# 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 $\ell_1+\cdots+\ell_{300}$.

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?

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.