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

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== Solution ==
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== 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>.
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 +
== Solution 2 ==
 +
Proceed with the same labeling as in Solution 1.
 +
<math>\angle B = \angle C = \angle A = \angle PDQ = 60^\circ</math>
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<math>\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CDQ + \angle C = 180^\circ</math>
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Therefore, <math>\angle PDB = \angle DQC</math>.
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Similarly, <math>\angle BPD = \angle QDC</math>.
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Now, <math>\bigtriangleup BPD</math> and <math>\bigtriangleup CDQ</math> are similar triangles.
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<math>\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}</math>
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Solving this system of equations yields <math>a = \frac{39}{5}</math> and <math>b = \frac{39}{7}</math>.
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Using the Law of Cosines on <math>APQ</math>:
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 +
<math>x^{2} = a^{2} + b^{2} - 2ab \cos{60}</math>
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 +
<math>x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{37}{5})</math>
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<math>x^{2} = \frac{39 \sqrt{39}}{35}</math>
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 +
The solution is <math>39 + 39 + 35 = \boxed{113}</math>.
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== 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 13:21, 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 1

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 $39 + 39 + 35 = \boxed{113}$.

Solution 2

Proceed with the same labeling as in Solution 1. $\angle B = \angle C = \angle A = \angle PDQ = 60^\circ$ $\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CDQ + \angle C = 180^\circ$ Therefore, $\angle PDB = \angle DQC$. Similarly, $\angle BPD = \angle QDC$. Now, $\bigtriangleup BPD$ and $\bigtriangleup CDQ$ are similar triangles. $\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}$ Solving this system of equations yields $a = \frac{39}{5}$ and $b = \frac{39}{7}$. 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
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