Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AIME I Problems/Problem 9"

(Changed the solution to an actual solution)
(→‎Solution)
Line 16: Line 16:
 
<math>a^{2} = (12 - a)^{2} + 9^{2} - 2 \times (12 - a) \times 9 \times \cos{60}</math>
 
<math>a^{2} = (12 - a)^{2} + 9^{2} - 2 \times (12 - a) \times 9 \times \cos{60}</math>
  
−
<math>a^{2} = 144 - 12a + a^{2} + 81 - 108 + 9a</math>
+
<math>a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a</math>
  
 
<math>a = \frac{39}{5}</math>
 
<math>a = \frac{39}{5}</math>
Line 24: Line 24:
 
<math>b^{2} = (12 - b)^{2} +3^{2} - 2 \times (12 - b) \times 3 \times \cos{60}</math>
 
<math>b^{2} = (12 - b)^{2} +3^{2} - 2 \times (12 - b) \times 3 \times \cos{60}</math>
  
−
<math>b^{2} = 144 - 12b + b^{2} + 9 - 36 + 3b</math>
+
<math>b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b</math>
  
 
<math>b = \frac{39}{7}</math>
 
<math>b = \frac{39}{7}</math>
Line 37: Line 37:
  
 
The solution is <math>39 + 39 + 35 = \boxed{113}</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 13:04, 29 March 2013

Problem 9

A paper equilateral triangle $ABC$ has side length 12. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance 9 from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.


Solution

Let $P$ and $Q$ be the points on $\overline{AB}$ and $\overline{AC}$, respectively, where the paper is folded.

Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it.

Let $a$, $b$, and $x$ be the lengths $AP$, $AQ$, and $PQ$, respectively.

We have $PD = a$, $QD = b$, $BP = 12 - a$, $CQ = 12 - b$, $BD = 9$, and $CD = 3$.

Using the Law of Cosines on $BPD$:

$a^{2} = (12 - a)^{2} + 9^{2} - 2 \times (12 - a) \times 9 \times \cos{60}$

$a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a$

$a = \frac{39}{5}$

Using the Law of Cosines on $CQD$:

$b^{2} = (12 - b)^{2} +3^{2} - 2 \times (12 - b) \times 3 \times \cos{60}$

$b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b$

$b = \frac{39}{7}$

Using the Law of Cosines on $APQ$:

$x^{2} = a^{2} + b^{2} - 2ab \cos{60}$

$x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{37}{5})$

$x^{2} = \frac{39 \sqrt{39}}{35}$

The solution is $39 + 39 + 35 = \boxed{113}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_9&oldid=51970"