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: 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 $\boxed{4}$

Answer: C 4