TITLE: For $A_i$ on $y=\sqrt{x}$ and $B_i$ on $x$-axis, with $\triangle B_{i-1}B_iA_i$ equilateral of side $\ell_i$. Find $\ell_1+\cdots+\ell_{300}$.
QUESTION [2 upvotes]: Let $O$ be the origin, $A_1,A_2,A_3,\ldots$ be distinct points on the curve $y=\sqrt{x}$ and $B_1,B_2,B_3,\cdots$ be points on the positive $X$-axis such that the triangles $OB_1A_1,B_1B_2A_2,B_2B_3A_3,\ldots$ are all equilateral triangles with side lengths $l_1,l_2,l_3,\cdots$ respectively. Find the value of $l_1+l_2+\ldots+l_{300}$.

My Attempt:
We have $OA_1:y=\tan (60^{\circ})x=\sqrt{3}x$ so
$$
A_1=\left(\frac{1}{3},\frac{\sqrt{3}}{3}\right)
\quad \text{and} \quad
B_1=\left(\frac{2}{3},0\right),\
l_1=\frac{2}{3}.
$$
Let $B_i=(x_i,0)$ then
\begin{align*}
B_iA_{i+1}:& y=\sqrt{3}(x-x_i) \\
\iff & \sqrt{3}(x-x_i)=\sqrt{x} \\
\iff & 3x^2-x(6x_i+1)+3x_i^2=0 \\
\iff & x=\frac{6x_i+1+\sqrt{12x_i+1}}{6}.
\end{align*}
and so $A_{i+1}:\left(\frac{6x_i+1+\sqrt{12x_i+1}}{6},\sqrt{\frac{6x_i+1+\sqrt{12x_i+1}}{6}}=\frac{\sqrt{3}(1+\sqrt{12x_i+1})}{6}\right)$ and
\begin{align*}
x_{i+1}
& =2\left(\frac{6x_i+1+\sqrt{12x_i+1}}{6}\right)-x_i \\
& =\frac{3x_i+1+\sqrt{12x_i+1}}{3}
\end{align*}
\begin{align*}
l_{i+1}
& = x_{i+1}-x_i \\
& = \frac{1+\sqrt{12x_i+1}}{3} \\
& = \frac{1+\sqrt{12(\ell_1+\ell_2+\ldots+\ell_i)+1}}{3}.
\end{align*}
We will prove, that $\ell_i=\frac{2i}{3}$
For $i=1$ it is true.
For $i=n+1$:
\begin{align*}
3 \ell_{n+1}
& = 1+\sqrt{12(\ell_1+\ell_2+\ldots+\ell_n)+1} \\
& = 1+\sqrt{8(1+2+\ldots+n)+1} \\
& = 1+\sqrt{(4n^2+4n+1)} \\
& = 2n+2.
\end{align*}
so
$$
\ell_{n+1}=
\frac{2(n+1)}{3}.
$$
Hence
$$
\ell_1+\ldots+\ell_{300}
=\frac{2}{3} (1+\ldots+300)
=30100.
$$
Am I true for this?

REPLY [1 votes]: Let $A_i=(x_i, y_i)$ and $B_i=(z_i, 0)$ then because $y_i>0$, $x_i-z_i>0$ and $z_{i+1}-x_i>0$ we must have
$$
\frac{y_i}{x_i-z_i}=\frac{y_i}{z_{i+1}-x_i}=\tan\frac{\pi}{3}=\sqrt{3}
$$
then
$$
z_i=y_i^2-\frac{\sqrt {3}y_i}{3}\\
z_{i+1}=y_i^2+\frac{\sqrt {3}y_i}{3}\\
l_i=\frac{2\sqrt{3}}{3}y_i
$$
Confronting them let $y_{i+1}=y_i+u_i$ with $u_i>0$ it follows
$$
y_{i+1}^2-\frac{\sqrt {3}y_{i+1}}{3}=y_i^2+\frac{\sqrt {3}y_i}{3}\Rightarrow u_i^2+2u_iy_i-\frac{\sqrt {3}u_i}{3}=\frac{2\sqrt{3}}{3}y_i\\
\Rightarrow u_i=-y_i+\frac{\sqrt 3}{6}+\sqrt{y_i^2+\frac{1}{12}-\frac{\sqrt 3}{3}y_i+\frac{2\sqrt{3}}{3}y_i}=-y_i+\frac{\sqrt 3}{6}+y_i+\frac{\sqrt 3}{6}=\frac{\sqrt 3}{3}
$$
then set $x_0=y_0=0$ we have
$$
y_{i+1}=y_i+\frac{\sqrt 3}{3}=\frac{\sqrt 3}{3}(i+1)\Rightarrow l_i=\frac{2\sqrt{3}}{3}\frac{\sqrt 3}{3}i=\frac{2}{3}i
$$
and this is an arithmetic succession and you can easily find its sum.
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