Difference between revisions of "2008 AMC 12B Problems/Problem 24"

(New page: ==Problem 24== Let <math>A_0=(0,0)</math>. Distinct points <math>A_1,A_2,\dots</math> lie on the <math>x</math>-axis, and distinct points <math>B_1,B_2,\dots</math> lie on the graph of <ma...)
 
(Solution)
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The relation above holds for <math>n=k</math> and for <math>n=k-1</math> <math>(k>1)</math>, so <cmath>\left(a_k\frac{\sqrt{3}}{2}\right)^2-\left(a_{k-1}\frac{\sqrt{3}}{2}\right)^2=</cmath>
 
The relation above holds for <math>n=k</math> and for <math>n=k-1</math> <math>(k>1)</math>, so <cmath>\left(a_k\frac{\sqrt{3}}{2}\right)^2-\left(a_{k-1}\frac{\sqrt{3}}{2}\right)^2=</cmath>
 
<cmath>=\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)</cmath>
 
<cmath>=\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)</cmath>
Or, <cmath>a_k-a_{k-1}=\frac23</cmath>Thus, <math>a_n=\frac{2n}{3}</math>, so <math>A_0A_n=a_n+a_{n-1}+\cdots+a_1=\frac{n(n+1)}{3}</math>. We want to find <math>n</math> so that <math>n^2<300<(n+1)^2</math>. <math>n=\boxed{17}</math> is our answer.
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Or, <cmath>a_k-a_{k-1}=\frac23</cmath> This implies that each segment of a successive triangle is <math>\frac23</math> more than the last triangle. To find <math>a_{1}</math>, we merely have to plug in <math>k=1</math> into the aforementioned recursion and we have <math>a_{1} - a_{0} = \frac23</math>. Knowing that <math>a_{0}</math> is <math>0</math>, we can deduce that <math>a_{1} = 2/3</math>.Thus, <math>a_n=\frac{2n}{3}</math>, so <math>A_0A_n=a_n+a_{n-1}+\cdots+a_1=\frac{2}{3} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{3}</math>. We want to find <math>n</math> so that <math>n^2<300<(n+1)^2</math>. <math>n=\boxed{17}</math> is our answer.
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==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2008|ab=B|num-b=23|num-a=25}}

Revision as of 20:55, 31 December 2009

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, \[a_k-a_{k-1}=\frac23\] This implies that each segment of a successive triangle is $\frac23$ more than the last triangle. To find $a_{1}$, we merely have to plug in $k=1$ into the aforementioned recursion and we have $a_{1} - a_{0} = \frac23$. Knowing that $a_{0}$ is $0$, we can deduce that $a_{1} = 2/3$.Thus, $a_n=\frac{2n}{3}$, so $A_0A_n=a_n+a_{n-1}+\cdots+a_1=\frac{2}{3} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{3}$. We want to find $n$ so that $n^2<300<(n+1)^2$. $n=\boxed{17}$ is our answer.

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions
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