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Difference between revisions of "1998 AIME Problems/Problem 12"

 
(problem/solution, by minsoens)
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== Problem ==
 
== Problem ==
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Let <math>ABC</math> be [[equilateral triangle|equilateral]], and <math>D, E,</math> and <math>F</math> be the [[midpoint]]s of <math>\overline{BC}, \overline{CA},</math> and <math>\overline{AB},</math> respectively.  There exist [[point]]s <math>P, Q,</math> and <math>R</math> on <math>\displaystyle \overline{DE}, \overline{EF},</math> and <math>\overline{FD}, \displaystyle</math> respectively, with the property that <math>P</math> is on <math>\overline{CQ}, Q</math> is on <math>\overline{AR}, \displaystyle</math> and <math>R</math> is on <math>\overline{BP}.</math>  The [[ratio]] of the area of triangle <math>ABC</math> to the area of triangle <math>PQR</math> is <math>a + b\sqrt {c}, \displaystyle</math> where <math>a, b</math> and <math>c</math> are integers, and <math>c</math> is not divisible by the square of any [[prime]].  What is <math>a^{2} + b^{2} + c^{2}</math>?
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 +
[[Image:1998_AIME-12.png]]
  
 
== Solution ==
 
== Solution ==
 +
Assign variables:
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<math>EP = FQ = x</math>
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<math>EQ = y</math>
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<math>PQ = k</math>
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Since <math>\displaystyle AE = \frac {1}{2}AB</math> and <math>AD = \frac {1}{2}AC</math>, <math>\triangle AED \sim \triangle ABC</math> and <math>ED \parallel BC</math>
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 +
[[Alternate Interior Angles]] Theorem says <math>\displaystyle \angle PEQ = \angle BFQ</math> and <math>\angle EPQ = \angle FBQ \displaystyle</math>
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 +
[[Vertical Angles]] Theorem says <math>\displaystyle \angle EQP = \angle FQB \displaystyle</math>
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So <math>\triangle EQP \sim \triangle FQB \displaystyle</math> and by [[CPCTC]] <math>\frac {EP}{EQ} = \frac {FB}{FQ}\Longrightarrow\frac {x}{y} = \frac {1}{x}\Longrightarrow x^{2} = y \displaystyle</math>
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Since <math>\displaystyle \triangle EDF</math> is equilateral, <math>EQ + FQ = EF = BF = 1\Longrightarrow x + y = 1</math>. Solving for <math>x</math> and <math>y</math> using <math>\displaystyle x^{2} = y</math> and <math>x + y = 1</math> gives <math>\displaystyle x = \frac {\sqrt {5} - 1}{2}</math> and <math>\displaystyle y = \frac {3 - \sqrt {5}}{2} \displaystyle</math>
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 +
Using the [[Law of Cosines]], we get
 +
<div style="text-align:center;"><math>k^{2}  =  x^{2} + y^{2} - 2xy\cos{\frac {\pi}{3}}</math>
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:<math> =  \left(\frac {\sqrt {5} - 1}{2}\right)^{2} + \left(\frac {3 - \sqrt {5}}{2}\right)^{2} - 2\left(\frac {\sqrt {5} - 1}{2}\right)\left(\frac {3 - \sqrt {5}}{2}\right)\cos{\frac {\pi}{3}}</math>
 +
:<math> = 7 - 3\sqrt {5}</math></div>
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We want the ratio of the squares of the sides, so <math>\displaystyle \frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}</math> so <math>a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = 083</math>
  
 
== See also ==
 
== See also ==
−
* [[1998 AIME Problems]]
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{{AIME box|year=1998|num-b=11|num-a=13}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 15:15, 8 September 2007

Problem

Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\displaystyle \overline{DE}, \overline{EF},$ and $\overline{FD}, \displaystyle$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR}, \displaystyle$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a + b\sqrt {c}, \displaystyle$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime. What is $a^{2} + b^{2} + c^{2}$?

1998 AIME-12.png

Solution

Assign variables: $EP = FQ = x$ $EQ = y$ $PQ = k$ Since $\displaystyle AE = \frac {1}{2}AB$ and $AD = \frac {1}{2}AC$, $\triangle AED \sim \triangle ABC$ and $ED \parallel BC$

Alternate Interior Angles Theorem says $\displaystyle \angle PEQ = \angle BFQ$ and $\angle EPQ = \angle FBQ \displaystyle$

Vertical Angles Theorem says $\displaystyle \angle EQP = \angle FQB \displaystyle$ So $\triangle EQP \sim \triangle FQB \displaystyle$ and by CPCTC $\frac {EP}{EQ} = \frac {FB}{FQ}\Longrightarrow\frac {x}{y} = \frac {1}{x}\Longrightarrow x^{2} = y \displaystyle$

Since $\displaystyle \triangle EDF$ is equilateral, $EQ + FQ = EF = BF = 1\Longrightarrow x + y = 1$. Solving for $x$ and $y$ using $\displaystyle x^{2} = y$ and $x + y = 1$ gives $\displaystyle x = \frac {\sqrt {5} - 1}{2}$ and $\displaystyle y = \frac {3 - \sqrt {5}}{2} \displaystyle$

Using the Law of Cosines, we get

$k^{2} = x^{2} + y^{2} - 2xy\cos{\frac {\pi}{3}}$
$= \left(\frac {\sqrt {5} - 1}{2}\right)^{2} + \left(\frac {3 - \sqrt {5}}{2}\right)^{2} - 2\left(\frac {\sqrt {5} - 1}{2}\right)\left(\frac {3 - \sqrt {5}}{2}\right)\cos{\frac {\pi}{3}}$
$= 7 - 3\sqrt {5}$

We want the ratio of the squares of the sides, so $\displaystyle \frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}$ so $a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = 083$

See also

1998 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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