AMC12 2002 B

AMC12 2002 B · Q19

AMC12 2002 B · Q19. It mainly tests Systems of equations, Manipulating equations.

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

若$a,b,$和$c$为正实数，且满足$a(b+c)=152,\ b(c+a)=162,$以及$c(a+b)=170$，则$abc$等于

(A) 672 672

(B) 688 688

(D) 720 720

(E) 750 750

Answer

Correct choice: (D)

正确答案：（D）

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

将三个方程相加得$2(ab+bc+ca)=152+162+170=484 \Longrightarrow ab+bc+ca=242$。分别用此式减去上述三个方程，依次得到$bc=90,\ ca=80,\ ab=72$。将它们相乘，$ab\cdot bc\cdot ca=a^2b^2c^2=90\cdot 80\cdot 72=720^2 \Longrightarrow abc=\boxed{720} \Rightarrow \mathrm{(D)}$。

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