2002 AMC 12B Problem 19

Problem 19 of 25HarderAlgebra

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

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Solution

Adding the three equations gives

2(ab+bc+ca)=484,

ab+bc+ca=242.

Subtracting each original equation from this yields

bc=90,

ca=80,

and

ab=72.

Multiplying,

(abc)^2=90\cdot80\cdot72=720^2,

and since

abc\gt0,

we get

abc=720.

Thus, the correct answer is D.

Concepts: system of equations · symmetry (algebra)

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.