All clock hands at equal degrees from each other

Question

My dad once asked me:

At what time will the second hand, minute hand, and hour hand on an analog clock all be 120° from each other?

It's a simple question, but I thought it was a fun one to figure out.

(Just keep in mind, each time the second hand moves, the minute and hour hand move just a little bit too.)

If the second hand ran smoothly (rather than jumping abruptly from tick to tick), would that change your answer?

Answer 1

At noon, all three hands are at $0$ degrees from each other.

Let's say the Moment of Truth is after the hour hand has moved around $x$ degrees. In that time, the minute hand moves $12x$ degrees and the second hand moves $720x$ degrees. So we want $x, 12x\pmod{360},$ $720x \pmod {360}$ to be separated by $120$ from each other.

From the minute and second hands we know $720x-12x\pm120$ is a multiple of $360$, which means $59x\pm10$ is a multiple of $30$. From the minute and hour hands we know $11x\pm120$ is a multiple of $360$. So $11x$ and $59x$ are both integers, which means $x$ is an integer (since 11 and 59 are coprime). But then $720x$ is a multiple of $360$, which means $x$ must be $120$ or $240 \pmod {360}$ and so $12x$ is another multiple of $360$ contradiction. So

there is no solution!

Answer 2

If the hour hand's speed is $v$, then it's $12v$ for the minute hand and $720v$ for the second hand. Since the second hand is always at a minute point, so are the other hands for them to be 120 degrees (20 minute points) away from each other. For the minute hand to be exactly at one of them, the second hand must point upright, but then one hand must point at 8 and the other 4, which is impossible!

If the second hand consistently moves, let the digital time be $h:m:s$. Then the hour hand is at $5h+m/12+s/720$, the min hand is at $m+s/60$ and the second hand is at $s$. For them to be 120 degs apart, the fractions representing all of them must have the same denominator. Since this can be the case for $s$ and $s/60$ only if $s=0$ (upright), we'll have to arrive at the same conclusion.

Answer 3

Funny, I thought about exactly this problem a couple of weeks ago.

I reasoned as follows: At 4:00, hours and minutes hands have the correct angle. That angle repeats 11 times in 12 hours. When the hour hand moves 1/11 turn, the minute hand moves 12/11 turns, that is 1 + 1/11. The 120° angle is preserved.

Similarily, when the minutes hand turns 1/11 (or 12/11), the seconds hand turns 60/11 turns, or 5 + 5/11 turns. So, when hours and minutes hands have the correct angle, the position of the seconds hand is a multiple of 1/11 turn from digit 12. But you need it at a multiple of 1/11 turn from digit 8. Since these 2 sets never overlap, an exact solution does not exist.

Answer 4

The best answer is "time to get a new clock" but the two best times are 9:05:25.452 and 2:54:34.548, with angles HM=119.83°, HS=120.00° and MS=120.17°. Python code:

import numpy as np
def bestN(a, b, c, N=1):
    e1 = np.abs(((b-a) % 60) -20)
    e2 = np.abs(((c-b) % 60) -20)
    e3 = np.abs(((a-c) % 60) -20)
    return np.argpartition(e1+e2+e3,N)[:N] # N indices with lowest error

x = np.linspace(0, 60*60*12, 60*60*12*1000, endpoint=False)
h,m,s = x/60/60, (x/60)%60, x%60 # modulo hours, minutes, seconds
[print('%02d:%02d:%.3f' % (h[i],m[i],s[i])) for i in bestN(h*60/12,m,s)]
[print('%02d:%02d:%.3f' % (h[i],m[i],s[i])) for i in bestN(h*60/12,s,m)]

Answer 5

Wrote a small mathematica program just to confirm what rand al'thor has already proved elegantly.

Let $\left(\theta_s,\theta_m,\theta_h\right)$ be the angle in degree made by the second, minute and hour hand respectively in $t$ seconds such that at $t=0$ all three are zero.

We know that at $t=1sec$ $\Rightarrow\left(\theta_s=6,\theta_m=\frac{1}{10},\theta_h=\frac{1}{120}\right)$.

Let $\left(\theta_{sm},\theta_{mh},\theta_{sh}\right)$ be the angle between sec-min, min-hr and sec-hr hands respectively.

Below is the function written in mathematica which takes in one argument which is the time passed in seconds and calculates these three angles at that sec and gives a True output only when all three are equal.

t[sec_] := 
Module[{},
\[Theta]s = (6)*sec;
\[Theta]m = (1/10)*sec;
\[Theta]h = (1/120)*sec;
\[Theta]sm = Mod[\[Theta]s - \[Theta]m, 360];
\[Theta]mh = Mod[\[Theta]m - \[Theta]h, 360];
\[Theta]sh = Mod[\[Theta]s - \[Theta]h, 360];
\[Theta]sm == \[Theta]mh == \[Theta]sh
];

So now we can run this function from $t=1s$ to $t=43200s$ (total seconds in 12 hrs) and check for each tick of the second hand if the given condition is satisfied. This can be done in mathematica as follows:-

DeleteCases[Table[If[t[i] == True, Sow[i]], {i, 1, 43200}], Null]
(*{43200}*)

The only value which the above command gives is 43200 which off-course is when we circle back to where we started. So to conclude we can say that there is no time between t=0 and t=43200 sec where the three angles are exactly equal to each other.

Answer 6

The two closest times I came up with are below, but even those weren't exact. I'll have to go back and figure out what time are better as we get a finer time, say if we had a continual running analog clock and measured 100ths of a second.

5:49:09 and 6:51:51

Answer 7

At 02:54:34 + 394/719 s and 09:05:25 + 325/719 s they'll be as far apart as possible (provided the clock isn't broken).