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 and wider 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
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 |
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