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

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<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15 </math>
 
<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15 </math>
  
== Solution ==
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== Solution 1 ==
  
 
<center><asy>
 
<center><asy>
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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 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 would 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>
+
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>
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 +
== Solution 2 (trig) ==
 +
The path that the rider takes along the Ferris wheel can be represented by a sinusoidal graph, where <math>x</math> represents the time in seconds. Since <math>x=0</math> is at the crest of the graph and not at the midline, we will use a cosine graph. Therefore, we will use the form:
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<cmath> f(x) = Acos(Bx - C) + D.</cmath>
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 +
The graph starts at the lowest point at <math>0</math> feet, then goes up to reach the highest point at <math>40</math> feet, then comes back down. Therefore, the amplitude is <math>20</math> (and negative since it starts at the bottom, not the top), and the vertical shift is <math>20</math>. There is no horizontal shift since the lowest point is at <math>x=0</math>. It takes <math>60</math> seconds to make one full revolution (the period), so <math>B = \frac{2\pi}{60} = \frac{\pi}{30}.</math> Now we have all the parts we need for the equation of our graph, and we can set it equal to the height we want, <math>10.</math>
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 +
<cmath>10 = -20cos(\frac{\pi}{30}x) + 20.</cmath>
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 +
We get to <math>\frac{1}{2} = cos(\frac{\pi}{30}x)</math> and remember that the problem is looking for the first instance of <math>f(x) = 10</math>, so <math>\frac{\pi}{30}x = \frac{\pi}{3}.</math> Solving, we get that <math>x = \boxed{\mathrm{(D) \ } 10}</math>.
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 +
~jp06132
  
 
== 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 11:57, 31 October 2021

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 1

[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 $\angle BOC = 60^\circ$

Solution 2 (trig)

The path that the rider takes along the Ferris wheel can be represented by a sinusoidal graph, where $x$ represents the time in seconds. Since $x=0$ is at the crest of the graph and not at the midline, we will use a cosine graph. Therefore, we will use the form: \[f(x) = Acos(Bx - C) + D.\]

The graph starts at the lowest point at $0$ feet, then goes up to reach the highest point at $40$ feet, then comes back down. Therefore, the amplitude is $20$ (and negative since it starts at the bottom, not the top), and the vertical shift is $20$. There is no horizontal shift since the lowest point is at $x=0$. It takes $60$ seconds to make one full revolution (the period), so $B = \frac{2\pi}{60} = \frac{\pi}{30}.$ Now we have all the parts we need for the equation of our graph, and we can set it equal to the height we want, $10.$

\[10 = -20cos(\frac{\pi}{30}x) + 20.\]

We get to $\frac{1}{2} = cos(\frac{\pi}{30}x)$ and remember that the problem is looking for the first instance of $f(x) = 10$, so $\frac{\pi}{30}x = \frac{\pi}{3}.$ Solving, we get that $x = \boxed{\mathrm{(D) \ } 10}$.

~jp06132

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AMC 10 Problems and Solutions

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