Spiraling Down

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Since $P_n$ is defined as the midpoint of $P_{n-4}P_{n-3},\,x_n=\dfrac{1}{2}(x_{n-4}+x_{n-3})$ for $n≥5.$

So we prove by induction that $\dfrac{1}{2}x_n+x_{n+1}+x_{n+2}+x_{n+3}=2$

The case $n=1$ is immediate, since $\dfrac{1}{2}\cdot 0+1+1+0=2$

Assume that the result holds for $n=k-1$ , that is, $\dfrac{1}{2}x_{k-1}+x_k+x_{k+1}+x_{k+2}=2$ . Then for $n=k$ ,

$\begin{aligned}\dfrac{1}{2}x_k+x_{k+1}+x_{k+2}+x_{k+3} & =\dfrac{1}{2}x_k+x_{k+1}+x_{k+2}+\dfrac{1}{2}(x_{k+3-4}+x_{k+3-3})\,\text{[by above]} \\ & =\dfrac{1}{2}x_{k-1}+x_k+x_{k+1}+x_{k+2}=2\,\text{[by the induction hypothesis]} \\ \end{aligned}$

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Similarly for $n≥5,\,y_n=\dfrac{1}{2}(y_{n-4}+y_{n-3})$ , so the same argument as above holds for $y$ , with $2$ replaced by

$\dfrac{1}{2}y_1+y_2+y_3+y_4=\dfrac{1}{2}\cdot 1+1+0+0=\dfrac{3}{2}$ .

So $\dfrac{1}{2}y_n+y_{n+1}+y_{n+2}+y_{n+3}=\dfrac{3}{2}$ for all $n$ .

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$\displaystyle\lim_{n\to\infty}\left(\dfrac{1}{2}x_n+x_{n+1}+x_{n+2}+x_{n+3}\right)=\dfrac{1}{2}\lim_{n\to\infty}x_n+\lim_{n\to\infty}x_{n+1}+\lim_{n\to\infty}x_{n+2}+\lim_{n\to\infty}x_{n+3}=2$

Since all the limits on the left hand side are same, we get $\displaystyle\dfrac{7}{2}\lim_{n\to\infty}x_n=2\,\implies\,\lim_{n\to\infty}x_n=\dfrac{4}{7}$

In the same way, $\displaystyle\dfrac{7}{2}\lim_{n\to\infty}y_n=\dfrac{3}{2}\,\implies\,\lim_{n\to\infty}y_n=\dfrac{3}{7}$

So, $\boxed{P_{\infty}=\left(\dfrac{4}{7},\dfrac{3}{7}\right)}$ .

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I "solved" this problem numerically, and I'm missing the tools for an exact solution. I figured that

$\begin{bmatrix} x_{4i+1} \\ x_{4i+2} \\ x_{4i+3} \\ x_{4i+4} \\ \end{bmatrix}=\begin{bmatrix} 0.5 & 0.5 & 0 & 0 \\ 0 & 0.5 & 0.5 & 0 \\ 0 & 0 & 0.5 & 0.5 \\ 0.25 & 0.25 & 0 & 0.5 \\ \end{bmatrix} ^i \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \\ \end{bmatrix}$

and

$\begin{bmatrix} y_{4i+1} \\ y_{4i+2} \\ y_{4i+3} \\ y_{4i+4} \\ \end{bmatrix}=\begin{bmatrix} 0.5 & 0.5 & 0 & 0 \\ 0 & 0.5 & 0.5 & 0 \\ 0 & 0 & 0.5 & 0.5 \\ 0.25 & 0.25 & 0 & 0.5 \\ \end{bmatrix} ^i \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix}$ .

Now it remaines to calculate $\displaystyle \lim _{n\to \infty } \begin{bmatrix} 0.5 & 0.5 & 0 & 0 \\ 0 & 0.5 & 0.5 & 0 \\ 0 & 0 & 0.5 & 0.5 \\ 0.25 & 0.25 & 0 & 0.5 \\ \end{bmatrix} ^n$ .

I raised it to high powers with a computer and obtained something which looked a lot like

$\begin{bmatrix} 1\over 7 & 2\over 7 & 2\over 7 & 2\over 7\\ 1\over 7 & 2\over 7 & 2\over 7 & 2\over 7\\ 1\over 7 & 2\over 7 & 2\over 7 & 2\over 7\\ 1\over 7 & 2\over 7 & 2\over 7 & 2\over 7\\ \end{bmatrix}$ .

Whence $\begin{bmatrix} x_{\infty} & y_{\infty} \\ x_{\infty} & y_{\infty} \\ x_{\infty} & y_{\infty} \\ x_{\infty} & y_{\infty} \\ \end{bmatrix}=\begin{bmatrix} 1\over 7 & 2\over 7 & 2\over 7 & 2\over 7\\ 1\over 7 & 2\over 7 & 2\over 7 & 2\over 7\\ 1\over 7 & 2\over 7 & 2\over 7 & 2\over 7\\ 1\over 7 & 2\over 7 & 2\over 7 & 2\over 7\\ \end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & 1\\ 1 & 0\\ 0 & 0\\ \end{bmatrix}=\begin{bmatrix} 4\over 7 & 3\over 7\\ 4\over 7 & 3\over 7\\ 4\over 7 & 3\over 7\\ 4\over 7 & 3\over 7\\ \end{bmatrix}$

Could this have been done rigourously w/o a computer? The Matrix' eigenvalues are far from nice so diagonalization is out of the question.

Anyhow, a very nice problem indeed, thank you!

I used a very tedious recurrence method to solve this. Given that ${ P }_{ n }$ is the midpoint of ${ P }_{ n-3 }$ and ${ P }_{ n-4 }$ , we have ${ P }_{ n }=\frac { 1 }{ 2 } \left( { P }_{ n-3 }+{ P }_{ n-4 } \right)$ . Expressing as a matrix, we get

$\left( \begin{matrix} { P }_{ n } \\ { P }_{ n-1 } \\ { P }_{ n-2 } \\ { P }_{ n-3 } \end{matrix} \right) =\begin{pmatrix} 0 & 0 & 0.5 & 0.5 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}\left( \begin{matrix} { P }_{ n-1 } \\ { P }_{ n-2 } \\ { P }_{ n-3 } \\ { P }_{ n-4 } \end{matrix} \right)$

The transformation matrix has the characteristic polynomial $2{ \lambda }^{ 4 }-\lambda -1=0$ ,where $\lambda$ are the eigenvalues. The eigenvalue $1$ can be easily factored out, leaving $2{ \lambda }^{ 3 }+2{ \lambda }^{ 2 }+2\lambda +1=0$ . Making the substitution $\lambda=t-\frac { 1 }{ 3 }$ , it can be simplified further to ${ t }^{ 3 }+\frac { 2 }{ 3 } t+\frac { 13 }{ 54 } =0$ . From here applying Cardano's formula gives the solutions $t=u+v$ and $t=\frac { -1 }{ 2 } \left( u+v \right) \pm \frac { \sqrt { 3 } }{ 2 } (u-v) i$ , where $u=\sqrt [ 3 ]{ \frac { -13 }{ 108 } +\frac { 1 }{ 12 } \sqrt { \frac { 11 }{ 3 } } }$ , $v=\sqrt [ 3 ]{ \frac { -13 }{ 108 } -\frac { 1 }{ 12 } \sqrt { \frac { 11 }{ 3 } } }$ . Converting back to $\lambda$ gives us the eigenvalues $\lambda =1$ , $\lambda =u+v-\frac { 1 }{ 3 }$ and $\lambda =\alpha \pm \beta i$ where $\alpha =\frac { -1 }{ 2 } \left( u+v \right) -\frac { 1 }{ 3 }$ and $\beta =\frac { \sqrt { 3 } }{ 2 } (u-v)$ .

After finding the corresponding eigenvectors for each eigenvalue, the transformation matrix can be expressed as

$\begin{pmatrix} 0 & 0 & 0.5 & 0.5 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}=\left( \begin{matrix} 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 3 } & { \left( \alpha +\beta i \right) }^{ 3 } & { \left( \alpha -\beta i \right) }^{ 3 } \\ 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 2 } & { \left( \alpha +\beta i \right) }^{ 2 } & { \left( \alpha -\beta i \right) }^{ 2 } \\ 1 & \left( u+v-\frac { 1 }{ 3 } \right) & \left( \alpha +\beta i \right) & \left( \alpha -\beta i \right) \\ 1 & 1 & 1 & 1 \end{matrix} \right) \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & \left( u+v-\frac { 1 }{ 3 } \right) & 0 & 0 \\ 0 & 0 & \left( \alpha +\beta i \right) & 0 \\ 0 & 0 & 0 & \left( \alpha -\beta i \right) \end{matrix} \right) { \left( \begin{matrix} 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 3 } & { \left( \alpha +\beta i \right) }^{ 3 } & { \left( \alpha -\beta i \right) }^{ 3 } \\ 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 2 } & { \left( \alpha +\beta i \right) }^{ 2 } & { \left( \alpha -\beta i \right) }^{ 2 } \\ 1 & \left( u+v-\frac { 1 }{ 3 } \right) & \left( \alpha +\beta i \right) & \left( \alpha -\beta i \right) \\ 1 & 1 & 1 & 1 \end{matrix} \right) }^{ -1 }$

Given that for a matrix $A=QD{ Q }^{ -1 }$ where D is the diagonal matrix of eigenvalues and Q is the matrix of corresponding eigenvectors, ${ A }^{ n }$ can be expressed as $Q{ D }^{ n }{ Q }^{ -1 }$ , we can thus express ${ P }_{ n }$ as such:

$\left( \begin{matrix} { P }_{ n } \\ { P }_{ n-1 } \\ { P }_{ n-2 } \\ { P }_{ n-3 } \end{matrix} \right) =\left( \begin{matrix} 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 3 } & { \left( \alpha +\beta i \right) }^{ 3 } & { \left( \alpha -\beta i \right) }^{ 3 } \\ 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 2 } & { \left( \alpha +\beta i \right) }^{ 2 } & { \left( \alpha -\beta i \right) }^{ 2 } \\ 1 & \left( u+v-\frac { 1 }{ 3 } \right) & \left( \alpha +\beta i \right) & \left( \alpha -\beta i \right) \\ 1 & 1 & 1 & 1 \end{matrix} \right) \left( \begin{matrix} { 1 }^{ n-7 } & 0 & 0 & 0 \\ 0 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ n-7 } & 0 & 0 \\ 0 & 0 & { \left( \alpha +\beta i \right) }^{ n-7 } & 0 \\ 0 & 0 & 0 & { \left( \alpha -\beta i \right) }^{ n-7 } \end{matrix} \right) { \left( \begin{matrix} 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 3 } & { \left( \alpha +\beta i \right) }^{ 3 } & { \left( \alpha -\beta i \right) }^{ 3 } \\ 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 2 } & { \left( \alpha +\beta i \right) }^{ 2 } & { \left( \alpha -\beta i \right) }^{ 2 } \\ 1 & \left( u+v-\frac { 1 }{ 3 } \right) & \left( \alpha +\beta i \right) & \left( \alpha -\beta i \right) \\ 1 & 1 & 1 & 1 \end{matrix} \right) }^{ -1 }\left( \begin{matrix} { P }_{ 7 } \\ { P }_{ 6 } \\ { P }_{ 5 } \\ { P }_{ 4 } \end{matrix} \right)$

Since $\left| u+v-\frac { 1 }{ 3 } \right| ,\left| \alpha \pm \beta i \right| <1$ , the limit as $n\rightarrow \infty$ gives $\left( \begin{matrix} { P }_{ \infty } \\ { P }_{ \infty } \\ { P }_{ \infty } \\ { P }_{ \infty } \end{matrix} \right) =\left( \begin{matrix} 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 3 } & { \left( \alpha +\beta i \right) }^{ 3 } & { \left( \alpha -\beta i \right) }^{ 3 } \\ 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 2 } & { \left( \alpha +\beta i \right) }^{ 2 } & { \left( \alpha -\beta i \right) }^{ 2 } \\ 1 & \left( u+v-\frac { 1 }{ 3 } \right) & \left( \alpha +\beta i \right) & \left( \alpha -\beta i \right) \\ 1 & 1 & 1 & 1 \end{matrix} \right) \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right) { \left( \begin{matrix} 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 3 } & { \left( \alpha +\beta i \right) }^{ 3 } & { \left( \alpha -\beta i \right) }^{ 3 } \\ 1 & { \left( u+v-\frac { 1 }{ 3 } \right) }^{ 2 } & { \left( \alpha +\beta i \right) }^{ 2 } & { \left( \alpha -\beta i \right) }^{ 2 } \\ 1 & \left( u+v-\frac { 1 }{ 3 } \right) & \left( \alpha +\beta i \right) & \left( \alpha -\beta i \right) \\ 1 & 1 & 1 & 1 \end{matrix} \right) }^{ -1 }\left( \begin{matrix} { P }_{ 7 } \\ { P }_{ 6 } \\ { P }_{ 5 } \\ { P }_{ 4 } \end{matrix} \right)$ $\left( \begin{matrix} { P }_{ \infty } \\ { P }_{ \infty } \\ { P }_{ \infty } \\ { P }_{ \infty } \end{matrix} \right) =\frac { 1 }{ 27{ u }^{ 3 }+27v^{ 3 }+108uv-64 } \left( \begin{matrix} -27 & -27 & 81uv-9 & 27{ u }^{ 3 }+27v^{ 3 }+27uv-1 \\ -27 & -27 & 81uv-9 & 27{ u }^{ 3 }+27v^{ 3 }+27uv-1 \\ -27 & -27 & 81uv-9 & 27{ u }^{ 3 }+27v^{ 3 }+27uv-1 \\ -27 & -27 & 81uv-9 & 27{ u }^{ 3 }+27v^{ 3 }+27uv-1 \end{matrix} \right) \left( \begin{matrix} { P }_{ 7 } \\ { P }_{ 6 } \\ { P }_{ 5 } \\ { P }_{ 4 } \end{matrix} \right)$ Now, putting in the values of the coordinates of ${ P }_{ 4 }$ to ${ P }_{ 7 }$ gives $\boxed{{ P }_{ \infty }=\left( \frac { 4 }{ 7 } ,\frac { 3 }{ 7 } \right) }$

x1,y1=0,1 x2,y2=1,1 x3,y3=1,0 x4,y4=0,0 c=0 while(c<=5000): x1,y1=(x1+x2)/2,(y1+y2)/2 x2,y2=(x2+x3)/2,(y2+y3)/2 x3,y3=(x3+x4)/2,(y3+y4)/2 x4,y4=(x4+x1)/2,(y4+y1)/2 c+=1 print (x1,y1) //Some Python!!!