Difference between revisions of "2013 AIME I Problems/Problem 9"
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− | == Solution == | + | == Solution 1 == |
Let <math>P</math> and <math>Q</math> be the points on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, where the paper is folded. | Let <math>P</math> and <math>Q</math> be the points on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, where the paper is folded. | ||
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The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Proceed with the same labeling as in Solution 1. | ||
+ | <math>\angle B = \angle C = \angle A = \angle PDQ = 60^\circ</math> | ||
+ | <math>\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CDQ + \angle C = 180^\circ</math> | ||
+ | Therefore, <math>\angle PDB = \angle DQC</math>. | ||
+ | Similarly, <math>\angle BPD = \angle QDC</math>. | ||
+ | Now, <math>\bigtriangleup BPD</math> and <math>\bigtriangleup CDQ</math> are similar triangles. | ||
+ | <math>\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}</math> | ||
+ | Solving this system of equations yields <math>a = \frac{39}{5}</math> and <math>b = \frac{39}{7}</math>. | ||
+ | Using the Law of Cosines on <math>APQ</math>: | ||
+ | |||
+ | <math>x^{2} = a^{2} + b^{2} - 2ab \cos{60}</math> | ||
+ | |||
+ | <math>x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{37}{5})</math> | ||
+ | |||
+ | <math>x^{2} = \frac{39 \sqrt{39}}{35}</math> | ||
+ | |||
+ | The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | ||
+ | |||
== 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}} |
Revision as of 14:21, 29 March 2013
Contents
Problem 9
A paper equilateral triangle has side length 12. The paper triangle is folded so that vertex
touches a point on side
a distance 9 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.
Solving this system of equations yields
and
.
Using the Law of Cosines on
:
The solution is .
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 |