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. | ||
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+ | ==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> | ||
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+ | 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
Contents
Problem 24
Let . Distinct points lie on the -axis, and distinct points lie on the graph of . For every positive integer is an equilateral triangle. What is the least for which the length ?
Solution 1
Let . We need to rewrite the recursion into something manageable. The two strange conditions, 's lie on the graph of and is an equilateral triangle, can be compacted as follows: which uses , where is the height of the equilateral triangle and therefore times its base.
The relation above holds for and for , so Or, This implies that each segment of a successive triangle is more than the last triangle. To find , we merely have to plug in into the aforementioned recursion and we have . Knowing that is , we can deduce that .Thus, , so . We want to find so that . is our answer.
Solution 2
Consider two adjacent equilateral triangles obeying the problem statement. For each, drop an altitude to the axis and denote the resulting heights and . From 30-60-90 rules, the difference between the base of these altitudes is
But the square root curve means that this distance is also expressible as (the coordinates are the squares of the heights). Setting these expressions equal and dividing throughout by leaves . So the difference in height of successive triangles is , meaning their bases are wider and wider each time. From here, one can proceed as in Solution 1 to arrive at .
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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