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

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[[Image:1988_AIME-12.png]]
 
[[Image:1988_AIME-12.png]]
  
== Solution ==
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== Solution 1 ==
 
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:  
 
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:  
  
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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>
 
<cmath>3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27</cmath>
 
<cmath>9(a + b + c) + 54 = abc=\boxed{441}</cmath>
 
<cmath>9(a + b + c) + 54 = abc=\boxed{441}</cmath>
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==Solution 2==
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Let the labels <math>A,B,C</math> be the weights of the vertices. First off, replace <math>d</math> with <math>3</math>. We see that the weights of the feet of the cevians are <math>A+B,B+C,C+A</math>. By mass points, we have that: <cmath>\dfrac{a}{3}=\dfrac{B+C}{A}</cmath> <cmath>\dfrac{b}{3}=\dfrac{C+A}{B}</cmath> <cmath>\dfrac{c}{3}=\dfrac{A+B}{C}</cmath>
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If we add the equations together, we get <cmath>\dfrac{a+b+c}{3}=\dfrac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B}{ABC}=\dfrac{43}{3}</cmath>
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If we multiply them together, we get <cmath>\dfrac{abc}{27}=\dfrac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B+2ABC}{ABC}=\dfrac{43}{3}+2=\dfrac{49}{3}</cmath>
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Multiplying both sides by <math>27</math>, we get that <math>abc=27\cdot \dfrac{49}{3}=\boxed{441}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 20:54, 13 September 2014

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 $d = 3$.

1988 AIME-12.png

Solution 1

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

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

Solution 2

Let the labels $A,B,C$ be the weights of the vertices. First off, replace $d$ with $3$. We see that the weights of the feet of the cevians are $A+B,B+C,C+A$. By mass points, we have that: \[\dfrac{a}{3}=\dfrac{B+C}{A}\] \[\dfrac{b}{3}=\dfrac{C+A}{B}\] \[\dfrac{c}{3}=\dfrac{A+B}{C}\]

If we add the equations together, we get \[\dfrac{a+b+c}{3}=\dfrac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B}{ABC}=\dfrac{43}{3}\]

If we multiply them together, we get \[\dfrac{abc}{27}=\dfrac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B+2ABC}{ABC}=\dfrac{43}{3}+2=\dfrac{49}{3}\]

Multiplying both sides by $27$, we get that $abc=27\cdot \dfrac{49}{3}=\boxed{441}$.

See also

1988 AIME (ProblemsAnswer KeyResources)
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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