Difference between revisions of "2008 AMC 12B Problems/Problem 24"
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Revision as of 09:55, 4 July 2013
Problem 24
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
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.
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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