Difference between revisions of "2002 AMC 10B Problems/Problem 24"
(→Solution) |
(→Solution) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 5: | Line 5: | ||
<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 == | + | == Solution 1 == |
<center><asy> | <center><asy> | ||
Line 16: | Line 16: | ||
path ferriswheel=Circle(O,r); | path ferriswheel=Circle(O,r); | ||
draw(ferriswheel); | draw(ferriswheel); | ||
− | draw(O--A); draw(O--C); draw(B--C); | + | 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); | ||
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> |
+ | |||
+ | == 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: | ||
+ | <cmath> f(x) = Acos(Bx - C) + D.</cmath> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <cmath>10 = -20cos(\frac{\pi}{30}x) + 20.</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~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 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 1
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
Solution 2 (trig)
The path that the rider takes along the Ferris wheel can be represented by a sinusoidal graph, where represents the time in seconds. Since is at the crest of the graph and not at the midline, we will use a cosine graph. Therefore, we will use the form:
The graph starts at the lowest point at feet, then goes up to reach the highest point at feet, then comes back down. Therefore, the amplitude is (and negative since it starts at the bottom, not the top), and the vertical shift is . There is no horizontal shift since the lowest point is at . It takes seconds to make one full revolution (the period), so 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,
We get to and remember that the problem is looking for the first instance of , so Solving, we get that .
~jp06132
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.