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Difference between revisions of "2002 AMC 12B Problems/Problem 19"

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\qquad\mathrm{(E)}\ 750</math>
 
\qquad\mathrm{(E)}\ 750</math>
 
== Solution ==
 
== Solution ==
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Adding up the three equations gives <math>2(ab + bc + ca) = 152 + 162 + 170 = 484 \Longrightarrow ab + bc + ca = 242</math>. Subtracting each of the above equations from this yields, respectively, <math>bc = 90, ca = 80, ab = 72</math>. Taking their product, <math>ab \cdot bc \cdot ca = a^2b^2c^2 = 90 \cdot 80 \cdot 72 = 720^2 \Longrightarrow abc = \boxed{720} \Rightarrow \mathrm{(D)}</math>.  
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Adding up the three equations gives <math>2(ab + bc + ca) = 152 + 162 + 170 = 484 \Longrightarrow ab + bc + ca = 242</math>. Subtracting each of the above equations from this yields, respectively, <math>bc = 90, ca = 80, ab = 72</math>. Taking their product, <math>ab \cdot bc \cdot ca = a^2b^2c^2 = 90 \cdot 80 \cdot 72 = 720^2 \Longrightarrow abc = \boxed{720} \Rightarrow \mathrm{(D)}</math>.
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== Solution 2 (Answer Choices) ==
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Expand the system and subtract equation 3 from equation 2 to get <math>ab-ac=-8</math>, and add it to the first equation to get <math>2ab=144</math> <math>\Longrightarrow</math> <math>ab=72.</math> Since we want <math>abc,</math> it's trivial that 72 is a factor of this. We can quickly see that none of the answer choices have the factor <math>72</math> except choice <math>\boxed{\mathrm{(D)}}.</math>
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~dbnl
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 00:08, 5 February 2025

Problem

If $a,b,$ and $c$ are positive real numbers such that $a(b+c) = 152, b(c+a) = 162,$ and $c(a+b) = 170$, then $abc$ is

$\mathrm{(A)}\ 672 \qquad\mathrm{(B)}\ 688 \qquad\mathrm{(C)}\ 704 \qquad\mathrm{(D)}\ 720 \qquad\mathrm{(E)}\ 750$

Solution

Adding up the three equations gives $2(ab + bc + ca) = 152 + 162 + 170 = 484 \Longrightarrow ab + bc + ca = 242$. Subtracting each of the above equations from this yields, respectively, $bc = 90, ca = 80, ab = 72$. Taking their product, $ab \cdot bc \cdot ca = a^2b^2c^2 = 90 \cdot 80 \cdot 72 = 720^2 \Longrightarrow abc = \boxed{720} \Rightarrow \mathrm{(D)}$.

Solution 2 (Answer Choices)

Expand the system and subtract equation 3 from equation 2 to get $ab-ac=-8$, and add it to the first equation to get $2ab=144$ $\Longrightarrow$ $ab=72.$ Since we want $abc,$ it's trivial that 72 is a factor of this. We can quickly see that none of the answer choices have the factor $72$ except choice $\boxed{\mathrm{(D)}}.$ ~dbnl

See also

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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