Difference between revisions of "2013 AIME I Problems/Problem 9"
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The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | ||
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+ | == Solution 3 (Coordinate Bash) == | ||
+ | See here: http://artofproblemsolving.com/community/c24001h1060659_2013_aime_i_problem_9 | ||
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=8|num-a=10}} | {{AIME box|year=2013|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:31, 29 March 2015
Problem 9
A paper equilateral triangle has side length
. The paper triangle is folded so that vertex
touches a point on side
a distance
from point
. The length of the line segment along which the triangle is folded can be written as
, where
,
, and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
.
Solution 1
Let and
be the points on
and
, respectively, where the paper is folded.
Let be the point on
where the folded
touches it.
Let ,
, and
be the lengths
,
, and
, respectively.
We have ,
,
,
,
, and
.
Using the Law of Cosines on :
Using the Law of Cosines on :
Using the Law of Cosines on :
The solution is .
Solution 2
Proceed with the same labeling as in Solution 1.
Therefore, .
Similarly, .
Now, and
are similar triangles, so
.
Solving this system of equations yields and
.
Using the Law of Cosines on :
The solution is .
Solution 3 (Coordinate Bash)
See here: http://artofproblemsolving.com/community/c24001h1060659_2013_aime_i_problem_9
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.