2021 WSMO Accuracy Round Problems/Problem 4
Problem 4
A 12-hour clock has a minute hand that is the same length as the second hand, and an hour hand half the length of the minute hand. In a day, the tip of the minute hand travels a distance of 
the tip of the second hand travels a distance of 
and the tip of the hour hand travels a distance of 
The value of 
can be expressed as 
, where 
and 
are relatively prime positive integers. Find 
.
Solution 1
WLOG, assume that the length of the minute hand is 2. This means that the length of the second hand is 2 and the length of the hour hand is 1. In a day, there are 24 hours, which means that the minute hand travels 
each day. Also, since there are 
hours, the hour hand travels 
each day. Finally, there are 
minutes in a day, which means that the second hand travels 
each day. Thus, the final answer is ![\[\frac{(24\cdot4\pi)\cdot(24\cdot4\pi)}{(4\pi)\cdot(24\cdot60\cdot4\pi)}=\frac{(\cancel{24}\cdot\cancel{4\pi})\cdot(24\cdot\cancel{4\pi})}{(\cancel{4\pi})\cdot(\cancel{24}\cdot60\cdot\cancel{4\pi})}=\frac{24}{60}=\frac{2}{5}\Longrightarrow 2+5=\boxed{7}.\]](http://latex.artofproblemsolving.com/b/d/1/bd112c3585b63eaa6a64d31d8e53624cf2d99bd4.png)
~pinkpig
Solution 2
Let the distance traveled by one revolution of the minute hand tip be 
. Note that 
, 
, and 
. Our desired expression becomes:
![\[\frac{(24C)^2}{(C)(60 \cdot 24 C)}\]](http://latex.artofproblemsolving.com/9/5/b/95b69cc6063aac9136b0b0f8ffa11177bd94be17.png)
![\[=\frac{24^2}{60 \cdot 24}\]](http://latex.artofproblemsolving.com/4/6/e/46ebb12f8313898f4bafb29b12dd105f4067b4b2.png)
![\[=\frac{2}{5}\]](http://latex.artofproblemsolving.com/8/e/5/8e58d969eac5b324838f47b3938f12b672f14f1d.png)
This gives us an answer of 
.
~BigKahuna227
