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Difference between revisions of "1988 AIME Problems/Problem 12"

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== Problem ==
 
== Problem ==
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Let <math>P</math> be an interior point of triangle <math>ABC</math> and extend lines from the vertices through <math>P</math> to the opposite sides.  Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> denote the lengths of the segments indicated in the figure.  Find the product <math>abc</math> if <math>a + b + c = 43</math> and <math>d = 3</math>.
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 +
{{image}}
  
 
== Solution ==
 
== Solution ==
 +
Call the [[cevian]]s AD, BE, and CF. Using area ratios (<math>\triangle PBC</math> and <math>\triangle ABC</math> have the same base), we have:
 +
 +
<math>\frac {d}{a + d} = \frac {[PBC]}{[ABC]}</math>
 +
 +
Similarily, <math>\frac {d}{b + d} = \frac {[PCA]}{[ABC]}</math> and <math>\frac {d}{c + d} = \frac {[PAB]}{[ABC]}</math>.
 +
 +
Then,
 +
<math>\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = \frac {[PBC]}{[ABC]} + \frac {[PCA]}{[ABC]} + \frac {[PAB]}{[ABC]} = \frac {[ABC]}{[ABC]} = 1</math>
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 +
The [[identity]] <math>\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = 1</math> is a form of [[Ceva's Theorem]].
 +
 +
Plugging in <math>d = 3</math>, we get
 +
 +
<cmath>\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1</cmath>
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<cmath>3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3)</cmath>
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<cmath>3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27</cmath>
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<cmath>9(a + b + c) + 54 = abc</cmath>
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<cmath>abc = 441</cmath>
  
 
== See also ==
 
== See also ==
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* [[1988 AIME Problems]]
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{{AIME box|year=1988|num-b=11|num-a=13}}
  
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{{AIME box|year=1988|num-b=11|num-a=13}}
+
[[Category:Intermediate Geometry Problems]]

Revision as of 17:36, 28 September 2007

Problem

Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$, $b$, $c$, and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Solution

Call the cevians AD, BE, and CF. Using area ratios ($\triangle PBC$ and $\triangle ABC$ have the same base), we have:

$\frac {d}{a + d} = \frac {[PBC]}{[ABC]}$

Similarily, $\frac {d}{b + d} = \frac {[PCA]}{[ABC]}$ and $\frac {d}{c + d} = \frac {[PAB]}{[ABC]}$.

Then, $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = \frac {[PBC]}{[ABC]} + \frac {[PCA]}{[ABC]} + \frac {[PAB]}{[ABC]} = \frac {[ABC]}{[ABC]} = 1$

The identity $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = 1$ is a form of Ceva's Theorem.

Plugging in $d = 3$, we get

\[\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1\] \[3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3)\] \[3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27\] \[9(a + b + c) + 54 = abc\] \[abc = 441\]

See also

1988 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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