Difference between revisions of "2015 AMC 10A Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. <math>\boxed{\textbf{(C)}}</math> | The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. <math>\boxed{\textbf{(C)}}</math> | ||
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| + | ==Solution 3== | ||
| + | The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is <math>20\pi</math>, so that is <math>20\pi</math> traveled on a <math>60\pi</math> clock circumference. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. <math>\boxed{\textbf{(C)}}</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}} | {{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:20, 4 February 2015
Problem
Clock - radius 20 cm Disk - radius 10 cm, externally tangent at 12 o'clock, arrow pointing up
If the disk rolls clockwise, where on the clock will it be when the arrow points up?
Solution
The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is 
.
Solution 2
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. 
Solution 3
The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is 
, so that is traveled on a 
clock circumference. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock.
See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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. 
