AMC12 2005 A

AMC12 2005 A · Q22

AMC12 2005 A · Q22. It mainly tests Coordinate geometry, 3D geometry (volume).

A rectangular box $P$ is inscribed in a sphere of radius $r$. The surface area of $P$ is 384, and the sum of the lengths of its $12$ edges is $112$. What is $r$?

一个长方体盒子 $P$ 内接于半径为 $r$ 的球中。$P$ 的表面积是 384，其 12 条棱的长度和是 112。$r$ 是多少？

(A) 8 8

(B) 10 10

(D) 14 14

(E) 16 16

Answer

Correct choice: (B)

正确答案：（B）

Solution

Box P has dimensions $l$, $w$, and $h$. Its surface area is \[2lw+2lh+2wh=384,\] and the sum of all its edges is \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\] The diameter of the sphere is the space diagonal of the prism, which is \[\sqrt{l^2 + w^2 +h^2}.\] Notice that \begin{align*}(l + w + h)^2 - (2lw + 2lh + 2wh) &= l^2 + w^2 + h^2 \\ &= 784 - 384 \\ &= 400,\end{align*} so the diameter is \[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.\] The radius is half of the diameter, so \[r=\frac{20}{2} = \boxed{\textbf{(B) } 10}.\]

盒子 $P$ 的长宽高分别为 $l$, $w$, $h$。其表面积为 \[2lw+2lh+2wh=384,\] 其所有棱长之和为 \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\] 球的直径等于长方体的体对角线长度，即 \[\sqrt{l^2 + w^2 +h^2}.\] 注意到 \begin{align*}(l + w + h)^2 - (2lw + 2lh + 2wh) &= l^2 + w^2 + h^2 \\ &= 784 - 384 \\ &= 400,\end{align*} 所以直径为 \[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.\] 半径是直径的一半，因此 \[r=\frac{20}{2} = \boxed{\textbf{(B) } 10}.\]

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