"Mercedes" Configuration of the Clock Hands - Is It Possible?

Question

At an evening party the clock on the wall stroke 8 o'clock. Several seconds later one of the guests commented: 'Look everybody, the time is now 08:00:20 and the clock hands form a "Mercedes" configuration: the hour hand points at 8, the minute hand - at 12, the second hand - at 4; the angles between the hands are equal!' I objected immediately, because in that moment the minute hand had moved ⅓ of a minute past 12, the hour hand was also little after 8 (assuming of course that each hand rotates at uniform speed - for example at 08:30:00 the hour hand is half-way between 8 and 9), so that the hands formed approximately, but not exactly, three 120° angles in that moment. The company agreed, but a question arose is there a moment when the exact configuration is possible, like 3 vectors, connecting the centroid of an equilateral triangle with its vertices?
I recalled another well-known problem: to prove that the 3 clock hands point at the same direction only at 12:00:00 noon and 12:00:00 midnight. The proof uses that 11 = 12 - 1 and 59 = 60 - 1 are relatively prime. The same consideration can help here too, but I would like to award the best answer to the easiest and shortest possible solution, submitted by our contributors.
Suggested category: 
Cars & Transportation > Car Makes > Mercedes-Benz

Mike · Accepted Answer

The hour hand position completely determines the positions of the minute hand and second hands.
Minute hand moves twelve times faster than the hour hand, and the second hand 720 times faster.
So they start equal at 12, from there you have k degrees, 12k degrees, and 720k degrees in H hours,  must have remainders when divided by 360 that are 120 apart, l, l+120, and l+240 in some order.  We can see then that in 2H hours, the angles would all double, and they would be 2l 2l+240 and 2l+480 which would also be a benz symbol.  now in 3h hours, you would have 3l, 3l+360, and 3l+720 which are all lined up, and can only be at 12:00  So h would be 4 or 8 which does not work.

Brice T · Answer

Indeed, if you count the time t in hours and the angles in turns, then:
- the angle made by the hours' hand is: a = t/12 (once around in 12 hours)
- the angle made by the minutes' hand is: b = t (once around every hour)
- the angle made by the seconds' hand is: c = 60t (60 times around per hour)

For the coinciding hands problem, you seek a value of t such that all angles are the same modulo an integer number of turns, i.e. you seek two integers n and m such that: b = a+n and c = b+m 
b=a+n, so n = b-a = t.11/12 or t = n.12/11
c = b+m, so m =b-c = t.59 or t = m/59
n, m must verify:
n.12/11 = m/59
59*12.n = 11.m
Since 11 is coprime with 59*12, this can be possible only if n is a multiple of 11, i.e. t is a multiple of 12.

For the "Mercedes clock" problem, you only need to add 1/3 of a turn to the formulation of the problem.

The minutes' hand must make a 120° angle with the hours' hand:
b = a+n+1/3
3(b-a) = 3n+1
11/4.t = 3n+1
t = (12n+4)/11

The seconds' hand must make another 120° angle with the minutes' hand:
c = b+m+1/3
3(c-b) = 3m+1
3*59.t = 3m+1
t = (3m+1)/177

We must have:
(12n+4)/11 = t = (3m+1)/177
2124.n + 708 = 33m + 11
33.m - 2124.n = 697
3.(11m - 708n) = 697
However, 697 is not multiple of 3, so there is no solution: the "Mercedes clock" does not exist.

You may also try with 240°, but the result will be the same. If after a time t, the minutes hand lead the hours' by 240° and the seconds' lead the minutes' by another 240°, then after a time 2t you would have the situation we just proved impossible, where there is 120° between hands.

Anonymous · Answer

Apparently it is impossible unless you're in space and approaching the speed of light.

On Earth, however...
We need to consider the 22 times where the hour and minute hand will be 120 or 240 degrees from each other.  These are, approximately:

120 (going clockwise)

00:21:49.1
01:27:16.4
02:32:43.6
03:38:10.9
04:43:38.2
05:49:05.5
06:54:32.7
08:00:00.0
09:05:27.3
10:10:54.5
11:16:21.8


240 (=120 counterclockwise)

00:43:38.2
01:49:05.5
02:54:32.7
04:00:00.0
05:05:27.3
06:10:54.5
07:16:21.8
08:21:49.1
09:27:16.4
10:32:43.6
11:38:10.9

Checking each, we can see that the second hand is never exactly 120 degrees from the hour or minute hand.  It's also interesting to note that after 8:00, the minute/second combination just repeats - it's all mod 11, so at 00:21:49.1, the minute hand is 4/11 of the way through the hour, then 5/11 at 1:27:16.4, and so on.  Therefore, we really only had to check how far apart the second and minute hand were at 1/11 rotation through 11/11 rotation.

2) I'll take your word for the second one, given the 1:12:720 ratio.  My brain is too fried after doing the first problem.  Maybe I'll get some sleep and take a crack at it, but at first glance, precision seems impossible outside of 12:00.

Fred · Answer

After working on this question, then returning to it, I find Mike's answer, which I like. Here's another, a little more along the lines suggested by Duke.

First note, as Mike does, that, compared to the hour hand (H)'s rate,
the minute hand, M, turns 12 x that rate;
the second hand, S, turns 720 x that rate.

Now follow the motions of M and S in the reference frame of H.
In that frame, M takes
12/11 h
to make one full sweep; S takes
720/719 min = 12/719 h
to do so.

Thus, every 12/11 hour, H and M align. Call this period a "long hour," or, LH. There are 11 of them in the 12-hour cycle in which all 3 hands repeat their mutual configuration.
One full sweep of S thus takes 12/719 h = 11/719 LH.

Now the "trigram" Mercedes-emblem configuration requires, at minimum, that M must be ⅓ turn ahead of, or behind, H. This will happen ⅓ and ⅔ LH past every LH, or 22 times in the 12-hour (= 11 LH) cycle. These all occur at integer multiples (but not at every integer multiple) of ⅓ LH = 4/11 h. But in each of these intervals, S goes
(⅓ LH) / (11/719 LH) = 719/33 = (22 - 7/33) revs.

And in order for S to be ±⅓ rev away from H, requires that for some integer (A) multiple of 719/33,
(719/33)A = B ± ⅓
where B is an integer.
But that can happen only if A is a multiple of 11, in which case, that multiple of
11(⅓ LH = 4/11 h) = 4 h
has passed since 12 o'clock, in which case, M and S align, and there will be no trigram.

EDIT (revised):
MathPhD brings up a point closely related to one I was considering broaching, but did not. I was going to pose it as a least-squares minimization of the deviation of the 3 angles from 120º.

Not having actually tried this, I realized, however, that the solution would be very close to an exact trigram, because the RMSD for 08:00:20 is already very small.

I now believe the solution for this form of the problem is close to that time + 1 LH. And there will be a "twin" solution that is a mirror image of the other -- 12:00:00 minus the other.
For 08:00:20, it would be 03:59:40, e.g. (for illustration only; these aren't quite exact solutions.)

EDIT2:
@ MathPhD: I think you can pare those 720 cases down right away, to 22 or 44, by eliminating the hour hand-minute hand (H-M) separations that are greater than the closest-to-120º cases, in which the second hand ≈bisects the ≈240º angle made by H & M.

EDIT3:
I think I've narrowed that "closest-to-trigram" problem down to 1 caser (along with its 12:00:00 complement).
Express all angles as fractions of a full turn. Start with all 11 of the [M=H+⅓] cases. (The 11 [M=H-⅓] cases will be exact mirror images of these; 12:00:00 complements.)
First take 08:00:00, where M=0, H=⅔, S=0

Now each successive case is 1 LH = (12/11)h = 1h (05&5/11)m = 1:05:27&3/11 later than the previous, in which H & M each advance 1/11, and S advances 5/11, and so, gains 4/11 on M. And because 11 is prime, each of the 11 cases has S advanced on M by each of the 11 residuals mod 11.
But we're trying to get it close to where it needs to be, M + ⅓. Closest to that is 4/11 = ⅓ + 1/33, and that is achieved at the second case:
08:00:00 + 1 LH = 09:05:27&3/11
To get the optimal case from that, reverse the S hand by just enough to make it bisect the 240º angle (noting that this angle will increase a little) made by H and M.
Tweak slightly to suit your particular choice of "optimal."

EDIT4:
S exactly bisects the M→H angle at
09:05:(366900/14509) = 09:05:25.2877524295...
and this, along with 12:00:00 - this = 02:54:34.7122475705...
are the solutions to MathPhD's question.

On another note, I'm favoring M3's argument at this point, as briefest & most elegant. Wuddaya think?

MathPhD · Answer

An interesting related question is to ask for the maximum of the minimum of the three angles between the hands.  Each of these angles will be given by linear equations in time (although which depends on several things).  Therefore, their derivatives are never zero and all minimums and maximums occur where two of these angles become equal.

It appears there are 720 situations that need to be examined.  I have not done this yet, but the answer is between 119 and 120.

Indica · Answer

Minute hand sweep out 6°/min and second hand 6°/sec. 

When these show x min, y sec (0≤x,y<60), angle between them = |6y−6(x+y/60)|

If this = 120 or 240 then 59y/60−x = ±20 or ±40   →   59y = 60(x±20) or 60(x±40)

Only solutions in range are x=20, y=0 and x=40, y=0   …   (i)

To complete star, hour hand must bisect other two and so be at mark 40 or 20 exactly

This contradicts (i) both of which indicate fractions of an hour.

M3 · Answer

consider the minute hand frozen at 0°

to maintain valid clock positions, the second hand is then moving at 354°/min, and the hour hand at 5.5°/min in the reverse direction

we need one of each of these hands to be at 120° & 240° positions
second hand: 120/354 & 240/354 --> 60/177 and 120/177 mins
hour hand : 120/5.5 & 240/5.5 --> 240/11 and 480/11 mins

since all the 4 times are distinct, 
we can't have a "Mercedes" Configuration

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"Mercedes" Configuration of the Clock Hands - Is It Possible?

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