2008 AMC 12B Problems/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 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 wider by $2/3$ units 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 an abundance of radicals in the recurrence. This motivates us to look at our discriminants in the quadratic formula that is 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 $n = \boxed{17}.$


Solution 4 (Pattern Observation)

We can iteratively calculate out the first few $A_i$ and $B_i$. We know that $A_0 = (0,0)$ and the line through $A_0$ and $B_1$ needs to make a $60^{\circ}$ angle with the x-axis (because the triangle is equilateral). The equation of a line that makes an angle $\theta$ with the x-axis and passes through the origin has equation $\tan(\theta)x$, so the line through $A_0$ and $B_1$ has equation $\sqrt{3}x$. We set this equal to $\sqrt{x}$ (to find where the two curves intersect) and, when solving, find that $x = \frac{1}{3}$. Therefore, $B_1 = (\frac{1}{3}, \frac{\sqrt{3}}{3})$. By equilateral triangle properties, we must then have that $A_1 = (\frac{2}{3}, 0)$. To find $B_2$, we find the equation of the line through $A_1$ that makes a $60^{\circ}$ angle with the $x$-axis. This is the same line as the one through the origin (which we already found) shifted $\frac{2}{3}$ to the right, so it has equation $y = \sqrt{3}(x - \frac{2}{3})$. Setting this equal to $\sqrt{x}$ and solving for $x$, we get $x = \frac{4}{3}$, so $B_2 = (\frac{4}{3}, \frac{2\sqrt{3}}{3})$. By equilateral triangle properties, we have that $A_2 = (2, 0)$. Repeating this process, we find that $B_3 = (3, \sqrt{3})$ and $A_3 = (4,0)$. At this point, we notice that the $y$-coordinate of each $B_i$ is $\frac{\sqrt{3}}{3}$ more than that of $B_{i-1}$, so $B_4 = (\frac{16}{3}, \frac{4\sqrt{3}}{3})$ and $A_4 = (\frac{20}{3}, 0)$. It may be helpful to continue calculating out the $A_i$ and $B_i$ if the pattern in the $x$-coordinates of the $A_i$ isn't visible yet. At this point, it can be noticed that the differences between the x coordinates of consecutive $A_i$ form an arithmetic sequence ($x_{A_1} - x_{A_0} = \frac{2}{3}$, $x_{A_2} - x_{A_1} = \frac{4}{3}$, etc.). Additionally, the numerator of each of the fractions is a consecutive even number, and the sum of the first $n$ even numbers is $n(n+1)$. Thus, we must find the least integral solution to $\frac{n(n+1)}{3} \geq 100$. This least solution is $\boxed{\textbf{(C)}~17}$.

~ cxsmi

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
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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