1977 AMC 12 第 27 题

先试着解答 1977 AMC 12 第 27 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1977 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

27.

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 55 英寸,到地板的距离为 1010 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 55 inches from each wall which that ball touches and 1010 inches from the floor, then the sum of the diameters of the balls is
2020 英寸
2020 inches
3030 英寸
3030 inches
4040 英寸
4040 inches
6060 英寸
6060 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
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 rr 的球,其球心为 (r,r,r)(r,r,r)
A sphere of radius rr tangent to all three planes has center (r,r,r)(r,r,r)
解答:
对于半径 rr,球心为 (r,r,r)(r,r,r),所给点为 (5,5,10)(5,5,10)。因此 2(5−r)2+(10−r)2=r2, 2(5-r)^2+(10-r)^2=r^2\text{,}化简得 r2−20r+75=0r^2-20r+75=0,即 (r−5)(r−15)=0(r-5)(r-15)=0。两个半径分别为 55 和 1515,所以直径之和为 2(5+15)=402(5+15)=40 英寸。
因此,正确答案是 C。
For radius r,r, the center is (r,r,r)(r,r,r) and the given point is (5,5,10).(5,5,10). Thus 2(5−r)2+(10−r)2=r2, 2(5-r)^2+(10-r)^2=r^2, which simplifies to r2−20r+75=0,r^2-20r+75=0, or (r−5)(r−15)=0.(r-5)(r-15)=0. The two radii are 55 and 15,15, so the sum of the diameters is 2(5+15)=402(5+15)=40 inches.
Therefore, the correct answer is C.
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