It's Good Friday!

I want to make perfect hot cross buns by superimposing two 12-hour clock faces on top of each other. Is this possible, and if so, how many ways are there of doing so?

In other words, how many pairs of times are there which, when displayed together using the hands of two 12-hour clocks, form a perfect right-angled cross?

(For example, the legs of the cross can't be perfectly vertical and horizontal because, while 3:00 and 6:00 would each give two legs of that cross, the other two would then be unattainable in each case.)

Disclaimer: I don't know the answer to this question, but I imagine that solving it will be a fun Easter puzzle challenge.

  • $\begingroup$ the minimum unit is the minute for the clock? I mean, in terms of angles between the clock hands. 3:00:00 and 3:00:22 is still considered a 90 degree angle? or is it the second? or even worse? $\endgroup$
    – Marius
    Apr 14, 2017 at 11:34
  • $\begingroup$ @Marius The angle between them has to be exact, yes: 89.9 or 90.1 degrees won't do. $\endgroup$ Apr 14, 2017 at 11:48
  • $\begingroup$ You might ask another question with 3 clocks now, but I think you will only get the 11 obvious configurations (3, 6, 9 and pi/11 rotations). $\endgroup$ Apr 14, 2017 at 12:19
  • $\begingroup$ Interesting that it can be solved in a lot of different ways. $\endgroup$
    – Nautilus
    Apr 15, 2017 at 11:49

5 Answers 5


I think I have the simplest answer.

The hands of a clock move at relative speeds of 1 (hour hand) and 12 (minute hand)

When one clock has the two hands at an angle of 90 or 180, we want the other clock to also have the two hands at an angle of 90 or 180.

Essentially, the hour hand on one clock will have to travel an angle of {90, 180 or 270} to match the other clock.

but in the same time, the minute hand would have traveled 12 * {90 or 180 or 270}, which is a multiple of 360, which means that it would be back where it started. hence the two clocks would have only 3 DISTINCT lines, which would not form a cross.

  • 1
    $\begingroup$ I'm accepting this answer as the most short, elegant, and unexpected solution. $\endgroup$ Jun 27, 2017 at 12:03

In the philosophy of Street fighting mathematics, whenever you can't use beautiful math, use Street fighting math !

So, let's divide into two cases:

  1. The angle in each of the clock is pi, and the hour and minute hand in each make a line. This configuration is clearly impossible, because you can imagine yourself as a bug sitting in the minute hand, and thus there are total 11 possibilities. Now, as 11 and 2 are coprime, no solution this case.

  2. The angle between hour and minute hands is pi/2, i.e they are perpendicular to each other. Repeat the analogous reasoning.

  • 1
    $\begingroup$ Huh? I don't understand this reasoning. It's certainly possible to have the hour and minute hand at an angle of pi or pi/2 apart from each other. $\endgroup$ Apr 14, 2017 at 11:53
  • $\begingroup$ @randal'thor You didn't read my answer carefully. I didn't meant to intend that ” it's impossible to have the hour and minute hand at an angle of pi or pi/2 apart from each other.”, but I meant that "Is impossible to overlay two such configuration such that they are mutually perpendicular, because the angle between two such configuration is multiple of pi/11, and you can't make pi/2 with them " $\endgroup$
    – Pripop
    Apr 14, 2017 at 11:58
  • $\begingroup$ OK, I understand what you're saying now. But are those 11 configurations necessarily going to be equally spaced (i.e. at gaps of pi/11)? $\endgroup$ Apr 14, 2017 at 12:00
  • $\begingroup$ @randal'thor Imagine from the POV of the bug and use symmetry. $\endgroup$
    – Pripop
    Apr 14, 2017 at 12:02
  • 1
    $\begingroup$ @Pripop I find "the POV of a bug" to be an incredibly unhelpful suggestion in deducing that. Unless someone is already intimately familiar with the pattern of logic you have in mind, I doubt they're going to figure it out without a bunch of extra research. Since this isn't a math site, I highly recommend making the logic explicit in your answer. $\endgroup$
    – jpmc26
    Apr 15, 2017 at 2:01

So, the first question is

when the two hands of a clock are a multiple of 90 degrees apart.


one goes round at 1/12 of a turn per hour, or 1/720 of a turn per minute; the other, at 1/60 of a turn per minute. So the (signed) angle between them changes at 1/60-1/720=11/720 of a turn per minute, so every 1/11 of a minute (note: there are 12x60x11=7920 of these per 12-hour period) the angle increases by 1/720 of a turn. Let's call 1/11 of a minute a "tick" and 1/720 of a full turn a "twist". So we have 7920 ticks per halfday, and 180 twists per 90-degree change, and the angle between the hands increases by one twist per tick. The hands start out both pointing upward, so every 180 ticks they will be a multiple of 180 degrees apart from one another. This happens 44 times per halfday.


between each pair of these the absolute angle of, say, the hour hand changes by 180 twists, which is to say 1/11 of a right angle. So these times fall into 11 classes of size 4, within each of which all the angles differ by multiples of 90 degrees. Each of these classes has the same structure: there is one time at which the hands are coincident, one at which the minute hand is 90 degrees ahead, one at which they are opposite, and one at which the minute hand is 90 degrees behind.


in each class, note that the minute hand points to the same place at all four times.


there is no way to put two of them together so as to cover all four arms of the cross

and the required answer is



It is...



If it's $a:b$, it appears as $5a+b/12:b$ on an analog clock. Regardless of the position of the hands, they are always $15r$ ($r$ being an integer) points ahead of or behind each other. Likewise, there are $3p$ ($p$ being an integer) hours between them.

The distance between the first clock's hands is $|5a-11b/12| = 15x$ with $x$ being an integer. Let the second clock indicate $a+3n:b+15m$ ($n$ and $m$ being integers, $|n|, |m| <4$). Then it appears as $5a+15n+(b+15m)/12:b+15m$. Since the angles between the hands are equal on both clocks, $15n-11*15m/12$ too is a multiple of 15, making $m$ divisible by 4, therefore 0. That means both minute hands must overlap, creating a contradiction.


For 90 degrees angles.

during 12h. There are 22 positions in which the clock hands form a 90 degree angle.
This happens because the minute hand does 12 rotation and the hour hand does 1. If we consider the hour hand as reference (stand still), the minute hand does 11 revolutions.
since 270 degrees is the same as 90 degrees, we get a total of 11 + 11 cases.
I looked online for these because I didn't have the patience to calculate them end ended up here.
The values are.
1. 0:16:22
2. 0:49:6
3. 1:21:50
4. 1:54:33
5. 2:27:17
6. 3:0:0
7. 3:32:44
8. 4:5:27
9. 4:38:11
10. 5:10:54
11. 5:43:39
12. 6:16:21
13. 6:49:6
14. 7:21:49
15. 7:54:33
16. 8:27:16
17. 9:0:0
18. 9:32:43
19. 10:5:28
20. 10:38:10
21. 11:10:55
22. 11:43:38


The idea would be to find 2 of these combinations that are exactly opposite to each other making a cross.

For 180 degrees angles.

The idea is the same but here it should be simpler because there are only 11 possible combinations. (180 degrees clock-wise is the same as 180 degrees counter-clock-wise).
Values for 180 degrees taken from here:
1. 0:32:43
2. 1:38:10
3. 2:43:38
4. 3:49:5
5. 4:54:32
6. 6:00:00
7. 7:05:27
8. 8:10:54
9. 9:16:21
10. 10:21:49
11. 11:27:16


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