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Difference between revisions of "1983 AHSME Problems/Problem 27"

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== Solution ==  
 
== Solution ==  
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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 <math>\arctan \left(\frac{1}{2}\right )</math>, since the stick is <math>1</math> m high and its shadow is <math>2</math> m long, so by considering a right-angled triangle, <math>\frac{r}{10}=\tan \left(\arctan \left(\frac{1}{2}\right )\right )=\tan \left (\frac{1}{2} \right )</math>. Since <math>\tan \left(\frac{\theta}{2} \right ) = \frac{1-\cos \left(\theta \right )}{\sin \left(\theta \right )}</math>, we find that <math>\frac{r}{10}=\sqrt{5}-2</math>. Hence <math>r=10\sqrt{5}-20</math>, so the answer is <math>\boxed{\textbf{(E)}\ 10\sqrt{5}-20}</math>.
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<asy>
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import geometry;
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import graph;
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unitsize(0.5 cm);
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pair O = (0,0), T = (0, -2.36067), S = (10, -2.36067), R = (1.05572, 2.11146);
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draw(circle(O, 2.3607));
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draw(T--S--R--O--cycle);
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draw(O--S);
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label("$T$", T, SW);
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label("$O$", O, W);
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label("$S$", S, SE);
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label("$R$", R, NE);
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</asy>
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Let <math>O</math> be the center of the sphere, <math>T</math> be the point where the sphere touches the ground, <math>S</math> be the tip of the sphere's shadow, <math>R</math> be the point where sun rays that reach the ground near <math>S</math> would approach the sphere, and <math>\theta = \angle RST</math> be the angle between the sun's rays and the ground.  
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Then <math>RS</math> and <math>RT</math> must both be tangent to the sphere, and therefore <math>RS = ST = 10</math> and <math>OS</math> bisects <math>\angle RST</math>.  So <math>\angle OST = \frac{\theta}{2}</math>.
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Since a <math>1</math> meter stick casts a <math>2</math> meter shadow, <math>\tan \theta = \frac{1}{2}</math>.  So <math>\cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1+\tan^2 \theta}} = \frac{1}{\sqrt{\frac{5}{4}}} = \frac{2}{\sqrt{5}}</math>, and <math>\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}</math>. So <math>\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</math>.
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Let <math>r = OT = OR</math> be the radius of the sphere.  Then <math>\tan\left(\frac{\theta}{2}\right) = \frac{r}{10}</math>, so <math>r = 10 \tan\left(\frac{\theta}{2}\right) = 10(\sqrt{5}-2) = \boxed{(\mathbf{E})\ 10\sqrt{5} - 20}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 14:22, 3 July 2025

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 $r = 10 \tan\left(\frac{\theta}{2}\right) = 10(\sqrt{5}-2) = \boxed{(\mathbf{E})\ 10\sqrt{5} - 20}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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