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...) |
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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. | + | 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 . 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
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.
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 |