Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 12B Problems/Problem 24"
Line 17: Line 17:
  
 
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>.
 
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>.
 +
 +
==Solution 3==
 +
Note that <math>A_{1}</math> is of the form <math>(2x,0)</math> for some <math>x</math>, and thus <math>B_{1}</math> is of the form <math>(x, x \sqrt{3}).</math> Then, we are told that <math>B_{1}</math> lies on the graph of <math>y = \sqrt{x}</math>, so <cmath>(x \sqrt{3})^{2} = x.</cmath> Solving for x, we get that <math>x = \frac{1}{3},</math> and so <math>A_{1} = (2/3,0)</math>.
 +
 +
Now, similarly to before, let <math>|A_{1}A_{2}|=2y.</math> Then, <math>B_{2} = (y+\frac{2}{3},y \sqrt{3})</math>, and so
 +
<cmath>(y \sqrt{3})^{2} = (y+\frac{2}{3}).</cmath>
 +
Solving using the quadratic formula gives
 +
<cmath>y = \frac{-(-1) \pm \sqrt{1^{2}-4(3)(-\frac{2}{3})}}{6} = \frac{1 \pm \sqrt{1^{2}+8}}{6} = \frac{2}{3}.</cmath>
 +
Then, <math>2y = \frac{4}{3},</math> so <math>A_{2} = (2,0).</math>
 +
In general, if <math>A_{n} = (a_{n},0)</math>
  
 
==See Also==
 
==See Also==

Revision as of 13:19, 2 August 2023

Problem

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 distance between the points where these altitudes meet the x-axis is \[\frac{h_{n+1}}{\sqrt{3}}+\frac{h_n}{\sqrt{3}} = \frac{h_{n+1}+h_n}{\sqrt{3}}\]

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 $n=\boxed{17}$.

Solution 3

Note that $A_{1}$ is of the form $(2x,0)$ for some $x$, and thus $B_{1}$ is of the form $(x, x \sqrt{3}).$ Then, we are told that $B_{1}$ lies on the graph of $y = \sqrt{x}$, so \[(x \sqrt{3})^{2} = x.\] Solving for x, we get that $x = \frac{1}{3},$ and so $A_{1} = (2/3,0)$.

Now, similarly to before, let $|A_{1}A_{2}|=2y.$ Then, $B_{2} = (y+\frac{2}{3},y \sqrt{3})$, and so \[(y \sqrt{3})^{2} = (y+\frac{2}{3}).\] Solving using the quadratic formula gives \[y = \frac{-(-1) \pm \sqrt{1^{2}-4(3)(-\frac{2}{3})}}{6} = \frac{1 \pm \sqrt{1^{2}+8}}{6} = \frac{2}{3}.\] Then, $2y = \frac{4}{3},$ so $A_{2} = (2,0).$ In general, if $A_{n} = (a_{n},0)$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_24&oldid=196355"