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2008 AMC 12B Problems/Problem 24

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Problem 24

Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ 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 $A_0A_n\geq100$ ?

$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$

Solution

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 $A_{n-1}B_nA_n$ 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, $a_k-a_{k-1}=\frac23$ Thus, $a_n=\frac{2n}{3}$ , so $A_0A_n=a_n+a_{n-1}+\cdots+a_1=\frac{n(n+1)}{3}$ . We want to find $n$ so that $n^2<300<(n+1)^2$ . $n=\boxed{17}$ is our answer.

See Also

2008 AMC 12B (Problems • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

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