Difference between revisions of "2002 AMC 10B Problems/Problem 24"

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path ferriswheel=Circle(O,r);
 
path ferriswheel=Circle(O,r);
 
draw(ferriswheel);
 
draw(ferriswheel);
draw(O--A); draw(O--C); draw(B--C);
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draw(O--A); draw(O--C); draw(B--C); draw(A--C);
 
pair[] ps={A,B,C,O}; dot(ps);
 
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("$O$",O,N); label("$A$",A,S); label("$B$",B,W); label("$C$",C,SE);
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</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 OBC = 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>
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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}}
 +
{{MAA Notice}}

Latest revision as of 14:06, 7 August 2018

Problem 24

Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ 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 $10$ vertical feet above the bottom?

$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15$

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); draw(A--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]

We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\triangle OBC$ is a $30-60-90$ triangle. That means $\angle BOC = 60^\circ$ and the ferris wheel has made $\frac{60}{360} = \frac{1}{6}$ of a revolution. Therefore, the time it takes to travel that much of a distance is $\frac{1}{6}\text{th}$ of a minute, or $10$ seconds. The answer is $\boxed{\mathrm{(D) \ } 10}$. Alternatively, we could also say that $\triangle ABC$ is congruent to $\triangle OBC$ by SAS, so $AC$ is 20, and $\triangle AOC$ is equilateral, and $\angle BOC = 60^\circ$

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
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

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