All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
Hands On A Clock (Posted on 2004-07-05) Difficulty: 3 of 5
Do the three hands on an analog clock (hours, minutes, seconds) ever divide the face of the clock into three equal segments, i.e. 120 degrees between each hand?

No Solution Yet Submitted by SilverKnight    
Rating: 3.5000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution soln: | Comment 19 of 23 |
First let us assume that our clock has 60 divisions. We will show that any time the hour hand and the minute hand are 20 divisions (120 degrees) apart, the second hand cannot be an integral number of divisions from the other hands, unless it is straight up (on the minute).

Let us use h for hours, m for minutes, and s for seconds. We will use =n to mean congruent mod n, thus 12 =5 7.

We know that m =60 12h, that is, the minute hand moves 12 times as fast as the hour hand, and wraps around at 60. We also have s =60 60m. This simplifies to s/60 =1 m, which goes to s/60 = frac(m) (assuming s is in the range 0 <= s < 60), which goes to s = 60 frac(m). Thus, if m is 5.5, s is 30.

Now let us assume the minute hand is 20 divisions ahead of the hour hand. So m =60 h + 20, thus 12h =60 h + 20, 11h =60 20, and, finally, h =60/11 20/11 (read 'h is congruent mod 60/11 to 20/11'). So all values of m are k + n/11 for some integral k and integral n, 0 <= n < 11. s is therefore 60n/11. If s is to be an integral number of units from m and h, we must have 60n =11 n. But 60 and 11 are relatively prime, so this holds only for n = 0. But if n = 0, m is integral, so s is 0.

Now assume, instead, that the minute hand is 20 divisions behind the hour hand. So m =60 h - 20, 12h =60 h - 20, 11h =60 -20, h =60/11 -20/11. So m is still k + n/11. Thus s must be 0.

But if s is 0, h must be 20 or 40. But this translates to 4 o'clock or 8 o'clock, at both of which the minute hand is at 0, along with the second hand.

Thus the 3 hands can never be 120 degrees apart


  Posted by neshal on 2004-08-02 07:41:10
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information