2002 AMC 12B Problems/Problem 19

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

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

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2002 AMC 12B Problems/Problem 19

Contents

Problem

Solution

Solution 2 (Answer Choices)

See also