2008 AMC 12B Problems/Problem 24
Problem
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 distance between the points where these altitudes meet the x-axis 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 by units each time. From here, one can proceed as in Solution 1 to arrive at .
Solution 3
Note that is of the form for some , and thus is of the form Then, we are told that lies on the graph of , so Solving for x, we get that and so .
Now, similarly to before, let Then, , and so Solving using the quadratic formula gives Then, so
In general, if for all integers , and for some real number , we have the following equation for : which give us when plugged into the quadratic formula gives but since must be positive, we have that and so Computing a few terms of using this method gives and .
Notice how all our 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 .
The discriminant of is Similarly, the discriminant of is and .
Note how our results keep coming out as the squares of the odd integers. Moreover, it seems that We will prove this with induction. The base case, , we have already verified.
Now, for the Inductive step, assume that for some integer . We will prove that this is true for as well.
Plugging this into our recurrence formula gives us Therefore, we have proved our claim. Now, we have that so we just need the least integer so that or Then, we see that is the smallest odd square larger than Therefore, we have , so
-Mr.Sharkman
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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