1983 AMC 12 第 27 题

先试着解答 1983 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 1983 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.

在一个晴天,一个大球体放在水平地面上。某一时刻,球体的影子从球体与地面的接触点起延伸 1010 米。同一时刻,一根米尺竖直放置,一端接触地面,投下长度为 22 米的影子。该球体的半径是多少米?(假设太阳光线互相平行,并将米尺视为线段。)
A large sphere is on a horizontal field on a sunny day. At a certain time the shadow of the sphere reaches out a distance of 1010 m from the point where the sphere touches the ground. At the same instant a meter stick (held vertically with one end on the ground) casts a shadow of length 22 m. What is the radius of the sphere in meters? (Assume the sun’s rays are parallel and the meter stick is a line segment.)
52\frac52
9−459-4\sqrt5
810−238\sqrt{10}-23
6−156-\sqrt{15}
105−2010\sqrt5-20
答案:E
知识点:球切线距离公式相似
小提示:
在竖直截面中,影子边界处的太阳光线与一个圆相切
In a vertical cross-section, the limiting sun ray is tangent to a circle
大提示:
米尺表明光线每水平前进 22 个单位就竖直上升 11 个单位
The meter stick shows that the ray rises 11 unit for every 22 horizontal units
解答:
以球体与地面的接触点为 (0,0)(0,0),圆心为 (0,r)(0,r)。影子边界处的太阳光线经过影子端点 (10,0)(10,0),斜率为 −12-\frac{1}{2},所以方程为 x+2y−10=0x+2y-10=0。相切意味着点 (0,r)(0,r) 到这条直线的距离等于 rr:10−2r5=r。 \frac{10-2r}{\sqrt5}=r\text{。}因此 r=102+5=105−20r=\frac{10}{2+\sqrt5}=10\sqrt5-20。
所以正确答案是 E。
Take the sphere’s ground-contact point as (0,0)(0,0) and its center as (0,r).(0,r). The limiting sun ray passes through the shadow endpoint (10,0)(10,0) and has slope −12,-\frac{1}{2}, so its equation is x+2y−10=0.x+2y-10=0. Tangency means the distance from (0,r)(0,r) to this line equals r:r: 10−2r5=r. \frac{10-2r}{\sqrt5}=r. Hence r=102+5=105−20.r=\frac{10}{2+\sqrt5}=10\sqrt5-20.
Therefore, the correct answer is E.
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