Difference between revisions of "2002 AMC 10B Problems/Problem 24"
(→Solution) |
(→Solution) |
||
| Line 25: | Line 25: | ||
</asy></center> | </asy></center> | ||
| − | We can let this circle represent the ferris wheel with center <math>O,</math> and <math>C</math> represent the desired point <math>10</math> feet above the bottom. Draw a diagram like the one above. We find out <math>\triangle OBC</math> is a <math>30-60-90</math> triangle. That means <math>\angle BOC = 60^\circ</math> and the ferris wheel has made <math>\frac{60}{360} = \frac{1}{6}</math> of a revolution. Therefore, the time it takes to travel that much of a distance is <math>\frac{1}{6}\text{th}</math> of a minute, or <math>10</math> seconds. The answer is <math>\boxed{\mathrm{(D) \ } 10}</math>. Alternatively, we | + | We can let this circle represent the ferris wheel with center <math>O,</math> and <math>C</math> represent the desired point <math>10</math> feet above the bottom. Draw a diagram like the one above. We find out <math>\triangle OBC</math> is a <math>30-60-90</math> triangle. That means <math>\angle BOC = 60^\circ</math> and the ferris wheel has made <math>\frac{60}{360} = \frac{1}{6}</math> of a revolution. Therefore, the time it takes to travel that much of a distance is <math>\frac{1}{6}\text{th}</math> of a minute, or <math>10</math> seconds. The answer is <math>\boxed{\mathrm{(D) \ } 10}</math>. Alternatively, we could also say that <math>\triangle ABC</math> is congruent to <math>\triangle OBC</math> by SAS, so <math>AC</math> is 20, and <math>\triangle AOC</math> is equilateral, and <math>\angle BOC = 60^\circ</math> |
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=B|num-b=23|num-a=25}} | {{AMC10 box|year=2002|ab=B|num-b=23|num-a=25}} | ||
Revision as of 16:41, 24 March 2013
Problem 24
Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius 
feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point 
vertical feet above the bottom?
Solution
![[asy] unitsize(1.5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair O=(0,0), A=(0,-20), B=(0,-10), C=(10sqrt(3),-10); real r=20; path ferriswheel=Circle(O,r); draw(ferriswheel); draw(O--A); draw(O--C); draw(B--C); pair[] ps={A,B,C,O}; dot(ps); label("$O$",O,N); label("$A$",A,S); label("$B$",B,W); label("$C$",C,SE); label("$10$",(O--B),W); label("$10$",(A--B),W); label("$20$",(O--C),NE); [/asy]](http://latex.artofproblemsolving.com/2/2/7/227bf6f2616507691c2d99276a4b892f93537403.png)
We can let this circle represent the ferris wheel with center 
and 
represent the desired point feet above the bottom. Draw a diagram like the one above. We find out 
is a 
triangle. That means 
and the ferris wheel has made 
of a revolution. Therefore, the time it takes to travel that much of a distance is 
of a minute, or seconds. The answer is 
. Alternatively, we could also say that 
is congruent to by SAS, so 
is 20, and 
is equilateral, and
See also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
