2008 AMC 12B Problem 24

Let $a_n = A_0A_n$ and $c_n = a_n - a_{n-1}$ be the base of the $n$th equilateral triangle. Its apex $B_n$ lies above the midpoint at height $\tfrac{\sqrt3}{2}c_n,$ and being on $y = \sqrt{x}$ gives $$\left(\tfrac{\sqrt3}{2}c_n\right)^2 = a_{n-1} + \tfrac{c_n}{2},$$ i.e. $$\tfrac34 c_n^2 = a_{n-1} + \tfrac{c_n}{2}.$$ Writing the same relation for the previous triangle and subtracting gives $c_n = c_{n-1} + \tfrac23,$ and with $c_1 = \tfrac23$ we get $c_n = \tfrac{2n}{3}.$ Summing, $$a_n = \sum_{k=1}^n \frac{2k}{3} = \frac{n(n + 1)}{3}.$$ We need $\tfrac{n(n + 1)}{3} \ge 100,$ i.e. $n(n + 1) \ge 300.$ Since $16 \cdot 17 = 272$ and $17 \cdot 18 = 306,$ the least such $n$ is $17.$ Thus, the correct answer is C.