Shollenbarger's Clock: Proving No Time Exists Where Hands Equidistant

[expectedvalue]

Shollenbarger's Clock:

Excluding 12:00, Is there ever a time on a clock where the second, minute and hour hands are equidistant from each other? (In other words, the three hands are all 120° apart? If yes, give the EXACT time(s) that this occurs. If no, prove that no such time exists. Good Luck!

 

[DM]

-------------------
If you include 12:00
-------------------

The second hand could be on 12:00, the minute hand on 4:00 and the hour hand on 8:00

Time would be 8:20

--------------------
If you exclude 12:00
--------------------

I personally don't think it's possible.

If any four hour marks distance stand for 120° (each hour mark is 30°), then 12:00 must be included.

Thus no such time exists!

 

[Parlyne]

-------------------
If you include 12:00
-------------------

The second hand could be on 12:00, the minute hand on 4:00 and the hour hand on 8:00

Time would be 8:20

But, at 8:20:00, the hour hand of a clock doesn't point directly at the 8. In fact, it will be one third of the way from 8 to 9. Thus, at 8:20:00 the hands will not all be $$120^\circ$$ apart.

--------------------
If you exclude 12:00
--------------------

I personally don't think it's possible.

If any four hour marks distance stand for 120° (each hour mark is 30°), then 12:00 must be included.

Thus no such time exists!

I think he was saying to exclude the time 12:00:00 from consideration, rather than the direction of pointing at the 12.

 

[expectedvalue]

Parlyne is correct. I wanted to exclude 12:00:00 from consideration as an answer, because all three hands are on top of each other. But one hand can be pointing at 12.

8:20:00 is not a valid solution:

From second to minute: 120°
From minute to hour: 130°
From hour to second: 110°

 

[DM]

What about:

hour hand on 4:00, minute hand on 12:00 and second hand on 8:00

Time = 4:00

 

[expectedvalue]

What about:

hour hand on 4:00, minute hand on 12:00 and second hand on 8:00

Time = 4:00

This time is: 4:00:40. This is not a valid solution:

From minute to hour: 119 2/3°
From hour to second: 119 2/3°
From second to minute: 120 2/3°

The hour hand cannot be at 4:00 or the minute hand at 12:00 when the second hand is at 8:00, because it would now be 40 seconds after 4:00, thus the hour hand and the second would have moved slightly.

Keep trying. (Fractions of a second are allowed as valid answers, if needed.)

 

[DM]

My last attempt of the day is:

Hour hand on 3:00

Minute hand on 0:59

Second hand on 0:00:39

Time = 3:59:39

 

[expectedvalue]

Time = 3:59:39

No, unfortunately 3:59:39 is not a valid answer:

From minute to hour: 121.925°
From hour to second: 114.175°
From second to minute: 123.9°

 

[geniusprahar_21]

its not possible...one way of proving it is assume the time as xhrs, yhrs, and z secs...and put appropriate conditions on them... then find angles and solve...

 

[DM]

its not possible...one way of proving it is assume the time as xhrs, yhrs, and z secs...and put appropriate conditions on them... then find angles and solve...

This is a very messy explanation. "Put appropriate conditions on them" doesn't give us the reason.

What conditions?

 

[gnomedt]

I too believe it's impossible. I haven't finished the proof, but the idea is that I make the hour, minutes and seconds hands of the clock into unit vectors which are a function of time, then take dot products and set them equal (because the dot product of any two of the aforementioned vectors must be |1|*|1|*cos[120 deg], i.e. 4 hour-marks apart) and finally show that there's no way that that's possible. There might be an easier way, but I like vectors.

Now here's another brainteaser, if this one turns out to be impossible: what time on Shollenbarger's Clock will make the hands the closest to 120 degrees?

 

[DaveC426913]

Correct me of I'm wrong, but no one said they have to be at exact integer times, right?

Don't these three hands line up at 120 degrees from each hour approximately 11 times in every 12 hour period? (I'll leave it to others to determine the exact times.)

 

[Parlyne]

Correct me of I'm wrong, but no one said they have to be at exact integer times, right?

Don't these three hands line up at 120 degrees from each hour approximately 11 times in every 12 hour period? (I'll leave it to others to determine the exact times.)

No. The hour and minute hands line up 11 times in each 12 hour period. The second hand will only be in line with these one of those 11 times.

 

[Parlyne]

I'm going to go out on a limb and say that there is no such time. To demonstrate, let's assume there is a time when the hands are all $$120^\circ$$ apart. Call the hour x, the minute y and the second z. Because of the format times are cited in, x and y must be integers.

Using these, we can express the angle at which each hand sits. The hour hand sits at $$\alpha = \left (\frac{x}{12} + \frac{y}{720} + \frac{z}{43200} \right)*360^\circ = 30x + \frac{y}{2} + \frac{z}{120}$$. The minute hand sits at $$\beta = \left ( \frac{y}{60} + \frac{z}{3600} \right )*360^\circ = 6y + \frac{z}{10}$$. And, the second hand sits at $$\gamma = 6z$$.

We set the second and minute hands $$120^\circ$$ apart by setting $$\beta = \gamma \pm 120$$. This gives the condition $$y = \frac{59}{60} z \pm 20$$. Since y must be an integer, z must be an integral multiple of 60/59. Thus, we can write $$z = \frac{60}{59}n$$ and $$y = n \pm 20$$, where n is an integer.

Finally, to put the hour hand in the correct position, we set $$\alpha = \gamma \pm 240$$. This reduces to the condition that $$x = \frac{11}{59} n \pm \frac{23}{2}$$.

We can see that, for x to be an integer, n must be valued such that $$\frac{11}{59} n$$ is an odd half integer. But, since 59 is odd and 11 and 59 are relatively prime, the only way for $$\frac{11}{59} n$$ to be an odd half integer is if n is an odd half integer multiple of 59 - $$n = \frac{2m+1}{2} 59$$. Since all factors in the numerator are odd, this can't be an integer. Thus, if n is an integer, x is not.

Thus, there are no x, y, and z satisfying the relationships above such that x and y are both integers.

 

[geniusprahar_21]

This is a very messy explanation. "Put appropriate conditions on them" doesn't give us the reason.

What conditions?

conditions like:
x,y,z should be integers...or atleast x and y should be integers(if u want to accept decimal seconds). further 0<= x <12... 0<= y <60 and 0 <= z < 60...with these conditions u'll never get an appropriate answer

 

[ceptimus]

It gets close enough for practical purposes at

02:54:34.576

 

[expectedvalue]

I tried to clarify this earlier...

Decimal seconds are allowed. The seconds do not have to be integers. But give your answers with fractions.

It gets close enough for practical purposes at

02:54:34.576

Unfortunately, this is a brain teaser and not a practical problem. So, 'close enough' is not equidistant.

 

[Jimmy Snyder]

The answer is that the three hands never form 120 degree angles w.r.t each other. I have an inelegant solution. There are in fact, 22 times in a 12 hour period that the hour and minute hands are 120 degrees apart. It suffices to determine where the second hand is at those times. I did so and found that the second hand is never at the right place at the right time. Here are the results of a computer run. The first column is the time. The second column is the angle between the hour and minute hands. The third column is the angle between the minute and second hands. Take the first pair of lines for instance. Sometime between 00:21:49 and 00:21:50, the hour and minute hands were 120 degrees apart. However, at no time during that interval was the second hand near to 120 degrees from the minute hand. There is some symmetry in the data so that it turns out that with a little more analysis, you wouldn't need to check all 22 times. The closest approach is at 2:54:32 (and at 3 other times).

00:21:49 -119.9917 -163.1000
00:21:50 -120.0833 -169.0000

00:43:38 -239.9833 33.8000
00:43:39 -240.0750 27.9000

01:27:16 -119.9667 67.6000
01:27:17 -120.0583 61.7000

01:49:05 -239.9583 264.5000
01:49:06 -240.0500 258.6000

02:32:43 -119.9417 -61.7000
02:32:44 -120.0333 -67.6000

02:54:32 -239.9333 135.2000
02:54:33 -240.0250 129.3000

03:38:10 -119.9167 169.0000
03:38:11 -120.0083 163.1000

03:59:59 -239.9083 5.9000
04:00:00 120.0000 0.0000
04:00:01 119.9083 -5.9000

04:43:38 -119.9833 33.8000
04:43:39 -120.0750 27.9000

05:05:27 120.0250 -129.3000
05:05:28 119.9333 -135.2000

05:49:05 -119.9583 264.5000
05:49:06 -120.0500 258.6000

06:10:54 120.0500 -258.6000
06:10:55 119.9583 -264.5000

06:54:32 -119.9333 135.2000
06:54:33 -120.0250 129.3000

07:16:21 120.0750 -27.9000
07:16:22 119.9833 -33.8000

07:59:59 -119.9083 5.9000
08:00:00 240.0000 0.0000
08:00:01 239.9083 -5.9000

08:21:49 120.0083 -163.1000
08:21:50 119.9167 -169.0000

09:05:27 240.0250 -129.3000
09:05:28 239.9333 -135.2000

09:27:16 120.0333 67.6000
09:27:17 119.9417 61.7000

10:10:54 240.0500 -258.6000
10:10:55 239.9583 -264.5000

10:32:43 120.0583 -61.7000
10:32:44 119.9667 -67.6000

11:16:21 240.0750 -27.9000
11:16:22 239.9833 -33.8000

11:38:10 120.0833 169.0000
11:38:11 119.9917 163.1000

 

[ceptimus]

The closest approach is at 2:54:32 (and at 3 other times).

Not quite. The second hand gets closer to the 120 degree position a couple of seconds after this time (and as the second hand goes round sixty times faster than the minute hand, the overall accuracy is better even though the minute hand is then a little off). That was the basis of my post, but apparently, it's not the answer the OP was looking for.

 

[Jimmy Snyder]

Not quite. The second hand gets closer to the 120 degree position a couple of seconds after this time.

You misinterpretted what I meant. Of the 22 times that the hour and minute hands are exactly 120 degrees apart, 2:54:32 finds the second hand at the closest to 120 degrees from the minute hand. Of course, there are other times that the second hand is exactly 120 degrees from the minute hand, but none of them are among the 22 times in my small universe.

 

[ceptimus]

Yes.

I was looking for the time when, if you call the angles between the three hands a, b, c, then the formula below has a minimum value:

abs(120 - a) + abs(120 - b) + abs(120 - c)

I think this occurs at roughly the time 02:54:34.576 perhaps a few milliseconds before or after.

I agree that with hands moving at a constant rate then at no time are all the angles 120 degrees.

 

[daveb]

Here is a proof (by contradiction) that it cannot happen. Suppose it can. Put the 3 hands in an x-y coordinate system, with the vectors for hours, minutes, and seconds represented respectively by h, m, and s. Since this is a coordinate system, it can be rotated and the hands are still 120 degrees from each other, and these must also be valid times. Rotate the system an angle theta such that the minutes hand points directly at 12. Then the hours hand must by necessity point directly to 4 (or 8), and the seconds hand directly to 8 (or 4). But neither of these are valid times where the three hands are 120 degrees apart. Therefore, the assumption that there is a time is false.

 

[Jimmy Snyder]

Since this is a coordinate system, it can be rotated and the hands are still 120 degrees from each other, and these must also be valid times.

Can you clarify why these must be valid times? For instance, at midnight all three hands are pointing to the 12. If you rotate this by 30 degrees, then all hands are pointing to the 1, but this is not a valid time.

 

[Promethium147]

Here is a proof (by contradiction) that it cannot happen. Suppose it can. Put the 3 hands in an x-y coordinate system, with the vectors for hours, minutes, and seconds represented respectively by h, m, and s. Since this is a coordinate system, it can be rotated and the hands are still 120 degrees from each other, and these must also be valid times. Rotate the system an angle theta such that the minutes hand points directly at 12. Then the hours hand must by necessity point directly to 4 (or 8), and the seconds hand directly to 8 (or 4). But neither of these are valid times where the three hands are 120 degrees apart. Therefore, the assumption that there is a time is false.

That's what I thought a first, but I remembered that the gears inside the clock that correspond to each hand are different, therefore the rotation of each hand is different, and you cannot simply rotate each hand x degrees to any position.

 

[PRDan4th]

The answer is that no time exists when the three hands are exactly 120 deg. apart. I agree with CEPTIMUS that it is very close at 2:54
:34.576, about +-1/6 degree error in two hands. This is also true at 2:54:34.548. Any time between these two times the angles are very close. I also agree with jimmysnyder that when the hour and minute hands are exactly 120 deg. (2:54:32.726) the other hand are off by about 10 deg. from 120 deg.

I cannot postulate a proof for this problem, can anyone?

PRDan4th

 

[Jimmy Snyder]

The answer is that no time exists when the three hands are exactly 120 deg. apart. ... I cannot postulate a proof for this problem, can anyone?

How do you know the answer if you can't prove it? Anyway, parlyne offered a proof in message #14 (but I have not checked it) and I offered one in post #18. What more do you want?

 

[Kurdt]

Ok here is how i tried.

Angle of hands given as a funtion of time:

$$\theta_x = \frac{2\pi}{60} t$$

$$\theta_y = \frac {2 \pi} {3600} t$$

$$\theta_z = \frac {2 \pi} {43200} t$$

where x,y,z denote second minute and hour hands respectively.

The relationship between the angles between the hands can be given as so:

$$\theta_x,y = \theta_x - \theta_y = 2 \pi t \frac{59} {3600}$$

and similarly for x,z and y,z pairs.

$$\theta_x,z = \theta_x - \theta_z = 2 \pi t \frac{719} {43200}$$

$$\theta_y,z = \theta_y - \theta_z = 2 \pi t \frac{11} {43200}$$

It is clear that these differences must be an integer multiple of $$\frac{2}{3} \pi$$ if they are to satisfy the condition set out in the original post. So take the y,z pair equation and place that equal to $$\frac{2}{3} \pi$$ as so.

$$2 \pi t \frac{11}{43200} = n \frac{2} {3} \pi$$

This gives the time as (if we assume t=0 is exactly 12:00),

$$t = n \frac{43200}{33} = n\times 1309.091s$$

Now every third integer will correspond to when the minute and hour hand coincide so we neglect them. What about the second hand? We can find its angle wrt the minute hand via the x,y angle equation above by entering the times we get when multiplying by the integer. The angle we obtain should be a multiple of $$\frac{2}{3} \pi$$.

$$m = 3 t \frac{59}{3600}$$

And after fiddling about in excell the only time m is an integer is when the second and minute hand coincide at four o'clock eight o'clock and twelve o'clock. Thats where my proof fizzles out :rofl:

Perhaps someone could tie it up for just a lowly silly physicist.

 

[Kurdt]

ok I'm fixing my latex I know its awful!

Edit: I see Parlyne tried a similar proof but it looks better than mine.