George Rebane

Here are a couple of problems that our high school students taking the annual merit scholarship TechTest may encounter. The solutions require no higher math, actually only a little bit of lower math, and can be puzzled out on the back of an envelope. You, however, may use a full sheet of paper.

Problem 1: What are all the feasible clock times at which the minute and hour hands lie concurrently (on top of each other) on a standard clock face? Calculate your answers to the nearest second.

Problem 2: What are all the feasible clock times at which the counter-clockwise angle from the minute hand to the hour hand makes a right angle (90 degrees) on a standard clock face? Calculate your answers to the nearest second.

Hint: It’s possible (advisable?) to puzzle out the approach to both problems in your head before making a single mark on paper. But in any event, always draw a picture.

Problem 2:

In a 12-hour clock cycle, there will be (11) 90-degree angle times. So each new 90-deg arrives at (1 hour, + 3600sec/11) apart, or about 1hr, 5 minutes, 27.27 seconds between times. The minute hand has to travel a little bit farther each revolution to 'catch up' to the hour hand to make a 90-deg angle.

Times:

9:00:00

10:05:27

11:10:55

12:16:22

1:21:49

2:27:16

3:32:44

4:38:11

5:43:38

6:49:05

7:54:33

then back to 9:00:00

Do you know how long it took to cycle through rotating the stem on my Timex Easy Reader watch to do this? They should've made that stem about an inch in diameter.

Haven't started #1 yet, but I suspect it will have the same 11 segments starting from 12:00:00.

Posted by: The Estonian Fox | 30 November 2023 at 04:27 AM

Forgot to add:

I volunteer in 7th & 8th grade math/science classes. They don't wear watches. And they don't know what an analog watch is - a couple kids have complained that they wish there were audio announcements of the time, so they didn't even have to look at the digital clock on the wall in the classroom. But when the bell rings - well, let's say that Mr. Pavlov would have been proud of their reaction times.

Posted by: The Estonian Fox | 30 November 2023 at 04:34 AM

Efox - 427am - Very good. Your times tie with my spreadsheet calculated times to within one second. When you post the Prob#1 answers, could you please also describe a little about your approach - i.e. how you formulated the problem - since there are several ways to climb that hill.

Posted by: George Rebane | 30 November 2023 at 11:39 AM

Problem #1, and the general solution-

Starting position:

The minute hand at time t=0 is at a 90o angle in front of the hour hand. The minute hand will again be 90o in front of the hour hand when the angle is 450o (360o + 90o).

Minute hand travels 0.1o /sec = (12/120)o/sec; hour hand travels (1/120)o/sec

So when [((12/120)o/sec x t + 90o) - ((1/120)o/sec x t)] = 450o , we have the next occurrence of a right angle.

Solve for t:

t (12/120 – 1/120)o =360o,

Or t (11/120)o = 360o - and here is where the segments of 11 comes in.

t = 3927.273 seconds later for the next 90o angle, or 1 hour, 5 minutes, 27.273 seconds between occurrences.

This holds for ANY starting point of hands on the clock. Say you start at 4:00:00 (120o angle between the hands). 1 hour, 5 minutes, 27.273 seconds later the hands will again be 120o apart. Starting point = 120o , ending point is 480o = 360o of travel.

So for overlapping (coincident) hands starting at 12:00:00, the next times will be

1:05:27

2:10:55

3:16:22

4:21:49

5:27:16

6:32:44

7:38:11

8:43:38

9:49:05

10:54:33, then 12:00:00 again.

Posted by: The Estonian Fox | 02 December 2023 at 12:59 PM

Efox 1259pm - Correctamente! Interesting that you took a slightly different tack than I, but both approaches tie. I'll post a pdf of my approach as an update.

Posted by: George Rebane | 02 December 2023 at 03:43 PM