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

Revision as of 13:53, 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 $A_{n} = (a_{n},0)$ for all integers $n$ , and $|A_{n}A_{n+1}| = 2x$ for some real number $x$ , we have the following equation for $x$ : $(x \sqrt{3})^{2} = x+a_{n},$ which give us when plugged into the quadratic formula gives $x = \frac{1 \pm \sqrt{1+12a_{n}}}{6},$ but since $x$ must be positive, we have that $x = \frac{1 + \sqrt{1+12a_{n}}}{6},$ and so $a_{n+1} = a_{n}+2x = a_{n}+\frac{1 + \sqrt{1+12a_{n}}}{3}.$ Computing a few terms of $a_{n}$ using this method gives $a_{3} = 4,$ $a_{4} = \frac{20}{3},$ and $a_{5} = 10$ .

Notice how all our $a_{n}$ terms so far are rational, even though there is a abundance of radicals in the recurrence. This motivates us to look at our discriminants in the quadratic formula that solved for $x$ .

The discriminant of $(x \sqrt{3})^{2} = x+a_{2}$ is $1+12 \cdot \frac{2}{3} = 9.$ Similarly, the discriminant of $(x \sqrt{3})^{2} = x+a_{3}$ is $1+12 \cdot a_{3} = 25,$ and $1+12 \cdot a_{4} = 49$ .

Note how our results keep coming out as the squares of the odd integers. Moreover, it seems that $1+12a_{n} = (2n+1)^{2}.$ We will prove this with induction. The base case, $n=2$ , we have already verified.

Now, for the Inductive step, assume that $1+12a_{n-1} = (2n-1)^{2}.$ for some integer $n-1$ . We will prove that this is true for $n$ as well.

Plugging this into our recurrence formula gives us $a_{n} = a_{n}+2x = a_{n-1}+\frac{1 + \sqrt{1+12a_{n-1}}}{3}$ $a_{n} = \frac{(2n-1)^{2}-1}{12} + \frac{1 + (2n-1)}{3}$ $a_{n} = \frac{4n^{2}-4n+1-1}{12} + \frac{8n}{12}$ $a_{n} = \frac{4n^{2}+4n+1-1}{12}$ $a_{n} = \frac{(2n+1)^{2}-1}{12}.$ Therefore, we have proved our claim. Now, we have that $|A_{0}A_{n}|=a_{n},$ so we just need the least integer $n$ so that $a_{n} > 100,$ or $(2n+1)^{2} > 1201.$ Then, we see that $35^{2} = 1225$ is the smallest odd square larger than $1201.$ Therefore, we have $2n+1=35$ , so $n = \boxed{17}.$

-Mr.Sharkman

@@ Line 59: / Line 59: @@
 Then, we see that <math>35^{2} = 1225</math> is the smallest odd square larger than <math>1201.</math> Therefore, we have <math>2n+1=35</math>, so <math>n = \boxed{17}.</math>
+-Mr.Sharkman
 ==See Also==
 {{AMC12 box|year=2008|ab=B|num-b=23|num-a=25}}

2008 AMC 12B (Problems • Answer Key • Resources)
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