# 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, $ab \cdot bc \cdot ca = a^2b^2c^2 = 90 \cdot 80 \cdot 72 = 720^2 \Longrightarrow abc = \boxed{720} \Rightarrow \mathrm{(D)}$.

## See also

 2002 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 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

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