Difference between revisions of "2015 AMC 10A Problems/Problem 14"

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$ \textbf{(A) }\mathrm{2 o'clock} \qquad\textbf{(B) }\mathrm{3 o'clock} \qquad\textbf{(C) }\mathrm{4 o'clock} \qquad\textbf{(D) }\mathrm{6 o'clock} \qquad\textbf{(E) }\mathrm{8 o'clock}
<math> \textbf{(A) }\mathrm{2 o'clock} \qquad\textbf{(B) }\mathrm{3 o'clock} \qquad\textbf{(C) }\mathrm{4 o'clock} \qquad\textbf{(D) }\mathrm{6 o'clock} \qquad\textbf{(E) }\mathrm{8 o'clock}</math>

Revision as of 00:49, 6 February 2015


The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?

[asy] size(170); defaultpen(linewidth(0.9)+fontsize(13pt)); draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1) { draw(0.9*dir(90-30*i)--dir(90-30*i)); label("$"+(string) i+"$",0.78*dir(90-30*i)); } dot(origin); draw(shift((0,1.87))*arrow); draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]

$\textbf{(A) }\mathrm{2 o'clock} \qquad\textbf{(B) }\mathrm{3 o'clock} \qquad\textbf{(C) }\mathrm{4 o'clock} \qquad\textbf{(D) }\mathrm{6 o'clock} \qquad\textbf{(E) }\mathrm{8 o'clock}$


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 $\boxed{\textbf{(C) }4 \ \mathrm{o'clock}}$ .

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. $\boxed{\textbf{(C)}}$

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 $20\pi$, so that is $20\pi$ traveled on a $60\pi$ arrow path. 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. $\boxed{\textbf{(C)}}$

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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

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