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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 14"
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== Problem ==
 
== Problem ==
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In [[triangle]] <math>ABC</math>, <math>\displaystyle AB = 308</math> and <math>\displaystyle AC=35</math>.  Given that <math>\displaystyle AD</math>, <math>\displaystyle BE,</math> and <math>\displaystyle CF,</math> [[intersect]] at <math>\displaystyle P</math> and are an [[angle bisector]], [[median of a triangle | median]], and [[altitude]] of the triangle, respectively, compute the [[length]] of <math>\displaystyle BC.</math>
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In [[triangle]] <math>ABC</math>, <math>AB = 308</math> and <math>AC=35</math>.  Given that <math>AD</math>, <math>BE,</math> and <math>CF,</math> [[intersect]] at <math>P</math> and are an [[angle bisector]], [[median of a triangle | median]], and [[altitude]] of the triangle, respectively, compute the [[length]] of <math>BC.</math>
  
 
[[Image:Mock AIME 2 2007 Problem14.jpg]]
 
[[Image:Mock AIME 2 2007 Problem14.jpg]]
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which yields finally <math>x = 277</math>.
 
which yields finally <math>x = 277</math>.
  
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==See Also==
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{{Mock AIME box|year=2006-2007|n=2|num-b=13|num-a=15}}
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*[[Mock AIME 2 2006-2007/Problem 13 | Previous Problem]]
 
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*[[Mock AIME 2 2006-2007/Problem 15 | Next Problem]]
 
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*[[Mock AIME 2 2006-2007]]
 
  
  
 
== Problem Source ==
 
== Problem Source ==
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4everwise thought of this problem after reading the first chapter of Geometry Revisited.
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4everwise thought of this problem after reading the first chapter of [http://www.amazon.com/exec/obidos/ASIN/0883856190/artofproblems-20 Geometry Revisited].
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 09:53, 4 April 2012

Problem

In triangle $ABC$, $AB = 308$ and $AC=35$. Given that $AD$, $BE,$ and $CF,$ intersect at $P$ and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of $BC.$

Mock AIME 2 2007 Problem14.jpg

Solution

Let $BC = x$.

By the Angle Bisector Theorem, $\frac{CD}{BD} = \frac{AC}{AB} = \frac{35}{308} = \frac{5}{44}$.

Let $CF = h$. Then by the Pythagorean Theorem, $h^2 + AF^2 = 35^2$ and $h^2 + BF^2 = x^2$. Subtracting the former equation from the latter to eliminate $h$, we have $BF^2 - AF^2 = x^2 - 35^2$ so $(BF + AF)(BF - AF) = x^2 - 1225$. Since $BF + AF = AB = 308$, $BF - AF = \frac{x^2 - 1225}{308}$. We can solve these equations for $BF$ and $AF$ in terms of $x$ to find that $BF = 154 + \frac{x^2 - 1225}{616} =$ and $AF = 154 - \frac{x^2 - 1225}{616}$.

Now, by Ceva's Theorem, $\frac{AE}{EC} \cdot \frac{CD}{DB} \cdot \frac{BF}{FA} = 1$, so $1 \cdot \frac{5}{44} \cdot \frac{BF}{AF} = 1$ and $5BF = 44AF$. Plugging in the values we previously found,

$5\cdot 154 + \frac{5(x^2 - 1225)}{616} = 44\cdot 154 - \frac{44(x^2 - 1225)}{616}$

so

$\frac{49}{616}(x^2 - 1225) = 39\cdot 154$

and

$x^2 - 1225 = 75504$

which yields finally $x = 277$.

See Also

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15


Problem Source

4everwise thought of this problem after reading the first chapter of Geometry Revisited.

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