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Difference between revisions of "2008 AMC 12B Problems/Problem 24"
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<math>\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21</math>
 
<math>\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21</math>
  
−
==Solution==
+
==Solution 1==
 
Let <math>a_n=|A_{n-1}A_n|</math>. We need to rewrite the recursion into something manageable. The two strange conditions, <math>B</math>'s lie on the graph of <math>y=\sqrt{x}</math> and <math>A_{n-1}B_nA_n</math> is an equilateral triangle, can be compacted as follows: <cmath>\left(a_n\frac{\sqrt{3}}{2}\right)^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots+a_1</cmath>
 
Let <math>a_n=|A_{n-1}A_n|</math>. We need to rewrite the recursion into something manageable. The two strange conditions, <math>B</math>'s lie on the graph of <math>y=\sqrt{x}</math> and <math>A_{n-1}B_nA_n</math> is an equilateral triangle, can be compacted as follows: <cmath>\left(a_n\frac{\sqrt{3}}{2}\right)^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots+a_1</cmath>
 
which uses <math>y^2=x</math>, where <math>x</math> is the height of the equilateral triangle and therefore <math>\frac{\sqrt{3}}{2}</math> times its base.
 
which uses <math>y^2=x</math>, where <math>x</math> is the height of the equilateral triangle and therefore <math>\frac{\sqrt{3}}{2}</math> times its base.
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<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> 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.
 
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.
 +
 +
==Solution 2==
 +
 +
Consider two adjacent equilateral triangles obeying the problem statement. For each, drop an altitude to the <math>x</math> axis and denote the resulting heights <math>h_n</math> and <math>h_{n+1}</math>. From 30-60-90 rules, the difference between the base of these altitudes is <cmath>\frac{h_{n+1}}{\sqrt{3}}+\frac{h_n}{\sqrt{3}} \Rightarrow \frac{h_{n+1}+h_n}{\sqrt{3}}</cmath>
 +
 +
But the square root curve means that this distance is also expressible as <math>h_{n+1}^2-h_n^2</math> (the <math>x</math> coordinates are the squares of the heights). Setting these expressions equal and dividing throughout by <math>h_{n+1}+h_n</math> leaves <math>h_{n+1}-h_n=\frac{1}{\sqrt{3}}</math>. So the difference in height of successive triangles is <math>\frac{1}{\sqrt{3}}</math>, meaning their bases are <math>2/3</math> wider and wider each time. From here, one can proceed as in Solution 1 to arrive at <math>n=\boxed{17}</math>.
  
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:35, 28 January 2018

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 1

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.

Solution 2

Consider two adjacent equilateral triangles obeying the problem statement. For each, drop an altitude to the $x$ axis and denote the resulting heights $h_n$ and $h_{n+1}$. From 30-60-90 rules, the difference between the base of these altitudes is \[\frac{h_{n+1}}{\sqrt{3}}+\frac{h_n}{\sqrt{3}} \Rightarrow \frac{h_{n+1}+h_n}{\sqrt{3}}\]

But the square root curve means that this distance is also expressible as $h_{n+1}^2-h_n^2$ (the $x$ coordinates are the squares of the heights). Setting these expressions equal and dividing throughout by $h_{n+1}+h_n$ leaves $h_{n+1}-h_n=\frac{1}{\sqrt{3}}$. So the difference in height of successive triangles is $\frac{1}{\sqrt{3}}$, meaning their bases are $2/3$ wider and wider each time. From here, one can proceed as in Solution 1 to arrive at $n=\boxed{17}$.

See Also

2008 AMC 12B (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

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