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1992 AIME Problems/Problem 14: Difference between revisions

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== Problem ==
== Problem ==
In triangle <math>ABC^{}_{}</math>, <math>\displaystyle A'</math>, <math>\displaystyle B'</math>, and <math>\displaystyle C'</math> are on the sides <math>\displaystyle BC</math>, <math>AC^{}_{}</math>, and <math>AB^{}_{}</math>, respectively. Given that <math>\displaystyle AA'</math>, <math>\displaystyle BB'</math>, and <math>\displaystyle CC'</math> are concurrent at the point <math>O^{}_{}</math>, and that <math>\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92</math>, find <math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}</math>.
In triangle <math>ABC^{}_{}</math>, <math>A'</math>, <math>B'</math>, and <math>C'</math> are on the sides <math>BC</math>, <math>AC^{}_{}</math>, and <math>AB^{}_{}</math>, respectively. Given that <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> are concurrent at the point <math>O^{}_{}</math>, and that <math>\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92</math>, find <math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}</math>.


== Solution ==
== Solution ==
{{solution}}
Using [[mass points]], let the weights of <math>A</math>, <math>B</math>, and <math>C</math> be <math>a</math>, <math>b</math>, and <math>c</math> respectively.
 
Then, the weights of <math>A'</math>, <math>B'</math>, and <math>C'</math> are <math>b+c</math>, <math>c+a</math>, and <math>a+b</math> respectively.
 
Thus, <math>\frac{AO^{}_{}}{OA'} = \frac{b+c}{a}</math>, <math>\frac{BO^{}_{}}{OB'} = \frac{c+a}{b}</math>, and <math>\frac{CO^{}_{}}{OC'} = \frac{a+b}{c}</math>.
 
Therefore:
<math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'} = \frac{b+c}{a} \cdot \frac{c+a}{b} \cdot \frac{a+b}{c}</math> <math>= \frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =</math>
 
<math>2+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}+\frac{ab(a+b)}{abc} = 2 + \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} </math> <math>= 2 + \frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'} = 2+92 = \boxed{094}</math>.


== See also ==
== See also ==

Revision as of 18:13, 12 June 2008

Problem

In triangle $ABC^{}_{}$, $A'$, $B'$, and $C'$ are on the sides $BC$, $AC^{}_{}$, and $AB^{}_{}$, respectively. Given that $AA'$, $BB'$, and $CC'$ are concurrent at the point $O^{}_{}$, and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$, find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$.

Solution

Using mass points, let the weights of $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively.

Then, the weights of $A'$, $B'$, and $C'$ are $b+c$, $c+a$, and $a+b$ respectively.

Thus, $\frac{AO^{}_{}}{OA'} = \frac{b+c}{a}$, $\frac{BO^{}_{}}{OB'} = \frac{c+a}{b}$, and $\frac{CO^{}_{}}{OC'} = \frac{a+b}{c}$.

Therefore: $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'} = \frac{b+c}{a} \cdot \frac{c+a}{b} \cdot \frac{a+b}{c}$ $= \frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =$

$2+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}+\frac{ab(a+b)}{abc} = 2 + \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}$ $= 2 + \frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'} = 2+92 = \boxed{094}$.

See also

1992 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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