Being an avid watch watcher, Hinea discovers that the hands of her watch are perpendicular at 3pm and at 9am.
She wonders how many times a day the hands of her watch are perpendicular to each other?
What answer does she come up with and why?
This problem involves taking a problem about time, looking at it geometrically, and solving it algebraically. The problem is about when hands on a watch are perpendicular. This is easily seen as a geometry problem as a diagram can be drawn to show the situation. Finding the algebra may be more challenging. This requires knowing about the relative speed of the hands of a watch and the significance of angle.
Hinea’s Watch's Hands, Geometry, Level 6 is a similar problem.
Being an avid watch watcher, Hinea discovers that the hands of her watch are perpendicular at 3pm and at 9am. She wonders how many times a day the hands of her watch are perpendicular to each other?
What answer does she come up with and why?
In the first instance, this problem can be attempted by trial and error using a real watch.
Suppose the hour hand moves Θo between two consecutive perpendicular occasions. Then the minute hand moves 180° + Θo.
The time taken is hours (the hour hand moves 1/12 x 360° = 30° per hour). But the minute hand moves 12 times as fast as the hour hand (the minute hand goes around 12 times to the hour hands once). So the time taken to move 180° + q ° is . Hence = . This gives q = and the time between two perpendiculars is 180/11 x 1/30 hours. Hence the number of perpendiculars in 24 hours is 24 / (180/360) = 44.
Note that there is a "logical" approach to this problem. Between one hour and the next, the hands of a clock have to be perpendicular twice. This seems to suggest that this would produce 48 (2 x 24) perpendicular events a day. However, at 3 o’clock the hands are perpendicular on the hour. So these occurrences take place between 2 and 3 o’clock as well as between 3 and 4 o’clock. So we lose one perpendicular event as a result. The same thing happens at 9 o'clock. Hence we need to subtract 4 perpendiculars from the original guess of 48 (one for each of 3am, 9am, 3pm and 9pm). So there are 44 perpendiculars in a day. (The answer to Hinea’s Watch's Hands problem is 22. You might consider with your students whether or not it is a coincidence that 44 = 2 x 22?)
Printed from https://meaningfulmaths.nt.edu.au/mmws/nz/resource/hinea-s-other-watch at 9:00pm on the 26th February 2024