AMC10 2018 B

AMC10 2018 B · Q4

AMC10 2018 B · Q4. It mainly tests Systems of equations, 3D geometry (surface area).

A three-dimensional rectangular box with dimensions $X$, $Y$, and $Z$ has faces whose surface areas are 24, 24, 48, 48, 72, and 72 square units. What is $X + Y + Z$?

一个三维长方体盒子尺寸为$X$、$Y$和$Z$，各个面的表面积为24、24、48、48、72和72平方单位。求$X + Y + Z$？

(A) 18 18

(B) 22 22

(D) 30 30

(E) 36 36

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Without loss of generality, assume that $X \le Y \le Z$. Then the geometric description of the problem can be translated into the system of equations, $XY=24$, $XZ=48$, and $YZ=72$. Dividing the second equation by the first yields $\frac{Z}{Y}=2$, so $Z=2Y$. Then $72=YZ=2Y^2$, so $Y^2=36$. Because $Y$ is positive, $Y=6$. It follows that $X=24 \div 6=4$ and $Z=72 \div 6=12$, so $X+Y+Z=22$.

答案（B）：不失一般性，设 $X \le Y \le Z$。那么该问题的几何描述可转化为方程组：$XY=24$，$XZ=48$，以及 $YZ=72$。用第二个方程除以第一个方程得 $\frac{Z}{Y}=2$，所以 $Z=2Y$。于是 $72=YZ=2Y^2$，从而 $Y^2=36$。因为 $Y$ 为正，$Y=6$。因此 $X=24 \div 6=4$，且 $Z=72 \div 6=12$，所以 $X+Y+Z=22$。

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