smc

Let $a_n=|A_{n-1}{A}_n|$. We need to rewrite the recursion into something manageable. The two strange conditions, $B$'s lie on the graph of $y=\sqrt{x}$ and $A_{n-1}{B}_n{A}_n$ is an equilateral triangle, can be compacted as follows: $${\left(a_n\frac{\sqrt{3}}{2}\right)}^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots +a_1$$ which uses $y^2=x$, where $x$ is the height of the equilateral triangle and therefore $\frac{\sqrt{3}}{2}$ times its base. The relation above holds for $n=k$ and for $n=k-1$ $(k>1)$, so $${\left(a_k\frac{\sqrt{3}}{2}\right)}^2-{\left(a_{k-1}\frac{\sqrt{3}}{2}\right)}^2=$$ $$=\left(\frac{a_k}{2}+a_{k-1}+a_{k-2}+\cdots +a_1\right)-\left(\frac{a_{k-1}}{2}+a_{k-2}+a_{k-3}+\cdots +a_1\right)$$ Or, $$a_k-a_{k-1}=\frac{2}{3}$$ This implies that each segment of a successive triangle is $\frac{2}{3}$ more than the last triangle. To find $a_1$, we merely have to plug in $k=1$ into the aforementioned recursion and we have $a_1-a_0=\frac{2}{3}$. Knowing that $a_0$ is $0$, we can deduce that $a_1=2/3$.Thus, $a_n=\frac{2n}{3}$, so $A_0{A}_n=a_n+a_{n-1}+\cdots +a_1=\frac{2}{3}\cdot \frac{n(n+1)}{2}=\frac{n(n+1)}{3}$. We want to find $n$ so that $n^2<300<(n+1{)}^2$. $n=\boxed{17}$ is our answer.