N Circle touching x-axis and fixed point

Question

If a circle $C_0$, with radius $1$ unit touches both the axes and as well as line ($ L_1 $) through $P(0,4)$, $L_1$ cut the $x$-axis at $(x_1 ,0)$ . Again a circle $C_1$ is drawn touching $x$-axis, line $L_1$ and another line $L_2$ through point $P$. $L_2$ intersects $x$-axis at $( x_2,0)$ and this process is repeated $n$ times. Then the value of : $$\lim_{n\to \infty }\frac{x_n}{2^n} $$ is

I am only able to find $C_1$ which is $(x-1)^2+(y-1)^2=1$ and $L_1$ which is $3x+4y=12$ but not able to proceed forward

Answer 1

Let points $C_n=(c_n,1)$ be the centers of the circles and $X_{n}=(x_n,0)$ the intersections of tangent lines with $x$-axis (see figure). As $C_{n-1}X_nC_n$ is a right triangle (because $C_{n-1}X_n$ and $X_nC_n$ are bisectors of adjacent angles), then by the geometric mean theorem the square of its altitude from $X_n$ (which has unit length) equals the product of the two parts into which the altitude divides the hypothenuse, that is: $(x_n-c_{n-1})(c_n-x_n)=1$. If we set $\delta_n=c_{n}-x_n$ this can be restated as $x_n-c_{n-1}=1/\delta_n$ and we obtain:

$$ \tag{1} c_{n}-c_{n-1}={1\over\delta_n}+\delta_n. $$

On the other hand, if $T_n$ is the tangency point in figure below, we have $PT_n=PX_n-\delta_n$ and (by the Pythagorean theorem) $PT_n^2=PC_{n}^2-1$, that is:

$$ \Big(\sqrt{(c_{n}-\delta_n)^2+16}-\delta_n\Big)^2=c_{n}^2+9-1, $$ which gives: $$ \tag{2} c_{n}={2\over\delta_n}-\delta_n. $$ Plugging this into $(1)$ gives then $$ c_{n-1}={1\over\delta_n}-2\delta_n \quad\text{and}\quad {1\over\delta_{n+1}}-2\delta_{n+1}={2\over\delta_n}-\delta_n, $$ that is: $$ \delta_{n+1}={1\over2}\delta_n. $$ As we can easily compute $\delta_1={1\over2}$ it follows that $$ \delta_n={1\over2^n} \quad\text{and}\quad x_n=c_n-\delta_n={2\over\delta_n}-2\delta_n=2^{n+1}-{1\over2^{n-1}}, $$ whence $\displaystyle\lim_{n\to\infty}{x_n\over 2^n}=2$.

Answer 2

This is a known configuration when $\triangle POX_n$ with the height $h=|OP|$ and inradius $r$ is split into $n$ triangles $\triangle POX_1,\triangle PX_1X_2,\dots\triangle PX_{n-1}X_n$, such that all their incircles have the same radius $r_n$.

The radius $r$ of the inscribed circle of $\triangle POX_n$ is found as

\begin{align} r&=\frac h2\,\left(1- \left(1-\frac{2r_n}h \right)^n \right) \tag{1}\label{1} . \end{align}
For $h=4,r_n=1$ we have

\begin{align} r&=2-2^{1-n} \tag{2}\label{2} . \end{align}

\begin{align} 2r&=|OP|+|OX_n|-|PX_n| =h+x_n-\sqrt{h^2+x_n^2} ,\\ x_n&= \frac{2r(h-r)}{h-2r} =2^{n+1}-2^{1-n} ,\\ \lim_{n\to \infty }\frac{x_n}{2^n} &=\frac{2^{n+1}-2^{1-n}}{2^n} =2 . \end{align}