Question

两个大小不同的球分别放在一间长方体房间的两个角落，每个球都与两面墙和地板相切。若每个球面上都有一点，它到该球所接触的两面墙的距离均为 $5$ 英寸，到地板的距离为 $10$ 英寸，则两球直径之和为

There are two spherical balls of different sizes lying in two corners of a rectangular room, each touching two walls and the floor. If there is a point on each ball which is $5$ inches from each wall which that ball touches and $10$ inches from the floor, then the sum of the diameters of the balls is

$20$ 英寸

$20$ inches

$30$ 英寸

$30$ inches

$40$ 英寸

$40$ inches

$60$ 英寸

$60$ inches

无法由所给信息确定

not determined by the given information

答案：C

知识点：球坐标几何二次方程

难度评级：2040

小提示：

将房间角点设为原点，把两面墙和地板视为三个坐标平面

Place the corner at the origin with the walls and floor as coordinate planes

大提示：

与三个坐标平面相切的半径为 $r$ 的球，其球心为 $(r,r,r)$

A sphere of radius $r$ tangent to all three planes has center $(r,r,r)$

解答：

对于半径 $r$ ，球心为 $(r,r,r)$ ，所给点为 $(5,5,10)$ 。因此 $2(5-r)^2+(10-r)^2=r^2\text{，}$ 化简得 $r^2-20r+75=0$ ，即 $(r-5)(r-15)=0$ 。两个半径分别为 $5$ 和 $15$ ，所以直径之和为 $2(5+15)=40$ 英寸。

因此，正确答案是 C。

For radius $r,$ the center is $(r,r,r)$ and the given point is $(5,5,10).$ Thus $2(5-r)^2+(10-r)^2=r^2,$ which simplifies to $r^2-20r+75=0,$ or $(r-5)(r-15)=0.$ The two radii are $5$ and $15,$ so the sum of the diameters is $2(5+15)=40$ inches.

Therefore, the correct answer is C.

1977 AMC 12 第 27 题

答案：C

难度评级：2040

小提示：

大提示：

解答：

其他年份的第 27 题