Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
 
(One intermediate revision by the same user not shown)
Line 13: Line 13:
  
 
== Solution ==  
 
== 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 <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>.
+
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>.
  
 
==See Also==
 
==See Also==

Latest revision as of 13:00, 29 January 2023

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

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 $1$ m high and its shadow is $2$ 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 $r=10\sqrt{5}-20$, so the answer is $\boxed{\textbf{(E)}\ 10\sqrt{5}-20}$.

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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_27&oldid=188341"