Difference between revisions of "1992 AIME Problems/Problem 14"

Revision as of 18:24, 4 July 2013

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

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

Revision as of 18:13, 12 June 2008 (view source) Xantos C. Guin (talk \| contribs) ← Older edit		Revision as of 18:24, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	[[Category:Intermediate Geometry Problems]]		[[Category:Intermediate Geometry Problems]]
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1992 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1992 AIME Problems/Problem 14"

Revision as of 18:24, 4 July 2013

Problem

Solution

See also