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 Ak , k≥4\displaystyle A_k\,,\,k \geq 4 was lying on segment Ak−3Ak−2\displaystyle A_{k-3}A_{k-2} such that it divides that segment in a constant ratio mn\large\frac{m}{n}= = 37\large\frac{3}{7} ∀  k≥4\displaystyle \,\forall\; k \geq 4

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

A1JJE\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 Ak−3Ak−2\displaystyle A_{k-3}A_{k-2} in a constant ratio, we divide ∠Ak−3Ak−1Ak−2\displaystyle\angle A_{k-3}A_{k-1}A_{k-2} in a constant ratio for all k≥4\displaystyle k \geq 4. Again AkA_k's converges to some point JJ and point EE is defined similarly. We have to find the ratio A1JJE\large\frac{A_1J}{JE}.

A new problem what I want to ask is:

Problem - Let there be a triangle A1A2A3\displaystyle A_1A_2A_3 as shown in the figure below. Lines AiAi+1\displaystyle A_iA_{i+1} are drawn with Ai+1\displaystyle A_{i+1} lying on segment Ai−2Ai−1\displaystyle A_{i-2}A_{i-1} such that for some real number r\displaystyle r in between 0\displaystyle 0 and 1\displaystyle 1

∠Ai−2AiAi+1∠Ai−2AiAi−1=ri≥3 , i∈N\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 △Ai−2Ai−1Ai\displaystyle\triangle A_{i-2}A_{i-1}A_i becomes smaller as i\displaystyle i is getting larger and ultimately converges to a point, say J\displaystyle J. Join A1\displaystyle A_1 and J\displaystyle J and extend it to meet A2A3\displaystyle A_2A_3 at E\displaystyle E.

Find the ratio A1JJE\displaystyle\frac{A_1J}{JE}

Unlike the previous problem, I think this problem is incomplete. Since this problem requires division of angles, angles of △A1A2A3\displaystyle \triangle A_1A_2A_3 must be given in order to find the ratio A1JJE\displaystyle\frac{A_1J}{JE}.

So for completeness of the problem suppose angles of △A1A2A3\displaystyle\triangle A_1A_2A_3 are given as A1=A , A2=B , A3=C\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 A1\displaystyle A_1 located at (0,0)\displaystyle (0,0), A2\displaystyle A_2 located at (a,0)\displaystyle (a,0) and A3\displaystyle A_3 located at (b,c)\displaystyle (b,c).

Let a=a12=\displaystyle a = a_{12} = length of side A1A2\displaystyle A_1A_2. So, a12>0\displaystyle a_{12} > 0.

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

b=A1A3cos⁡A1\displaystyle b = A_1A_3\cos A_1

c=A1A3sin⁡A1\displaystyle c = A_1A_3\sin A_1

Now, A1A3sin⁡A2=A1A2sin⁡A3\displaystyle \frac{A_1A_3}{\sin A_2} = \frac{A_1A_2}{\sin A_3} (sine rule of triangle)

⇒A1A3=A1A2sin⁡A2sin⁡A3=a12sin⁡A2sin⁡A3\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,

b=a12sin⁡A2cos⁡A1sin⁡A3\displaystyle b = \frac{a_{12}\sin A_2\cos A_1}{\sin A_3}

c=a12sin⁡A2sin⁡A1sin⁡A3\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 ApAq\displaystyle A_pA_q is denoted by apq\displaystyle a_{pq}.

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

  • For simplicity in writing we denote

∠Ai−2AiAi−1\hspace{20pt} \displaystyle\angle A_{i-2}A_iA_{i-1} with Ai\displaystyle A_i for all i≥3\displaystyle i \geq 3

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

si=AiAi+3AiAi+1=ai i+3ai i+1\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 i∈N\displaystyle i \in \mathbb{N}

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

An=(xn,yn)\displaystyle A_n = (x_n, y_n) for all n∈N\displaystyle n \in \mathbb{N}

Then we are given

x1=y1=0\displaystyle x_1 = y_1 = 0

x2=a12  ,  y=0\displaystyle x_2 = a_{12}\;,\;y = 0

x3=a12sin⁡A2cos⁡A1sin⁡A3  ,  y3=a12sin⁡A2sin⁡A1sin⁡A3\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,

(x4,y4)=(1−s1)(x1,y1)+s1(x2,y2)\displaystyle (x_4,y_4) = (1 - s_1)(x_1,y_1) + s_1(x_2,y_2)

and in general,

(xn,yn)=(1−sn−3)(xn−3,yn−3)+sn−3(xn−2,yn−2)\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

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

yn=(1−sn−3)yn−3+sn−3yn−2\displaystyle y_n = (1 - s_{n-3})y_{n-3} + s_{n-3}y_{n-2} for all n≥4\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 xn\displaystyle x_n there is another sequence sn\displaystyle s_n that is need to be solved.

Taking n=1\displaystyle n = 1,

s1=A1A4A4A2=a14a12=area(△A1A3A4)area(△A1A3A2)=12a13a34sin⁡(rA3)12a13a23sin⁡A3=sin⁡A2sin⁡(rA3)sin⁡A4sin⁡A3\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 n=i>1\displaystyle n = i > 1,

si=AiAi+3AiAi+1=ai i+3ai i+1=area(△AiAi+2Ai+3)area(△AiAi+2Ai+1)=12ai i+2ai+2 i+3sin⁡(rAi+2)12ai i+2ai+1 i+2sin⁡Ai+2=sin⁡[(1−r)Ai+1]sin⁡(rAi+2)sin⁡Ai+3sin⁡Ai+2\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 sn\displaystyle s_n defined as

s1=sin⁡A2sin⁡(rA3)sin⁡A3sin⁡A4\hspace{20pt}\displaystyle s_1 = \frac{\sin A_2\sin(rA_3)}{\sin A_3\sin A_4}

and sn=sin⁡[(1−r)An+1]sin⁡(rAn+2)sin⁡An+2sin⁡An+3\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≥2\displaystyle n \geq 2

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

A1 , A2 , A3\displaystyle A_1\,,\,A_2\,,\,A_3 are the three initial angles given. It is easy to see that

A4=A1+rA3\displaystyle A_4 = A_1 + rA_3

A5=A2+rA4\displaystyle A_5 = A_2 + rA_4

seems like that next term will be

A6=A3+rA5\displaystyle A_6 = A_3 + rA_5

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

A6=(1−r)A3+rA5\displaystyle A_6 = (1-r)A_3 + rA_5

and in general

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

So we get the sequence An\displaystyle A_n defined as

A1=A , A2=B , A3=C\displaystyle A_1 = A\,,\,A_2 = B\,,\,A_3 = C and

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

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

So what we have done from starting is summarised below

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

An=(xn,yn)\displaystyle A_n = (x_n,y_n) for all n∈N\displaystyle n\in\mathbb{N}

Definition of sequence xn\mathbf{x_n}

x1=0    ,    x2=a12  ,  x3=a12sin⁡A2cos⁡A1sin⁡A3\displaystyle x_1 = 0\;\;,\;\;x_2 = a_{12}\;,\;x_3 = \frac{a_{12}\sin A_2\cos A_1}{\sin A_3}

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

Definition of sequence yn\mathbf{y_n}

y1=0    ,    y2=0  ,  y3=a12sin⁡A2sin⁡A1sin⁡A3\displaystyle y_1 = 0\;\;,\;\;y_2 = 0\;,\;y_3 = \frac{a_{12}\sin A_2\sin A_1}{\sin A_3}

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

Definition of sequence sn\mathbf{s_n}

s1=sin⁡A2sin⁡(rA3)sin⁡A3sin⁡A4\displaystyle s_1 = \frac{\sin A_2\sin(rA_3)}{\sin A_3\sin A_4}

sn=sin⁡[(1−r)An+1]sin⁡(rAn+2)sin⁡An+2sin⁡An+3    \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≥2n \geq 2

Definition of sequence An\mathbf{A_n}

A1=A    ,    A2=B  ,  A3=C\displaystyle A_1 = A\;\;,\;\;A_2 = B\;,\;A_3 = C

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

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

We need to find the coordinates of point J\displaystyle J to get the required ratio A1JJE\displaystyle\frac{A_1J}{JE}. Let coordinates of point J\displaystyle J be (xJ,yJ)\displaystyle(x_J, y_J). Then we have,

xJ=lim⁡n→∞xn\displaystyle x_J = \lim_{n \to \infty}x_n and yJ=lim⁡n→∞yn\displaystyle y_J = \lim_{n \to \infty}y_n

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

We see that sequence An\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 33 with constant cofficients.

Let An=λn , n≥3\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,

λn=(1−r)λn−3+rλn−1\lambda^n = (1-r)\lambda^{n-3} + r\lambda^{n-1}

⇒λ3=(1−r)+rλ2\Rightarrow \lambda^3 = (1-r) + r\lambda^2

⇒λ3−rλ2−(1−r)=0\Rightarrow \lambda^3 - r\lambda^2 - (1-r) = 0

There are the following three roots to this equation:

λ1=1  ,      λ2=−1−r2+(3+r)(1−r)2i  ,      λ3=−1−r2−(3+r)(1−r)2i      \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=−1i = \sqrt{-1}

So we get,

An=α1n+βλ2n+γλ3nA_n = \alpha 1^n + \beta\lambda_2^n + \gamma\lambda_3^n for all n≥3n \geq 3

To get the values of α,β,γ\alpha, \beta, \gamma we use the values of A3 , A4 , A5A_3\,,\,A_4\,,\,A_5 in terms of A , B , CA\,,\,B\,,\,C.

We have

A3=C  ,    A4=A+rC  ,    A5=B+r(A+rC)=B+rA+r2CA_3 = C\;,\;\;A_4 = A + rC\;,\;\;A_5 = B + r(A + rC) = B + rA + r^2C

So,

C=A3=α+βλ23+γλ33⋯Eq. 1C = A_3 = \alpha + \beta\lambda_2^3 + \gamma\lambda_3^3\hspace{20pt}\cdots\text{Eq. 1}

A+rC=A4=α+βλ24+γλ34⋯Eq. 2A + rC = A_4 = \alpha + \beta\lambda_2^4 + \gamma\lambda_3^4\hspace{20pt}\cdots\text{Eq. 2}

B+rA+r2C=A5=α+βλ25+γλ35⋯Eq. 3B + 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,

[1λ23λ331λ24λ341λ25λ35][αβγ]=[CA+rCB+rA+r2C]\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 α,β,γ\alpha, \beta, \gamma from these equations in terms of rr by writing λ2\lambda_2 and λ3\lambda_3 in terms of rr we get, (I am not writing the intermediate steps, it is very tedious for writing here)

α=A+B+C3−2r=180°3−2r[∵A,B,C\displaystyle\alpha = \frac{A + B + C}{3 - 2r} = \frac{180\degree}{3 - 2r}\hspace{20pt}[\because A, B, C are the angles of triangle]]

β=[A2(2r−3)+(1+r−r2)B2(1−r)2(2r−3)+(r3−r2+r−2)C2(1−r)2(2r−3)]+[(1+2r−r2)A2(1−r)(2r−3)(3+r)(1−r)+(r2+r−3)B2(1−r)(2r−3)(3+r)(1−r)−3rC2(2r−3)(3+r)(1−r)]i\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

γ=[A2(2r−3)+(1+r−r2)B2(1−r)2(2r−3)+(r3−r2+r−2)C2(1−r)2(2r−3)]−[(1+2r−r2)A2(1−r)(2r−3)(3+r)(1−r)+(r2+r−3)B2(1−r)(2r−3)(3+r)(1−r)−3rC2(2r−3)(3+r)(1−r)]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 γ=β‾  \displaystyle \gamma = \overline{\beta}\;, where z‾\displaystyle\overline{z} denotes conjugate of z\displaystyle z.

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

A1=A  ,    A2=B\displaystyle A_1 = A\;,\;\;A_2 = B

An=180°3−2r+β(−1−r2+(3+r)(1−r)2i)n+β‾(−1−r2−(3+r)(1−r)2i)n\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 n≥3\displaystyle n \geq 3

with β\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 A1A2A3A_1A_2A_3 and any special value of rr 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.

#Geometry

Note by Shikhar Srivastava
11 months ago

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Don't ask me!

Yajat Shamji - 11 months 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 lim⁡n→∞∠An−2AnAn−1=π3−2rlim⁡n→∞∠AnAn−1An−2=(1−r)π3−2rlim⁡n→∞∠AnAn−2An−1=(1−r)π3−2r\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 AnA_n (note that ∣λ2∣=∣λ3∣<1|\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 xx and yy directions, but I'm not sure that's the case here.

Chris Lewis - 11 months ago

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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 xnx_n. These limiting angles sums to 180°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 AnA_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.

Shikhar Srivastava - 11 months ago

I started with the specific case of r=12r = \frac{1}{2}, which means each angle is bisected, and chose a right angle isosceles triangle as a convenient triangle to use with A1(0,0)A_1(0, 0), A2(2,0)A_2(2, 0), and A3(1,1)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 (xn,yn)=(12(xn−3+xn−2),12(xn−3+xn−2))(x_n, y_n) = (\frac{1}{2}(x_{n - 3} + x_{n - 2}), \frac{1}{2}(x_{n - 3} + x_{n - 2})) for n≥3n \geq 3 and (x1,y1)=(0,0)(x_1, y_1) = (0, 0), (x2,y2)=(2,0)(x_2, y_2) = (2, 0), (x3,y3)=(1,1)(x_3, y_3) = (1, 1). After some computation, JJ is J(65,25)J(\frac{6}{5}, \frac{2}{5}), EE is E(32,12)E(\frac{3}{2}, \frac{1}{2}), A1J=2105A_1J = \frac{2\sqrt{10}}{5}, JE=1010JE = \frac{\sqrt{10}}{10}, and A1JJE=4\frac{A_1J}{JE} = 4 .

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

David Vreken - 11 months ago

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Brilliant, those are definitely "nice" parameters!

I'd been looking at r=12r=\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=34r=\frac34 and initial points (in order) (0,0),(2,0),(1,tan⁡30∘)(0,0),(2,0),(1,\tan 30^{\circ}), which heads to a final point at (97,37)(\frac97,\frac{\sqrt3}{7}).

Chris Lewis - 11 months ago

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do you mean (1,tan⁡30°)(1, \tan 30°)?

David Vreken - 11 months ago

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@David Vreken – I did! Thanks, edited now.

Chris Lewis - 11 months ago

With a little bit of calculation, I found that the coordinates A1(0,0)A_1(0, 0), A2(2,0)A_2(2, 0), and A3(1,tan⁡(1−r3−2r⋅180°))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′=sin⁡(r3−2r⋅180°)2⋅cos⁡(1−r3−2r⋅180°)⋅sin⁡(13−2r⋅180°)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".

David Vreken - 11 months ago

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@David Vreken – It is a nice result.

Shikhar Srivastava - 11 months ago

I was trying to find some result with r=12r = \frac{1}{2} (without any special triangle) but haven't got much. It was nice idea to take right isosceles triangle with r=12r = \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.

Shikhar Srivastava - 11 months ago

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

David Vreken - 11 months ago

I noticed that the ratio A1JJE\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 A1(0,0)A_1(0, 0), A2(2,0)A_2(2, 0), and A3(1,1)A_3(1, 1) above with r=12r = \frac{1}{2} makes A1JJE=4\frac{A_1J}{JE} = 4, however, using A1(0,0)A_1(0, 0), A2(1,0)A_2(1, 0), and A3(0,1)A_3(0, 1) instead (like in the original question's solution) with r=12r = \frac{1}{2} makes A1JJE=3\frac{A_1J}{JE} = 3. I believe that means that the original question must also include the extra parameter that the height of △A1A2A3\triangle A_1A_2A_3 must equal the base of △A1A2A3\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, A1JJE=4\frac{A_1J}{JE} = 4 for r=12r = \frac{1}{2} irregardless of the size and shape of the triangle.

David Vreken - 11 months ago

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If you are using r=12r = \frac{1}{2} in this question then the ratio will change with respect to vertices of triangle. But if you are r=12r = \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 rr as

A1JJE=21−r\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 r1−r\frac{r}{1 - r} instead of 37\frac{3}{7} in that question and solve further whether by algebraic sequential method or by geometry, you will get the ratio as

A1JJE=21−r\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 A1=(0,0)  , A2=(a,0)  , A3=(b,c)A_1 = (0,0)\;,\,A_2 = (a,0)\;,\,A_3 = (b,c), with a,c>0a,c > 0 and b∈Rb\in\mathbb{R}, it represents all triangle possible (in order). For ex - triangle with A1=30°  , A2=60°  , A3=90°A_1 = 30\degree\;,\,A_2 = 60\degree\;,\,A_3 = 90\degree and another triangle with A1=60°  , A2=90°  , A3=30°A_1 = 60\degree\;,\,A_2 = 90\degree\;,\,A_3 = 30\degree with same size are considered different.

We have

x1=y1=0  , x2=a  , y2=0  , x3=b  , y3=cx_1 = y_1 = 0\;,\,x_2 = a\;,\,y_2 = 0\;,\,x_3 = b\;,\,y_3 = c and

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

And we get the solution of general term of xnx_n as:

When r=34\displaystyle r = \frac{3}{4},

xn=49(a+b)+(−4a+12an+32b−24bn9)(−12)n\displaystyle x_n = \frac{4}{9}\big(a + b\big) +\bigg(\frac{-4a + 12an + 32b - 24bn}{9}\bigg)\bigg(\frac{-1}{2}\bigg)^n

yn=49c+(32c−24cn9)(−12)n\displaystyle y_n = \frac{4}{9}c +\bigg(\frac{32c - 24cn}{9}\bigg)\bigg(\frac{-1}{2}\bigg)^n

So, J=(4(a+b)9,4c9)\displaystyle J = \bigg(\frac{4(a + b)}{9},\frac{4c}{9}\bigg)

Using A1\displaystyle A_1 and J\displaystyle J we get,

E=(a+b2,c2)\displaystyle E = \bigg(\frac{a + b}{2}, \frac{c}{2}\bigg)

So, A1J=49(a+b)2+c2\displaystyle A_1J = \frac{4}{9}\sqrt{(a + b)^2 + c^2}

and JE=(12−49)(a+b)2+c2=118(a+b)2+c2\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 A1JJE=8=21−34=21−r\displaystyle \frac{A_1J}{JE} = 8 = \frac{2}{1 - \frac{3}{4}} = \frac{2}{1 - r}

When r≠34\displaystyle r \neq \frac{3}{4},

xn=a+b3−r+β1(−12+124r−3)n+γ1(−12−124r−3)n\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

yn=c3−r+β2(−12+124r−3)n+γ2(−12−124r−3)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 β\beta and γ\gamma, because ∣−12±124r−3∣<1\displaystyle \bigg|-\frac{1}{2} \pm \frac{1}{2}\sqrt{4r - 3}\bigg| < 1

So, J=(a+b3−r,c3−r)\displaystyle J = \bigg(\frac{a + b}{3 - r},\frac{c}{3 - r}\bigg)

Using A1\displaystyle A_1 and J\displaystyle J we get,

E=(a+b2,c2)\displaystyle E = \bigg(\frac{a + b}{2}, \frac{c}{2}\bigg)

So, A1J=13−r(a+b)2+c2\displaystyle A_1J = \frac{1}{3 - r}\sqrt{(a + b)^2 + c^2}

and JE=(12−13−r)(a+b)2+c2=1−r2(3−r)(a+b)2+c2\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 A1JJE=13−r1−r2(3−r)=21−r\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 A1A2A3A_1A_2A_3 we get the same result of the ratio A1JJE\frac{A_1J}{JE}.

Shikhar Srivastava - 11 months ago

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I'm confused. If A1JJE=21−r\frac{A_1J}{JE} = \frac{2}{1 - r}, then for the original question where r=37r = \frac{3}{7}, A1JJE=21−37=72\frac{A_1J}{JE} = \frac{2}{1 - \frac{3}{7}} = \frac{7}{2}, but the answer is A1JJE=207\frac{A_1J}{JE} = \frac{20}{7}.

David Vreken - 11 months ago

Nevermind, I'm mixing up variables. The original question actually has r=33+7=310r = \frac{3}{3 + 7} = \frac{3}{10}, which does come out to A1JJE=21−r=207\frac{A_1J}{JE} = \frac{2}{1 - r} = \frac{20}{7}.

David Vreken - 11 months ago

Combining some results the equations r′=sin⁡(r3−2r⋅180°)2⋅cos⁡(1−r3−2r⋅180°)⋅sin⁡(13−2r⋅180°)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 A1JJE=21−r′\frac{A_1J}{JE} = \frac{2}{1 - r'} found in this thread and simplifying, if r=∠Ai−2AiAi+1∠Ai−2AiAi−1r = \frac{\angle A_{i-2}A_iA_{i+1}}{\angle A_{i-2}A_iA_{i-1}}, then:

A1JJE=2(sin⁡(r3−2r⋅180°)sin⁡(2−r3−2r⋅180°)+1)\frac{A_1J}{JE} = 2\bigg(\frac{\sin (\frac{r}{3 - 2r} \cdot 180°)}{\sin (\frac{2 - r}{3 - 2r} \cdot 180°)} + 1\bigg)

.

David Vreken - 11 months ago

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

Shikhar Srivastava - 11 months ago

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

Shikhar Srivastava - 11 months ago

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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 3200032000 centres on this site so there's a decent chance of finding a match! It could well help with the r=12r=\frac12 case (and perhaps others).

Chris Lewis - 11 months ago
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