2015 AMC 10A Problems/Problem 14

Problem

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+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]$

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

Solution 1

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 $\boxed{\textbf{(C) }4 \ \text{o' clock}}$ . Similarly, the arrow would be pointing downward at 6:00. It would already have completed three 180 degree turns. Therefore, two 180 degree turns would be completed at 4:00.

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

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 $20\pi$ , 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. $\boxed{\textbf{(C) 4 o' clock}}$

Solution 4

Suppose that the small disk also had a clock face on it, and that both disks were toothed wheels, free to rotate around their centers. The part of the picture where they engage would look like this: $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label("12",1.56*dir(90)); label("1",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label("6",.74*dir(-90)+(0,3)); label("5",.74*dir(-60)+(0,3)); label("4",.74*dir(-30)+(0,3)); [/asy]$ The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\circ$ clockwise (i.e. 2 hours). $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label("$30^\circ$",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*"12",1.56*dir(120)); label(rotate(30)*"1",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label("$60^\circ$",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*"6",.74*dir(-150)+(0,3)); label(rotate(-60)*"5",.74*dir(-120)+(0,3)); label(rotate(-60)*"4",.74*dir(-90)+(0,3)); [/asy]$ However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\circ$ clockwise to see what happens when the small cog rolls around it. $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label("12",1.56*dir(90)); label("1",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label("$90^\circ$",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*"6",.74*dir(-180)+c); label(rotate(-90)*"5",.74*dir(-150)+c); label(rotate(-90)*"4",.74*dir(-120)+c); [/asy]$ It turns out that, when the point of tangency moves $30^\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\boxed{\textbf{(C) }\text{4 o' clock}}$

Solution 5

We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We now draw corresponding arrows at each hour and fold it back again.

Williamgolly

Edited by ab2024

Solution 6

We can approach this problem with angle measures. As the circumference of the disk is $10\pi,$ and the clock $20 \pi,$ we have that 30 degrees, or the angular measure between hours, of the disk is only 15 degrees of the clock. This yields that every two hour ticks that the clock rotates, on the third one, the ticks will meet. However, the disk must rotate 360 degrees in order to come back to its original position, so the angular measure that the disk has covered relative to the clock is simply $12 \cdot 15 \cdot \frac{2}{3},$ or $120^\circ$ from the 12 starting point, so $\frac{120}{30} = \boxed{\textbf{(C) 4}}.$

~ab2024

Solution 7

If the big clock were a flat plane, then the smaller clock could travel $\dfrac{40\pi}{20\pi}=2$ full revolutions.

But we also need to account for rotation. If we mark a red dot on the bottom of the small clock/bottom of the arrow, and then drag it around the clock, the direction of the arrow would still change. After traveling around the big clock, the small clock would travel $1$ full rotation.

Considering these two movements, the small clock travels 3 full rotations around the big clock so the arrow is next pointing upwards at $\dfrac{12 \text{ o'clock}}{3}=\boxed{\textbf{(C)}\text{ 4 o'clock}}$ .

~BakedPotato66

Solution 8

The center of rotation is in the center of the smaller circle, but extends to the center of the larger circle. That means the circumference of the circle in relation to the arrow is $60 \pi$ . The other circle is $20 \pi$ and so that is $\frac{1}{3}$ . So $\frac{12}{3} = 4$ which is $\boxed{C}$ .

Video Solution 1

https://youtu.be/hPEgOYA4Xq8

~savannahsolver

Video Solution 2

Video Solution: https://www.youtube.com/watch?v=kCsQbTi3iLU

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 10 Problems and Solutions

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