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1983 AHSME Problems/Problem 27

Revision as of 18:41, 27 January 2019 by Sevenoptimus (talk | contribs) (Cleaned up the solution and improved its readability)

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 $10 \ \text{m}$ 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 $2 \ \text{m}$. 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

Consider the angle that the shadow makes with the ground. Since the sun's rays are parallel, it's the same as the angle made with the shadow of the stick. We know that the angle of the stick is $\arctan \left(\frac{1}{2}\right )$, since the stick is $1 \ \text{m}$ high and its shadow is $2 \ \text{m}$ long, so by considering a right-angled triangle, $\frac{r}{10}=\tan \left(\arctan \left(\frac{1}{2}\right )\right )=\tan \left (\frac{1}{2} \right )$. Since $\tan \left(\frac{\theta}{2} \right ) = \frac{1-\cos \left(\theta \right )}{\sin \left(\theta \right )}$, we find that $\frac{r}{10}=\sqrt{5}-2$. Hence $r=10\sqrt{5}-20$, and our answer is $\fbox{\textbf{E}}$.

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