2021 WSMO Accuracy Round Problems/Problem 4

Revision as of 15:04, 29 January 2022 by Bigkahuna227 (talk | contribs) (Created page with "==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 minut...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $m,$ the tip of the second hand travels a distance of $s,$ and the tip of the hour hand travels a distance of $h.$ The value of $\frac{m^2}{hs}$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find $a+b$.

Solution

Let the distance traveled by one revolution of the minute hand tip be $C$. Note that $m=24C$, $s = 60 \cdot 24 C$, and $h=2(\frac{C}{2}) = C$. Our desired expression becomes: \[\frac{(24C)^2}{(C)(60 \cdot 24 C)}\] \[=\frac{24^2}{60 \cdot 24}\] \[=\frac{2}{5}\]

This gives us an answer of $2 + 5 = \boxed{7}$.

~BigKahuna227