2r - 3)\sqrt{(3 + r)(1 - r)}} + \frac{(r^2 + r - 3)B}{2(1 - r)(2r - 3)\sqrt{(3 + r)(1 - r)}} - \frac{3rC}{2(2r - 3)\sqrt{(3 + r)(1 - r)}}\Bigg]i$

$\displaystyle\gamma = \Bigg[\frac{A}{2(2r-3)} + \frac{(1 + r - r^2)B}{2(1 - r)^2(2r - 3)} + \frac{(r^3 - r^2 + r - 2)C}{2(1 - r)^2(2r - 3)}\Bigg] - \Bigg[\frac{(1 + 2r - r^2)A}{2(1 - r)(2r - 3)\sqrt{(3 + r)(1 - r)}} + \frac{(r^2 + r - 3)B}{2(1 - r)(2r - 3)\sqrt{(3 + r)(1 - r)}} - \frac{3rC}{2(2r - 3)\sqrt{(3 + r)(1 - r)}}\Bigg]i$

We see that $\displaystyle \gamma = \overline{\beta}\;$, where $\displaystyle\overline{z}$ denotes conjugate of $\displaystyle z$.

So we get the general term of sequence $\displaystyle A_n$ as:

$\displaystyle A_1 = A\;,\;\;A_2 = B$

$\hspace{50pt}\displaystyle A_n = \frac{180\degree}{3 - 2r} + \beta\Bigg(-\frac{1 - r}{2} + \frac{\sqrt{(3 + r)(1 - r)}}{2}i\Bigg)^n + \overline{\beta}\Bigg(-\frac{1 - r}{2} - \frac{\sqrt{(3 + r)(1 - r)}}{2}i\Bigg)^n$ for all $\displaystyle n \geq 3$

with $\displaystyle \beta$ as given above.

I have solved this question till here and now I am not able to proceed further. If anyone has any idea how to proceed now, please contribute your ideas into this. It is going very complex and large expressions everywhere, so one can use some special triangle $A_1A_2A_3$ and any special value of $r$ to reduce the expressions and solve further. Please help if anyone has any idea how to proceed further or any other approach to solve this problem.

Note by Shikhar Srivastava
1 year ago

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- 1 year ago

- 1 year ago

Wow, interesting question! I've mostly looked at this numerically so far, but aside from the angle relation you found, I haven't actually found any examples where a "nice" starting triangle (integer side lengths, integer coordinate vertices, equilateral, right-angled) leads to a "nice" (recognisable rational or algebraic) endpoint.

It's certainly a nice property that the angles converge as per your argument; in fact we have \begin{aligned} \lim_{n \to \infty} \angle A_{n-2} A_{n} A_{n-1} &=\frac{\pi}{3-2r} \\ \lim_{n \to \infty} \angle A_{n} A_{n-1} A_{n-2} &=\frac{(1-r)\pi}{3-2r} \\ \lim_{n \to \infty} \angle A_{n} A_{n-2} A_{n-1} &=\frac{(1-r)\pi}{3-2r} \end{aligned}

which come from your expression for $A_n$ (note that $|\lambda_2|=|\lambda_3|<1$, so only the first term matters in the limit). I was surprised the triangle tends towards an isosceles, non-equilateral one (perhaps I shouldn't have been, though).

I wonder if a different coordinate system might be helpful - something more natural to a triangle, like trilinear or barycentric? (I'm not very used to these but it might be worth investigating.) It made sense to use Cartesian coordinates in the first problem as the operations involved could easily be resolved independently in the $x$ and $y$ directions, but I'm not sure that's the case here.

- 1 year ago

This is a nice result you have deduced. I haven't noticed much on behaviour of angles might be because I was focused on finding general term of $x_n$. These limiting angles sums to $180\degree$ without using the angle sum property of triangles in finding those limiting angles, so I think it is a good indication that I haven't made any mathematical mistake in any expression.

After solving this question with cartesian coordinates, I have tried this a bit with trilinear but in that case also, sequence of coordinate of $A_n$ was coming to be recurrence with variable coefficients as in this case. I don't know barycentric. I will search for it to see whether it is useful to this question or not.

- 1 year ago

I started with the specific case of $r = \frac{1}{2}$, which means each angle is bisected, and chose a right angle isosceles triangle as a convenient triangle to use with $A_1(0, 0)$, $A_2(2, 0)$, and $A_3(1, 1)$, because then the new points are always midpoints of previous segments.

A recursion relationship can be set up with the midpoints so that $(x_n, y_n) = (\frac{1}{2}(x_{n - 3} + x_{n - 2}), \frac{1}{2}(x_{n - 3} + x_{n - 2}))$ for $n \geq 3$ and $(x_1, y_1) = (0, 0)$, $(x_2, y_2) = (2, 0)$, $(x_3, y_3) = (1, 1)$. After some computation, $J$ is $J(\frac{6}{5}, \frac{2}{5})$, $E$ is $E(\frac{3}{2}, \frac{1}{2})$, $A_1J = \frac{2\sqrt{10}}{5}$, $JE = \frac{\sqrt{10}}{10}$, and $\frac{A_1J}{JE} = 4$ .

I still need to work on a general case for $r$, though.

- 1 year ago

Brilliant, those are definitely "nice" parameters!

I'd been looking at $r=\frac12$ as well (the bisection certainly helps), but your example just made me realise something else: if the starting triangle is the right "shape" (that is, isosceles with the angles in my earlier post), all the triangles formed are similar.

Another nice(ish) example I've found this way is with $r=\frac34$ and initial points (in order) $(0,0),(2,0),(1,\tan 30^{\circ})$, which heads to a final point at $(\frac97,\frac{\sqrt3}{7})$.

- 1 year ago

do you mean $(1, \tan 30°)$?

- 1 year ago

I did! Thanks, edited now.

- 1 year ago

With a little bit of calculation, I found that the coordinates $A_1(0, 0)$, $A_2(2, 0)$, and $A_3(1, \tan (\frac{1-r}{3-2r} \cdot 180°))$ will lead to triangles that are similar and will cut each side at a ratio of $r' = \frac{\sin (\frac{r}{3-2r} \cdot 180°)}{2 \cdot \cos (\frac{1-r}{3-2r} \cdot 180°) \cdot \sin (\frac{1}{3-2r} \cdot 180°)}$, then making it a similar problem to the original "converging triangle problem".

- 1 year ago

It is a nice result.

- 1 year ago

I was trying to find some result with $r = \frac{1}{2}$ (without any special triangle) but haven't got much. It was nice idea to take right isosceles triangle with $r = \frac{1}{2}$. This has reduced the question to simply bisecting previous segment. But this actually converts this question to my previous question.

I always want to solve this question using basic elementary geometry as the solution of previous question. But I think that will not work here.

- 1 year ago

Yes, you're right, it does just convert this question to your previous question.

- 1 year ago

I noticed that the ratio $\frac{A_1J}{JE}$ does change depending on where its vertices are, in both this question and its original question. For example, taking my right isosceles triangle with $A_1(0, 0)$, $A_2(2, 0)$, and $A_3(1, 1)$ above with $r = \frac{1}{2}$ makes $\frac{A_1J}{JE} = 4$, however, using $A_1(0, 0)$, $A_2(1, 0)$, and $A_3(0, 1)$ instead (like in the original question's solution) with $r = \frac{1}{2}$ makes $\frac{A_1J}{JE} = 3$. I believe that means that the original question must also include the extra parameter that the height of $\triangle A_1A_2A_3$ must equal the base of $\triangle A_1A_2A_3$ to make the given solution correct.

Edit: Nevermind, I had mixed up some variables, so what I said here is not true. Shikhar Srivastava is correct, $\frac{A_1J}{JE} = 4$ for $r = \frac{1}{2}$ irregardless of the size and shape of the triangle.

- 1 year ago

If you are using $r = \frac{1}{2}$ in this question then the ratio will change with respect to vertices of triangle. But if you are $r = \frac{1}{2}$ in the previous question then that ratio does not depend on the vertices of triangle, not even on size and shape of triangle. It just depend on $r$ as

$\displaystyle\frac{A_1J}{JE} = \frac{2}{1 - r}$ for any starting triangle.

You can see my solution to that question, the solution is purely based on elementary geometry and no where used the size and shape of triangle. If you put $\frac{r}{1 - r}$ instead of $\frac{3}{7}$ in that question and solve further whether by algebraic sequential method or by geometry, you will get the ratio as

$\displaystyle\frac{A_1J}{JE} = \frac{2}{1 - r}$.

I have just solved that question by algebraic method using cartesian coordinates and got the following results:

When $A_1 = (0,0)\;,\,A_2 = (a,0)\;,\,A_3 = (b,c)$, with $a,c > 0$ and $b\in\mathbb{R}$, it represents all triangle possible (in order). For ex - triangle with $A_1 = 30\degree\;,\,A_2 = 60\degree\;,\,A_3 = 90\degree$ and another triangle with $A_1 = 60\degree\;,\,A_2 = 90\degree\;,\,A_3 = 30\degree$ with same size are considered different.

We have

$x_1 = y_1 = 0\;,\,x_2 = a\;,\,y_2 = 0\;,\,x_3 = b\;,\,y_3 = c$ and

$x_n = (1 - r)x_{n-3} + rx_{n-2}$ for all $n \geq 4$

And we get the solution of general term of $x_n$ as:

When $\displaystyle r = \frac{3}{4}$,

$\displaystyle x_n = \frac{4}{9}\big(a + b\big) +\bigg(\frac{-4a + 12an + 32b - 24bn}{9}\bigg)\bigg(\frac{-1}{2}\bigg)^n$

$\displaystyle y_n = \frac{4}{9}c +\bigg(\frac{32c - 24cn}{9}\bigg)\bigg(\frac{-1}{2}\bigg)^n$

So, $\displaystyle J = \bigg(\frac{4(a + b)}{9},\frac{4c}{9}\bigg)$

Using $\displaystyle A_1$ and $\displaystyle J$ we get,

$\displaystyle E = \bigg(\frac{a + b}{2}, \frac{c}{2}\bigg)$

So, $\displaystyle A_1J = \frac{4}{9}\sqrt{(a + b)^2 + c^2}$

and $\displaystyle JE = \bigg(\frac{1}{2} - \frac{4}{9}\bigg)\sqrt{(a + b)^2 + c^2} = \frac{1}{18}\sqrt{(a + b)^2 + c^2}$

Hence $\displaystyle \frac{A_1J}{JE} = 8 = \frac{2}{1 - \frac{3}{4}} = \frac{2}{1 - r}$

When $\displaystyle r \neq \frac{3}{4}$,

$\displaystyle x_n = \frac{a + b}{3 - r} +\beta_1\bigg(-\frac{1}{2} + \frac{1}{2}\sqrt{4r - 3}\bigg)^n + \gamma_1\bigg(-\frac{1}{2} - \frac{1}{2}\sqrt{4r - 3}\bigg)^n$

$\displaystyle y_n = \frac{c}{3 - r} +\beta_2\bigg(-\frac{1}{2} + \frac{1}{2}\sqrt{4r - 3}\bigg)^n + \gamma_2\bigg(-\frac{1}{2} - \frac{1}{2}\sqrt{4r - 3}\bigg)^n$

It is not needed to calculate $\beta$ and $\gamma$, because $\displaystyle \bigg|-\frac{1}{2} \pm \frac{1}{2}\sqrt{4r - 3}\bigg| < 1$

So, $\displaystyle J = \bigg(\frac{a + b}{3 - r},\frac{c}{3 - r}\bigg)$

Using $\displaystyle A_1$ and $\displaystyle J$ we get,

$\displaystyle E = \bigg(\frac{a + b}{2}, \frac{c}{2}\bigg)$

So, $\displaystyle A_1J = \frac{1}{3 - r}\sqrt{(a + b)^2 + c^2}$

and $\displaystyle JE = \bigg(\frac{1}{2} - \frac{1}{3 -r}\bigg)\sqrt{(a + b)^2 + c^2} = \frac{1 - r}{2(3 - r)}\sqrt{(a + b)^2 + c^2}$

Hence $\displaystyle \frac{A_1J}{JE} = \frac{\frac{1}{3 - r}}{\frac{1 - r}{2(3 - r)}} = \frac{2}{1 - r}$

Hence we see that for any triangle $A_1A_2A_3$ we get the same result of the ratio $\frac{A_1J}{JE}$.

- 1 year ago

I'm confused. If $\frac{A_1J}{JE} = \frac{2}{1 - r}$, then for the original question where $r = \frac{3}{7}$, $\frac{A_1J}{JE} = \frac{2}{1 - \frac{3}{7}} = \frac{7}{2}$, but the answer is $\frac{A_1J}{JE} = \frac{20}{7}$.

- 1 year ago

Nevermind, I'm mixing up variables. The original question actually has $r = \frac{3}{3 + 7} = \frac{3}{10}$, which does come out to $\frac{A_1J}{JE} = \frac{2}{1 - r} = \frac{20}{7}$.

- 1 year ago

Combining some results the equations $r' = \frac{\sin (\frac{r}{3-2r} \cdot 180°)}{2 \cdot \cos (\frac{1-r}{3-2r} \cdot 180°) \cdot \sin (\frac{1}{3-2r} \cdot 180°)}$ and $\frac{A_1J}{JE} = \frac{2}{1 - r'}$ found in this thread and simplifying, if $r = \frac{\angle A_{i-2}A_iA_{i+1}}{\angle A_{i-2}A_iA_{i-1}}$, then:

$\frac{A_1J}{JE} = 2\bigg(\frac{\sin (\frac{r}{3 - 2r} \cdot 180°)}{\sin (\frac{2 - r}{3 - 2r} \cdot 180°)} + 1\bigg)$

.

- 1 year ago

One point need to be added in this is that it is true for starting triangle with angles $\displaystyle \frac{180\degree}{3 - 2r}\;,\,\frac{180\degree(1 - r)}{3 - 2r}\;,\, \frac{180\degree(1 - r)}{3 - 2r}$.

- 1 year ago

Thank you both of you @Chris Lewis @David Vreken for contributing your ideas and deriving some wonderful results out of this question.

- 1 year ago

Thanks for sharing the question. I like this as a forum to discuss these sort of meta-questions.

By the way, have you heard of the Encyclopedia of Triangle Centres? It has a search facility which might be of interest here. I haven't had a chance to look in detail yet but the idea is that if you calculate the trilinear coordinates of a point you're interested in, you can check if it's a "known" triangle centre. There are about $32000$ centres on this site so there's a decent chance of finding a match! It could well help with the $r=\frac12$ case (and perhaps others).

- 1 year ago