[Guest]

I searched this one in the forum but couldn't find it. Sorry if it's a repost. The question is this:

Take a standard analog clock (the one with three hands, for hours, minutes and seconds). Assume the seconds hand turns evenly, instead of "jumping" from one second to the next. Is there any time of the day when the angle between any two of the hands is exactly 120 degrees? (i.e: They form the Mercedes Benz logo). If so, what time is that, and if not why not?

Edited March 16, 2011 by skolnick

[curr3nt]

Do the hour and minute hands jump or move evenly?

[Guest]

Do the hour and minute hands jump or move evenly?

Assume all the three hands turn evenly.

[Guest]

I don't even know how to start with this one! At first I think, well if it's three hands, then one third past the hour, and then one third this, but wait the hour hand moves one third, so one sixth of one twelfth, or is it ninth? Oh crap, I gotta start over!

[curr3nt]

the angles of the hands at given times...

angle of the second hand = 6s

angle of the minute hand = 6m + s/10

angle of the hour hand = 30h + (6m + s/10)/12

where the following are true

0 <= s < 60

0 <= m < 60 - m is an integer

0 <= h < 12 - h is an integer

Given 720 values of m an h is there an s where the difference between angles is 120?

[Guest]

There's a logical and a mathematical answer... I'm struggling with both.

Anyway, here's something to think about:

for every 1° degree the hour hand goes, the minute hand goes 12°

for every 1° degree the minute hand goes, the second hand goes 360°

[Guest]

the angles of the hands at given times...

angle of the second hand = 6s

angle of the minute hand = 6m + s/10

angle of the hour hand = 30h + (6m + s/10)/12

where the following are true

0 <= s < 60

0 <= m < 60 - m is an integer

0 <= h < 12 - h is an integer

Given 720 values of m an h is there an s where the difference between angles is 120?

take into account that s m and h are real and positive, not just integers.

[curr3nt]

take into account that s m and h are real and positive, not just integers.

hours and minute values can only be integers

11.5 hours is actually 11 hours and 30 minutes.

Any decimal value of hours can and should be converted into minutes and any decimal value of minutes can and should be converted into seconds.

to continue where I left off with those equations...

one of the following six sets has to be true to meet your condition

( 30h + m/2 + s/120 ) - ( 6m + s/10 ) = 120

( 6m + s/10 ) - 6s = 120

( 30h + m/2 + s/120 ) - 6s = 120

6s - ( 6m + s/10 ) = 120

( 6m + s/10 ) - 6s = 120

6s - ( 30h + m/2 + s/120 ) = 120

( 6m + s/10 ) - ( 30h + m/2 + s/120 ) = 120

( 30h + m/2 + s/120 ) - 6s = 120

6s - ( 6m + s/10 ) = 120

( 6m + s/10 ) - ( 30h + m/2 + s/120 ) = 120

6s - ( 30h + m/2 + s/120 ) = 120

( 30h + m/2 + s/120 ) - ( 6m + s/10 ) = 120

--OR--

s = 3600h/11 - 60m - 14400/11 AND s = 60m/59 - 1200/59

s = 3600h/719 - 60m/719 - 14400/719 AND s = 60m/59 + 1200/59

s = 60m/59 - 1200/59 AND s = 3600h/719 + 60m/719 + 14400/719

s = 3600h/11 - 60m + 14400/11 AND s = 3600h/719 - 60m/719 - 14400/719

s = 60m/59 + 1200/59 AND s = 3600h/11 - 60m + 14400/11

s = 3600h/719 + 60m/719 + 14400/719 AND s = 3600h/11 - 60m - 14400/11

Assuming I did my algebra right there are no values of m and h that make any of those sets true.

[Guest]

hours and minute values can only be integers

11.5 hours is actually 11 hours and 30 minutes.

Any decimal value of hours can and should be converted into minutes and any decimal value of minutes can and should be converted into seconds.

to continue where I left off with those equations...

one of the following six sets has to be true to meet your condition

( 30h + m/2 + s/120 ) - ( 6m + s/10 ) = 120

( 6m + s/10 ) - 6s = 120

( 30h + m/2 + s/120 ) - 6s = 120

6s - ( 6m + s/10 ) = 120

( 6m + s/10 ) - 6s = 120

6s - ( 30h + m/2 + s/120 ) = 120

( 6m + s/10 ) - ( 30h + m/2 + s/120 ) = 120

( 30h + m/2 + s/120 ) - 6s = 120

6s - ( 6m + s/10 ) = 120

( 6m + s/10 ) - ( 30h + m/2 + s/120 ) = 120

6s - ( 30h + m/2 + s/120 ) = 120

( 30h + m/2 + s/120 ) - ( 6m + s/10 ) = 120

--OR--

s = 3600h/11 - 60m - 14400/11 AND s = 60m/59 - 1200/59

s = 3600h/719 - 60m/719 - 14400/719 AND s = 60m/59 + 1200/59

s = 60m/59 - 1200/59 AND s = 3600h/719 + 60m/719 + 14400/719

s = 3600h/11 - 60m + 14400/11 AND s = 3600h/719 - 60m/719 - 14400/719

s = 60m/59 + 1200/59 AND s = 3600h/11 - 60m + 14400/11

s = 3600h/719 + 60m/719 + 14400/719 AND s = 3600h/11 - 60m - 14400/11

Assuming I did my algebra right there are no values of m and h that make any of those sets true.

Hi!

I simply assumed that since the hour, minutes and seconds hands can be in any position (not just the minutes positions) then any real 0 <= r < 360 will work as the angle, not only integers.

Regards.

[Guest]

Has anybody guessed yes? At 8:00:20

[fabpig]

But at 8 00.20 the min hand is just past the 12. Might be a good starting point tho...

Has anybody guessed yes? At 8:00:20

Edited March 17, 2011 by fabpig

[Guest]

But at 8 00.20 the min hand is just past the 12. Might be a good starting point tho...

SORRY fabpig But when it will be 8:20, The hour hand will me a bit more than 120 degrees so it may not work

[Guest]

According to me it is impossible that any two hands are at 120 degrees

because if the min. hand would be at any position other than 12 the hour hand will not be at 120 and similarly is sec.hand is not at 12 the minute hand will be at a different position ,But since two can't be at same place i'ts impossible

[superprismatic]

Just for fun, I wrote a little program to see how close

I can get to the desired configuration. The best I came

up with is 9:05:25.4375 where the hour and minute hands

are separated by 119.83°, the hour and second hands are

separeated by 120.09°, and the minute and second hads are

separated by 120.08°. I tried all times in .0001 second

increments. Pretty close, huh?

[k-man]

I don't really have the time right now to calculate, but if someone wants to try this, here is a suggestion to how this may be approached

Every hour (except 3:00-3:59 and 7:00-7:59) there are 2 times when the hour and minute hands form exactly 120 degrees angle. During those exception hours it happens only once, so all you need to do is to calculate a total of 22 precise times when that happens. Then just look where the second hand is at those times and whether it forms 120 degree angles with the other 2 hands. For example, the easiest of those 22 times are exactly 4:00:00 and 8:00:00. At those times, the hour and minute hands form exactly 120 degrees angle, but the second hand doesn't, so those can be discarded. All that's left is to calculate the 20 remaining times. Good luck!

[Guest]

Has anybody guessed yes? At 8:00:20

Hi!

This answer is not correct, because in 20 seconds the minutes hand turns 2 degrees (a minute spans 6 degrees), and the hour hand turns 1/6th of a degree, therefore the angles between the hands would not be exactly 120 degrees.

Regards.

[Guest]

The proof is quite simple if, instead of worrying about the motion of three hands, we take the hour hand as our frame of reference. The relative angular speeds then of the other two hands are then 11t/120 deg/s (minute hand) and 719t/120 deg/s (second hand) for t in seconds (t = 0 at 12:00:00). The minute hand then sweeps through 120 degrees relative to the hour hand in 14,400/11 seconds and the second hand sweeps 120 degrees relative to the hour hand in 14,400/719 seconds. So, among other things, we must find two integers, m and n such that m/11 = n/719. Since 11 and 719 are both prime the only solution is m = 11k and n = 719k (k = 0, 1, 2, ....), which corresponds to all three hands pointing at 12.

Edited March 17, 2011 by d3k3

[Guest]

Since It's a discussion and nobody knows the correct answer I think We should try voting or any other method to reach the conclusion :unsure:

[Guest]

Has anybody guessed yes? At 8:00:20

this is wrong, seeing as it is 20 seconds past 8 and the minute hand keeps moving, it will still not be exactly on the 12 and therefore there will not be exactly 120 degrees between each hand of the clock.

[Guest]

I'm getting bored by this discussion, so please let us reach the conclusion as fast as possible

[EventHorizon]

I'm getting bored by this discussion, so please let us reach the conclusion as fast as possible

Did you notice d3k3's post? He solved it.

[superprismatic]

Did you notice d3k3's post? He solved it.

So did curr3nt in post #8. He did it in a different way, but he was first.

[Guest]

20.0000001 seconds after midnight the second hand is 120 degrees past the hour hand. In fact, the second hand is 120 ahead and after both the hour and minute hands at one point on nearly every rotation (1 min)

[curr3nt]

Using...

angle of the second hand = 6s

angle of the minute hand = 6m + s/10

angle of the hour hand = 30h + m/2 + s/120

at 0:0:20.0000001

second hand is at 120.0000006

minute hand is at 2.00000001

hour hand is at 0.1666666675

20.0000001 seconds after midnight the second hand is 120 degrees past the hour hand. In fact, the second hand is 120 ahead and after both the hour and minute hands at one point on nearly every rotation (1 min)