# You cannot have 120 degrees of separation, but how close can you get?

If you take a look at a standard three-hand gear-driven analog clock, there is never(see footnote) a time of day where the hands of the clock divide the face equally into 120 degree segments, but you can get close.

What's the nearest set of angles you can get, and what time does it occur at?

Bear in mind that on an analog clock, gears are used to provide the motion of the hands. For every 360 degree rotation of the seconds hand, the minute hand proportionally moves through 6 degrees. For every passing minute, the hour hand moves half a degree (30 degrees per hour). Hands do not "stick" to a division and then instantly jump; all hands (even the seconds hand on a clock that ticks) describe an arc and thus move through fractions of a degree. A clock showing 8:20pm will have a seconds hand at 0 degrees, a minute hand at 120 degrees and an hour hand at 250 degrees thanks to the extra 10 degrees the 20 minutes added on.

Though you can (and should) use any number of decimal places in your workings, in the interests of clarity and simplicity of answering, please present answers that involve at most 2 decimal places in terms of fractions of a second / fractions of a degree. As per good mathematical practices, use more decimal places of accuracy in your workings than in your answer to prevent accumulated rounding errors having a significant impact.

In terms of clarity on whether one answer is better than another, the closer that all angles can be to 120, the better an answer is. Angles of 119, 120 and 121 could be expressed as 120 with an average absolute deviation of 0.6 degree ((1+0+1)/3). Answers of 119.6, 119.6 and 120.8 have a deviation of 0.53 degree((0.4+0.4+0.8)/3) and would be considered "closer" to 120

Note: a similar question has been asked before here, but it was phrased as "is there ever a time of day?" and the answer is "no". This question asks how close you can get and what time does it occur at.

Fair warning: the related question on this site does include an unsubstantiated (no working) but spoiler obscured answer to the "what time?" aspect though the other question didn't ask for it

The answer that @jarnbjo found is demonstrably the best possible.

I wrote in a comment on that answer,

"Because of the continuous motion, and the way [OP has] defined 'best', I'm pretty sure the exact (infinite decimal places) best will be when the hour and second hand meet at exactly 120°. Any slight deviation from this will move the second hand away more than the minute hand moves closer..."

and I stand by this, to elaborate:

OP defines "best" by averaging the absolute values of the differences of the three angles from 120°. If we put the hour and second hand at exactly 120° and the minute hand at 120° plus x, then the differences are (0, +x, -x) which averages to 2x/3. If we try to move it a tiny bit to bring the minute hand deviation closer to zero, the second hand will move away from zero 60 times as fast.
Now OP's "continuity" thing was a bit confusing, but I interpret it as follows: either they are moving at a constant speed; or there are one-second ticks, but even so, for a mechanical clock, the motion from one tick to the next takes a finite time and is continuous, and the three hands are moving in a 1:12:720 ratio. So you do still get a 120° conjunctions of hour and second hands somewhere, and the three hands will be in the same positions as they would be in the constant speed case. (As for time, it would be within one second of when it would be in the constant speed case, though exactly what fraction of a second, that will depend on the nature of the ticking.)

In what follows, let's just assume constant speed.

So now I am going to

divide the circle into 2157 equal slices (about 0.16690°)

which sounds like a pretty random thing to do, but

the minute hand moves 12 times faster than the hour hand, and the second hand moves 60 times faster than the minute hand. Now imagine we start from noon when all hands are aligned.
- When the hour hand has moved ahead one whole slice, the second hand has moved 720 slices, so it is 719 slices ahead of the hour hand, which is exactly one third of a circle, or 120° exactly.
- When the hour hand has moved ahead two whole slices, the second hand has moved 1440 slices, and it is exactly 120° behind.
- When the hour hand has moved ahead three whole slices, the second hand has moved 2160 = 2157+3 slices, so it is exactly aligned with the hour hand.
And the pattern repeats every three whole slices the hour hand moves.

Therefore we know that the solution involves

whole numbers of slices

Now a tiny bit of number-theoretic voodoo,

$$1961 = \frac{1}{11} \mod 2157$$ I mean that 1961 is the only integer (between 1 and 2157) such that, when you multiply by 11, divide by 2157, and take the remainder, you get 1. It exists and is unique because 11 and 2157 do not have a common divisor.
The result is that, when doing arithmetic 'modulo 2157', multiplying by 1961 is like dividing by 11.

This matters because

When the hour hand moves ahead one slice, the minute hand moves ahead twelve, so the difference increases by eleven.
So if you want to know where the hour hand is when the minute hand is exactly X slices ahead of the hour hand, take X, multiply by 1961, divide by 2157 and take the remainder. This will be the only answer that involves a whole number of slices.

Now OP told us that the perfect 120° conjunction doesn't exist. But what does OP know! Let's try to find it

Using the above formula, for X = 719 (120°) we get the hour hand at 1438 slices ahead of noon and for X = 1438 (240°) we get 719. Well that's no good, this is 4 o'clock and 8 o'clock exactly, there will indeed be 120° between the hour and minute hands, and 120° between the hour and second hands, but 0° between the minute and second hands. We want the minute and second hands on opposite sides of the hour hand!

Well, I guess OP was right. Ok, next best thing.

Since we know whole numbers of slices are involved, one slice off is the best we can hope for. So X = 719±1 or 1438±1. But 720 and 1437 are divisible by 3, that will give us the second and hour hands aligned.
You can compute X=718 gives us the hour hand at 1634 slices ahead of noon. You can then compute the minute hand is at 195, and the second hand at 915, and see that indeed the minute hand is 718 slices ahead of the hour hand, and the second hand 719 slices behind.

And the mirror image solution

X=1439 gives 523, 1962, and 1242 slices for hour, minute, and second hands.

Converting this into hours, minutes, and seconds

- 9 hours, 5 minutes, 25 325/719 seconds (325/719 = 0.4520...)
- 2 hours, 54 minutes, 34 394/719 seconds (394/719 = 0.5480...)

Which is what @jarnbjo found. Yay!

• That's awesome. (though are you sure the answer isn't 42?) – Mohirl Dec 6 '18 at 18:14

After a little bit of coding, I found this solution which I believe is correct:

5:49:10 The angles here, if I'm not mistaken, are 120.32º for the hours-minutes hands, 119.1º for the minutes-seconds hands, and 120.57º for the seconds-hours hands, which deviates from the 120-120-120 in a total of 0.57º + 0.32º + 0.9º = 1.79º.

• Please read my answer and tell me what you think . – Rishi Oct 14 at 7:12

The best solution I can find is round about 9:05:25.452 with the hour hand at 272.71210° and the second hand at 152.71210° exactly 120° apart. With the minute hand at 32.54520°, it deviates from the unreachable 'all 120°' solution by 0.16690°. There is a similar solution around 02:54:34.547 with the hour, minute and second hands at 87.28790°, 327.45480° and 207.28790°.

• Don't forget that the question called for 2 decimal places.. – Caius Jard Dec 4 '18 at 13:36
• @CaiusJard I was not quite sure what you meant with that statement. In the paragraph before, you clearly state that you are looking for a solution for a clock with continuously moving (and not sticking) hands. These specifications contradict each other. – jarnbjo Dec 4 '18 at 13:44
• It was intended to prevent a runaway number of decimal places in the answer, i.e. one person saying "It's 12:34:56.789" as an answer, and another saying "actually it's 12:34:56.78874235328745634" (and a third quoting even more dp to achieve a more accurate answer). I appreciate that it quantizes the movement of the hands but limits have to be set so we can know when to stop calculating :) Of course, if you've found a solution whereby 12:34:56.78050 is the best, and both 12:34:56.780499999999 / 7805000000001 are worse etc then feel free to mention, but the actual answer in 2 dp is 12:34:56.78 – Caius Jard Dec 4 '18 at 13:59
• @CaiusJard Then you have to specify the problem more precisely. It actually makes a difference if the second hand is moving continously, or if it moves in discrete steps and if so, if the steps are 0.01s (equivalent to 0.06°) or 0.01°. If the second hand moves discretely, what about the minute and hour hands? Do they move discretely as well and if so, how large are the steps? – jarnbjo Dec 4 '18 at 14:30
• @CaiusJard Because of the continuous motion, and the way you have defined "best", I'm pretty sure the exact (infinite decimal places) best will be when the hour and second hand meet at exactly 120°. Any slight deviation from this will move the second hand away more than the minute hand moves closer. I haven't checked jarnbjo's calculations, but I'm guessing they are correct and this is the final answer. – deep thought Dec 5 '18 at 2:57

I think

There is not a particular solution . If there is one solution then there are such many solutions . In other words if angle between hands is a , b and c then it can occur at 12 times ( for proof see below) . Just fix the hands and carefully rotate the clock you get the visual.

But

All those close solutions will occur when angle between hour and minute is almost 120° . Let take simplest case , it is exactly 4 and second is about $$x$$ seconds more then 40 . Now difference between angles are $${x \over 5}-{x \over 3600}$$ , $${x \over 60} - {x \over 5}$$ , $${x \over 3600}-{x \over 60}$$ . Now for many small values of $$x$$ you can get those approximations and same pairs of differences will occur 12 times. As your definition of minimum occur at $$x=0$$.

Edited:

Proof of statement

Assume simplest case , the angle between hands is zero . Now one clear solution is 00:00:00 . But will it happen again . Assume it happens at x:y:z '0 clock. $$x$$ is [0,12). Difference between hands are $$x+{y \over 60}-({y \over 5}+{z \over 60})$$ $$={y \over 5}+{z \over 60}-{z \over 5}$$$$=x+{y \over 60}-{z \over 5}=0$$

Solutions are $$y={660x \over 133}$$ and $$z={720x \over 133}$$ and $$x$$ is from 0 to 12. So we get same angles 12 times.

I came up with the same puzzle myself and wrote up my solution before finding your web site. Even though my metric for a best answer was different, I got the exact same answers:

• 9 hours, 5 minutes, 25 325/719 seconds
• 2 hours, 54 minutes, 34 394/719 seconds

My derivation is quite difference. Is there a way to send a pdf attachment?

• Hi @user63192, welcome to Puzzling SE! Take the tour if you haven’t already! You can take screenshots of the PDF and add them to your post, or you can upload it to a cloud storage service like Google Drive and put the link here. Hope that helps! – PiIsNot3 Oct 14 at 2:58