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

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==Problem 9==
 
==Problem 9==
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A paper equilateral triangle <math>ABC</math> has side length 12. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance 9 from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>.
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A paper equilateral triangle <math>ABC</math> has side length <math>12</math>. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance <math>9</math> from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>.
  
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<asy>
 +
import cse5;
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size(12cm);
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pen tpen = defaultpen + 1.337;
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real a = 39/5.0;
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real b = 39/7.0;
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pair B = MP("B", (0,0), dir(200));
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pair A = MP("A", (9,0), dir(-80));
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pair C = MP("C", (12,0), dir(-20));
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pair K = (6,10.392);
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pair M = (a*B+(12-a)*K) / 12;
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pair N = (b*C+(12-b)*K) / 12;
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draw(B--M--N--C--cycle, tpen);
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draw(M--A--N--cycle);
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fill(M--A--N--cycle, mediumgrey);
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pair shift = (-20.13, 0);
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pair B1 = MP("B", B+shift, dir(200));
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pair A1 = MP("A", K+shift, dir(90));
 +
pair C1 = MP("C", C+shift, dir(-20));
 +
draw(A1--B1--C1--cycle, tpen);</asy>
  
 
== Solution 1 ==
 
== Solution 1 ==

Revision as of 12:44, 30 December 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 $m+n+p$.

[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]

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 $APQ$:

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

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

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

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

Solution 2

Proceed with the same labeling as in Solution 1.

$\angle B = \angle C = \angle A = \angle PDQ = 60^\circ$

$\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CQD + \angle C = 180^\circ$

Therefore, $\angle PDB = \angle CQD$.

Similarly, $\angle BPD = \angle QDC$.

Now, $\bigtriangleup BPD$ and $\bigtriangleup CDQ$ are similar triangles, so

$\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}$.

Solving this system of equations yields $a = \frac{39}{5}$ and $b = \frac{39}{7}$.

Using the Law of Cosines on $APQ$:

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

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

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

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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