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); | + | 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 | + | 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 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
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 |
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