1983 AHSME Problems/Problem 27

Problem 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 $10$ 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 $2$ 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.)

$\textbf{(A)}\ \frac{5}{2}\qquad \textbf{(B)}\ 9 - 4\sqrt{5}\qquad \textbf{(C)}\ 8\sqrt{10} - 23\qquad \textbf{(D)}\ 6-\sqrt{15}\qquad \textbf{(E)}\ 10\sqrt{5}-20$

Solution

$[asy] import geometry; import graph; unitsize(0.5 cm); pair O = (0,0), T = (0, -2.36067), S = (10, -2.36067), R = (1.05572, 2.11146); draw(circle(O, 2.3607)); draw(T--S--R--O--cycle); draw(O--S); label("$T$", T, SW); label("$O$", O, W); label("$S$", S, SE); label("$R$", R, NE); [/asy]$

Let $O$ be the center of the sphere, $T$ be the point where the sphere touches the ground, $S$ be the tip of the sphere's shadow, $R$ be the point where sun rays that reach the ground near $S$ would approach the sphere, and $\theta = \angle RST$ be the angle between the sun's rays and the ground.

Then $RS$ and $RT$ must both be tangent to the sphere, and therefore $RS = ST = 10$ and $OS$ bisects $\angle RST$ . So $\angle OST = \frac{\theta}{2}$ .

Since a $1$ meter stick casts a $2$ meter shadow, $\tan \theta = \frac{1}{2}$ . So $\cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1+\tan^2 \theta}} = \frac{1}{\sqrt{\frac{5}{4}}} = \frac{2}{\sqrt{5}}$ , and $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$ . So $\tan \left(\frac{\theta}{2}\right) = \frac{1-\cos \theta}{\sin \theta} = \frac{1 - \frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = \sqrt{5}-2$ .

Let $r = OT = OR$ be the radius of the sphere. Then $\tan\left(\frac{\theta}{2}\right) = \frac{r}{10}$ , so $r = 10 \tan\left(\frac{\theta}{2}\right) = 10(\sqrt{5}-2) = \boxed{(\mathbf{E})\ 10\sqrt{5} - 20}$ .

1983 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1983 AHSME Problems/Problem 27

Problem 27

Solution

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