Question

Let $A_0 = (0, 0).$ Distinct points $A_1, A_2, \ldots$ lie on the $x$ -axis, and distinct points $B_1, B_2, \ldots$ lie on the graph of $y = \sqrt{x}.$ For every positive integer $n,$ $A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n \ge 100?$

$13$

$15$

$17$

$19$

$21$

Solution:

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 $\left(\tfrac{\sqrt3}{2}c_n\right)^2 = a_{n-1} + \tfrac{c_n}{2}, \quad\text{i.e.}\quad \tfrac34 c_n^2 = a_{n-1} + \tfrac{c_n}{2}.$

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.

2008 AMC 12B Problem 24

Solution:

Problem 24 in Other Years