1983 AMC 12 Problem 27

Attempt Problem 27 of the 1983 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1983 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

27.

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
Answer: E
Concepts:spheretangent linedistance formulasimilarity
Small hint:
In a vertical cross-section, the limiting sun ray is tangent to a circle
Big hint:
The meter stick shows that the ray rises 11 unit for every 22 horizontal units
Solution:
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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