Watches in advertisements tend to be set to 10:10 (unless you're Apple). The reason seems to be that the hands are symmetrical when they're at 10:10.

Except they're not. The hour hand moves continuously, not discretely; that is, at 12:30, for instance, the hour hand will not be pointed straight up, but rather halfway between the 12 and the 1.

Well, to make things even more confusing, the minute hand doesn't move discretely, either.

Now, I don't know about you, but I hate things being imprecise. So where should the second hand be such that the hour and minute are symmetrical? And I mean exactly where should it be. If that requires a fraction, or an infinite series, or whatever, so be it. But no rounding.

And just because I'm nitpicky, would you mind working out the same math for all symmetrical pairs, one for each hour? (That is, roughly 1:55, 2:50, 3:45, etc. 12:00 and 6:00 are too easy and are therefore not required for our purposes.)

I'll tell you all in advance: the second hand won't be symmetrical. That just drives me nuts, but I guess I'll have to live with it. Such is life. But please tell me: is there any time for which the three hands are rotated at the same interval around the clock (that is to say, 120º or 2/3πrad between each of the three hands)?

Oh, and since I'm still OCD, would you mind showing your work? You know, I want to make sure it's exact...

  • 1
    $\begingroup$ My OCD demands I point out that on many (most?) timepieces, the second-hand does not move continuously, but discretely. Now what?! :) $\endgroup$
    – Rubio
    Jul 3, 2017 at 16:42
  • $\begingroup$ @Rubio I'm the one writing the post - I can only be nitpicky about my watches. :) Just go with it - it's 100% solvable even with that criterion. $\endgroup$
    – DonielF
    Jul 3, 2017 at 17:01
  • $\begingroup$ Something something 1/11th something something $\endgroup$
    – Dr Xorile
    Jul 3, 2017 at 18:55
  • $\begingroup$ What is symmetrical? $\endgroup$
    – paparazzo
    Jul 3, 2017 at 20:16
  • 1
    $\begingroup$ How are 12:30 and 6:30 'easily' symmetrical? $\endgroup$
    – boboquack
    Jul 3, 2017 at 23:19

3 Answers 3


First question:

At noon, both hands point upwards, which is symmetrical. The next symmetrical state is a little after 0:55, the next after that just after 1:50, and so on. If you count them, it turns out there are 13 such symmetrical positions, until it repeats 12 hours later.

These positions are evenly spaced. The hour hand and minute hand move the same amount each time. Therefore, the symmetrical times occur every $\frac{12}{13}$ hours, or every $55$ minutes $23 \frac{1}{13}$ seconds.

Just to check: the hour hand moves $30$ degrees per hour, so it does $\frac{12}{13} *30 = \frac{1}{13} *360$ degrees between symmetrical times. The minute hand moves 12 times as fast, so it goes $\frac{12}{13} *360 = (1-\frac{1}{13}) * 360 = -\frac{1}{13} * 360 \mod 360$, which is the same amount as the hour hand but in the opposite direction, as required.

The second question:

This can be done in a similar way. This time however, the positions in which the minute and hour hands have 120 degrees between them occur 11 times in a span of 12 hours, i.e. every $\frac{12}{11}$ hours. The hour hand moves at $30$ degrees per hour, the minute hand at $360$ degrees per hour. The difference grows at $360-30=330$ degrees per hour. At noon the hands are together, so at $\frac{120}{330} = \frac{4}{11}$ hours after noon they are at $120$ degrees for the first time. Therefore, the 11 times are $\frac{4+12k}{11}$ hours after noon for $k=0,1,...10$.

Again, the hour hand moves $30$ degrees per hour, so after $\frac{4+12k}{11}$ hours it has moved $\frac{4+12k}{11}*30$ degrees.

The minute hand moves $360$ degrees per hour, so after $\frac{4+12k}{11}$ hours it has moved $\frac{4+12k}{11}*360 \equiv 120 +\frac{4+12k}{11}*30 \mod 360$ degrees.

The second hand moves $60$ times as fast, so after $\frac{4+12k}{11}$ hours it has moved $\frac{4+12k}{11}*360*60$ degrees. Unfortunately for no whole value of $k=0,1,...,10$ is this equal to $240 +\frac{4+12k}{11}*30 \mod 360$, so there is no solution.


Let's pretend everything moves continuously, and measure time in hours, where noon or midnight is exactly hour zero. And our unit of angles will be the revolution. Then

at time t, the minute hand is at angle t-n where n is the integer part of t, and the hour hand is at angle t/12. These are symmetrical when t-n = 1-t/12, or when 13/12 t = n+1, or when t = 12/13 (n+1), for t=0,1,...,12. Now, at time t the second hand, which moves 60x faster than the minute hand, is at angle 60t minus the appropriate integer. So the requested angles are 720/13 times 1,2,...,13, modulo 1. Multiply these by 2pi or 360 to get angles in radians or degrees, if you want.

As for the final question:

we require one of two conditions, depending on the order of the hands. Specifically, we need 60t = t+a+m and t/12 = t+b-n for some integer m,n, where either a=1/3 and b=2/3 or a=2/3 and b=1/3. That is, we need 59t = a+m and 11t = 12n-12b. The first of these requires that 177t be an integer. The second requires that 11t be an integer. Since 177 and 11 have no common factor, this requires that t be an integer. But then a = 59t-m is an integer, which is impossible since a is either 1/3 or 2/3. So no, the three hands never divide the clock face into three equal parts.

  • 1
    $\begingroup$ Very good. I worked out the problem from a slightly different angle (so to speak), but correct answer nonetheless. Just for clarification, would you mind adding in the times themselves? $\endgroup$
    – DonielF
    Jul 3, 2017 at 17:43

I did this based on number of seconds from noon, Because Reasons™.

The angles work out such that:
s / 10 mod 360 = angle of minute hand (in degrees, clockwise from noon)
360 - ( s / 120 ) = angle of hour hand (in degrees, counterclockwise from noon)
We want those to be equal; setting them equal to each other and solving gives solutions of: $$s=\frac{43200n}{13}$$ for n=0,...,11

Solving these gives decimal times as follows:


  • $\begingroup$ Much closer to how I did it, surprisingly, and also correct. $\endgroup$
    – DonielF
    Jul 3, 2017 at 18:14

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