AMC12 2014 B

AMC12 2014 B · Q14

AMC12 2014 B · Q14. It mainly tests Linear equations, 3D geometry (volume).

A rectangular box has a total surface area of 94 square inches. The sum of the lengths of all its edges is 48 inches. What is the sum of the lengths in inches of all of its interior diagonals?

一个长方体盒子总表面积为94平方英寸。所有棱长的和为48英寸。其所有体内对角线长度的和是多少英寸？

(A) $8\sqrt{3}$ $8\sqrt{3}$

(B) $10\sqrt{2}$ $10\sqrt{2}$

(D) $20\sqrt{2}$ $20\sqrt{2}$

(E) $40\sqrt{2}$ $40\sqrt{2}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Denote the edge lengths by $x$, $y$, and $z$. The surface area is $2(xy+yz+zx)=94$ and the sum of the lengths of the edges is $4(x+y+z)=48$. Therefore $144=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=x^2+y^2+z^2+94$, so $x^2+y^2+z^2=50$. By the Pythagorean Theorem applied twice, each of the 4 internal diagonals has length $\sqrt{50}$, and their total length is $4\sqrt{50}=20\sqrt{2}$. A right rectangular prism with edge lengths 3, 4, and 5 satisfies the conditions of the problem.

答案（D）：设棱长为 $x$、$y$、$z$。表面积为 $2(xy+yz+zx)=94$，所有棱长之和为 $4(x+y+z)=48$。因此 $144=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=x^2+y^2+z^2+94$，所以 $x^2+y^2+z^2=50$。将勾股定理应用两次，4 条内部对角线每条长度为 $\sqrt{50}$，它们的总长度为 $4\sqrt{50}=20\sqrt{2}$。棱长为 3、4、5 的长方体满足题目条件。

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