School-Safe Puzzle Games

## The Complex Clock Puzzle

OK, so here we have it: Bilbao’s 4th and final clock challenge.

Our general audience will be pleased because it is entirely feasible for you to come up with an approximate solution to the puzzle.  To get an exact answer, however, you’ll need to be a math ace.

So for all who asked for harder puzzles (hope you’re still out there Tommy!), here we go:

What is the exact time when the three hands of a clock (hour, minute, second) are closest to being evenly separated (i.e. 120 degrees)

Will list those who can figure out the exact answer here, as they come in.

Update: After discussing things with Bilbao, there is one exact answer to this puzzle, and many incredibly good approximations. So far, no one 1 person has gotten the exact answer. What we do have, however, are several ‘almost’ exact valid solutions, but it seems they are not being reached by the method that gives the exact answer.

While many of these approximations are so close to the exact value that all but your math professor would consider them valid, we’re going to hold out hope that someone will come up with the exact method & answer. We’ll even throw in a Smartkit tshrit if you can do it.

### 66 Comments to “The Complex Clock Puzzle”

1. aaronlau | Profile

Did a quick trial and error method.
It will be at 09:05 and 25 sec, also at 02:54 and 31 sec.
It will be two answers due to mirror image of the movement.
Will confirm my ans and be more exact when I have more time to work it out.

2. diegote | Profile

I arrived at this result using iteration. At the end I did a little correction based on that the secondss hand jumps from one second to the next, it’s not a continuous movement.

I think it happens at 11:16:20, but not completely sure because of my weak method.

3. aaronlau | Profile

Drilled down to a more accurate answer of:
1) 9:05 and 25.45244 seconds
2) 2:54 and 34.54756 seconds

4. bach | Profile

i would say 9:05:27 and 3/11ths of a second

find all times where hr and min hand are 20 hashes apart, choose the one where second hand is roughly 20 hashes away from the hr and min hand.

5. bach | Profile

actually.. there is a better answer. will post later using guess and check.

6. michaelc | Profile

Setting hours and minutes exactly 1/3 (120°) apart, the closest time I’ve found so far is ~9:05:27.27.

Or 9 hours, 5 minutes, 27 3/11 seconds.

This time isn’t exactly evenly seperated, so I’m not 100% sure it is the closest, or the ball park that were playing in. For that, I’ll have to give it more thought.

However,in this case, the hour and minute hand are at exactly 120°, or 240° (1/3 and 2/3). The second hand is “off” by 1/33 or 10.9°.

Getting that as close as I can, I see,

9 hours, 5 minutes, 24 324/349 seconds.

An even closer result putting the second hand much closer, and thus reducing the error in the hour an minute hands.

Best I got so far…

7. hex | PUZZLE MASTER | Profile

It seems there is no exact solution. The closest solutions would be:
h m s Delta1 Delta2 Delta3 avg
0 21 41.82 119.94 120.73 119.33 120
0 43 23.64 120.12 118.55 121.33 120
1 26 47.27 119.76 122.91 117.33 120
1 49 29.09 119.8 122.36 117.83 119.996667
2 32 52.73 120.08 119.09 120.83 120
2 54 34.55 119.98 120.18 119.83 119.996667
3 37 58.18 119.89 121.27 118.83 119.996667
8 22 1.82 119.89 121.27 118.83 119.996667
9 5 25.45 119.98 120.18 119.83 119.996667
9 27 7.27 120.08 119.09 120.83 120
10 10 30.91 119.8 122.36 117.83 119.996667
10 33 12.73 119.76 122.91 117.33 120
11 16 36.36 120.12 118.55 121.33 120
11 38 18.18 119.94 120.73 119.33 120
The times that give an average delta of 120 would be a finer set of results.
Took 6 darn pages of calculations to figure out!

8. hex | PUZZLE MASTER | Profile

Forgot to mention that the finer set of results above depends on the average delta. This is not totally correct, as we would have to compute the least squares sum to get a better solution:
2:54:34.55 and 9:05:25.45 would be the best solutions

9. michaelc | Profile

Whoops! I’ve sent the rocket into the ground! >

I meant 9 hours 5 minutes, 25 275/697 seconds.

Maybe that’s the best guess so far.

10. michaelc | Profile

Another solution that is as equally close as the last one I gave is
~2:54:34.52 where the seconds equal 34 38/73.

I’ll play with this thing more later.

11. hex | PUZZLE MASTER | Profile

Make some minor refinements:
2:54:34.57 and 9:05:25.42

12. hex | PUZZLE MASTER | Profile

Make some minor refinements:
2:54:34.58 and 9:05:25.42

13. trunicated | Profile

Since I’d rather not do math, and would rather use this as a programming exercise, I went ahead and made a clock simulation. It has a threshold so that I can approximate milliseconds. Here’s the pastebin of the code, so you can try it yourself!

What I ended up getting was

05:49:09

with a threshold of 108 ticks (there were 43200 ticks around the face of the clock). This means that our “perfect moment” occurred somewhere between 05:49:08.85 and 05:49:09.15.

I think that’s a pretty good approximation :-P

14. hex | PUZZLE MASTER | Profile

I have not been listed yet as having got the exact answer (which is not an exact solution having 120 degrees exactly). What are the criteria of an exact answer?
If it is requiring a formula, then:
2:54:34.576271… -> seconds = 3600/59x(m/60-1/3)= 34.576271…
9:05:25.423729… -> seconds = 3600/59x(m/60+1/3)= 25.423729…
The above answers make the 3 hands separated by 120.169492…, 120, 119.830508… degrees, which is the closest possible as per my calculations.

15. bach | Profile

i was originally going to do this guess and check, but tried a different approach

first step- find a close time to the desired time, thus setting a workable range of numbers for the equation. i found this approximate time by finding times from 12-11 at which the minute and hour hand were 20 hashes apart, then choosing the time where the second hand was also approximately 20 hashes apart from the minute and hour hand. this time was 9:05:27.27

third step- work backwards to find the equation

here is the equation-

x=seconds elapsed after 9:05
y=closeness to equal angles(my operative definition of this can be inferred from equation, who knows if it’s the best one)

y=abs((x-((x/60)+5))-20)+abs(((65+(x/60))-(45+(x/72)))-20)+abs((((x/72)+45)-x)-20)

fourth step- find out the minimum y value

16. bach | Profile

hahah.. should have said find out minimum x value.

17. Jimmy Anders | PUZZLE MASTER | Profile

My answer is at 02:54:34 + 3145978/5600903 s and also at
09:05:25 + 2454925/5600903 s.

Let a_1 be the angle between the minute and second hands,
a_2 := the angle between the minute and hour hands, and
a_3 := the angle between the hour and second hands.

Taking “time when the three hands… are closest to being evenly separated” to mean the “clock triple” that minimizes:
d( (a_1, a_2, a_3), (120, 120, 120)), where d( x, y) is the normal distance function, I then proceeded this way.

Supposing that the correct time would be very close to one where the hour and minute hand were at 120 degrees, I made a list of the times when the minute and hour hands were at 120 degrees from eachother (using the techniques of the previous problems) and then picked the time where the second hand was closest to being in the correct position to use as a starting point for my calculations.

There were two such times:
t_1:= 02:54:32 + 8/11 s, and t_2 := 09:05:27 + 3/11 s.

Since these times are mirror images of eachother, they are both equally as good. For either time we start out with:
(a_1, a_2, a_3) = (1440/11, 120, 1200/11)
Now let x be the number of seconds after t_1 (or before t_2).
Then x seconds after/before t_1/t_2:

a_1 = 1440/11 – 6*x + [1/10]*x,
a_2 = 120 – [1/10]*x + [1/120]*x, and
a_3 = 1200/11 + 6*x – [1/120]*x.

This is calculated using the fact that the S hand moves 6 degrees per second, the M hand at 1/10 and the H hand 1/120, and then choosing +/- depending on whether the motion of that hand increases or decreases the angle.
Now to minimize that distance formula. Since the square root function is always increasing, it suffices to minimize what’s under the radical in said formula, which is:
[a_1 – 120]^2 + [a_2 – 120]^2 + [a_3 – 120]^2.

After substituting/expanding/combining like terms, I get:

[509173/43200]*x^2 – [1427/33]*x.

A fact from college algebra (which is simple enough to prove) is that the x that minimizes this expression is:

x = – b/[2*a] = [1427/33] / [2 * 509173/43200] = 10274400/5600903.
So, the desired time is both at x seconds after t_1 and x seconds before t_2. These times are 02:54:34 + 3145978/5600903 s and 09:05:25 + 2454925/5600903 s.

18. Kllr Wolf | Profile

a quick guess is 1:50:30. I would have to spend a bit more time than I have to try to come up with something more exact

19. bach | Profile

just slightly changed my equation.. this one makes more sense. the other one’s minimum looked funny, considering what it was supposed to represent. rk, sorry for all these extra posts. feel free to delete the useless ones. i’m pretty sure this is my final answer.

revised equation:

y=abs((((5+(x/60))+60)-(((5+(x/60))/12)+45))-20)+abs((x-(5+(x/60)))-20)+abs(((((5+(x/60))/12)+45)-x)-20)

at the minimum, x=25.451419 y=.055651

20. bach | Profile

ok, since i don’t know calc my last minimum was approximated by looking at a graph. after much grief, i plugged in the equation to my ti83 and just had it find the minimum for me. soooo.

x=25.452014
y=.05563291

9:05:25:452014

21. Roel | Profile

Ok this was a difficult one. I thought about solving it mathematically as follows (details see previous clock riddles):

Formula for angle between hour and minute handle:
H_TO_M = TIME*11*360/(12*60*60)

Formula for angle between hour and second handle:
H_TO_S = TIME*719*360/(12*60*60)

Formula for angle between minute and second handle:
M_TO_S = TIME*708*360/(12*60*60)

And then I would find the minimum of the function:
|120-H_TO_M| + |120-H_TO_S| + |120-M_TO_S|

I couldn’t solve this mathematically so I did it with Excel for every second (43200 rows) and sorted everything.

The best 10 results were:
5:49:9 (1.80 degrees total difference between handles)
6:10:51 (1.80 degrees total difference between handles)
3:37:58 (2.37 degrees total difference between handles)
8:22:2 (2.37 degrees total difference between handles)
9:27:7 (3.12 degrees total difference between handles)
2:32:53 (3.12 degrees total difference between handles)
11:38:18 (3.60 degrees total difference between handles)
0:21:42 (3.60 degrees total difference between handles)
8:0:20 (4.00 degrees total difference between handles)
3:59:40 (4.00 degrees total difference between handles)

Mathematically the three functions H_TO_M, H_TO_S and M_TO_S are triangle waves. The general formula for triangle waves is

y = arccos(sin(x))

But I wasn’t going to start calculating the maximum of three of those functions. I guess one would do that with derivatives.

22. bizarette18 | PUZZLE MASTER | Profile

9.05.25 325/719
2.54.34 394/719

23. Jimmy Anders | PUZZLE MASTER | Profile

Oh, I just realized this morning that I left off the constant term in my expansion at the end. It doesn’t change my answer, I just forgot to explain the omission before (I never bothered to compute it). Whoops!

24. RK | Founder | Profile

Bizarette18 actually got the exact correct answer before I made the update, but it was overlooked by accident. Wow wow wow, way to go Bizarette!!!

25. michaelc | Profile

Here’s another stab at it…

In a 12 hour period, there’s 22 times that the hour and minute hand fall at 120°-240° from each other. If there would be an exact time for all 3 hands to be 120° from each other, it would happen on 1 of those 22 occasions.

Looking at the 22 times possible that the hour and minute hand fall 120 – 240°, and seeing how close the second hand gets to 120° in any of those times, you can see there is no perfect solution that exists. Which is what makes finding an exact solution of the “closest” possible time so difficult I think.

Anyway, there are 2 times that the second hand falls 10.9° (1/33 round) from the 120° (1/3) desired mark. The rest of the times, the second hand is much farther off.

So, those 2 times are 9 hours, 5 minutes, and 300/11 seconds, and 2 hours, 54 minutes, and 360/11 seconds. From here I would assume we would find the closest time possible by adjusting the second hand to make the angles closest to 120°, and that is ignoring the other 20 times in 12 hours that the second hand was farther away from 120° from the hour and minute hand.

Now, this is the tricky part, finding an closest solution that is not going to be exact! There has to be some criteria which says, this time is closer than this time.

My method was input the values into a spreadsheet, and finding out the best value for the second hand such that the standard deviation of the angles was the least possible value. If the standard deviation was 0, the angles would exactly 120° apart. Maybe there’s much better criteria, that I’m not aware of at this point?

Anyway, when you do this, you get the std dev for the one at 2 is just ever so slightly lower than the one at 9.

My final answer for this problem…

2 hours, 5 minutes, 34.56169 seconds.

If someone has a better method, (and goodness knows there must be one!) I’d love to be enlightened.

26. bach | Profile

is there really only one method to find the answer? seems like it would depend on how you want to call “closest to being evenly separated”.

my operational definition of closeness to equal angles was the following-

distance in terms of ticks between…
hr and minute hand= (s)
min and sec hand= (t)
hr and sec hand= (u)

absolute value of…
(s)-20
(t)-20
(u)-20

the lowest possible sum of these 3 absolute values would imo yield the time at which the hands are most equidistant. it’s minus 20 because 20 ticks apart would be equidistant.

then, couldn’t you just make an equation to model possible times and find the global minimum? that’s what i did, at least.

27. aaronlau | Profile

Starting from 9:05.00 the equation is
Error^2 = (360t/43200)^2 + (360t/43200 – 360t/60 + 152.5)^2 + (360t/3600 – 360t/60 + 150)^2
Therefore, simplifying and differentiating for the lowest Error
t = 25.43506824 seconds [Ans]

28. aaronlau | Profile

Nice question.
Ops, left out one term.

Error^2 = (360t/43200 – 360t/3600)^2 + (360t/43200 – 360t/60 + 152.5)^2 + (360t/3600 – 360t/60 + 150)^2

I already simplified the angles or else the equation would have been even longer.

29. Migrated | Profile

Please prove me wrong, but after some thought, isn’t it possible to work out the correct answer, and then to simply snapshot that framework and rotate the three hands as if they were welded, in which case you can get almost an infinite number of answers. Is there any flaws in thinking this way?

30. APEX.JP | Profile

The exact time when the three hands of a clock are closest to being evenly separated is 9:05:25.44

Well done Bizarette18

31. bach | Profile

migrated- i think that because of the varying rates of each of the hands, you cannot simply pick and choose where they will be. there is a finite set of possible hand location combinations.

32. RK | Founder | Profile

from here on out, we’ll be listing only those who get the exact answer.

Bizarette18, we’ll make good on our tshirt offer, shoot me an email at info@smart-kit.com.

33. bizarette18 | PUZZLE MASTER | Profile

I thought perhaps I ought to give an explanation, since I didn’t bother when I originally submitted my answer. First I got approximate answers by finding all 22 times when the hour and min hands were 20 min apart, and then took the two times when the second hands were closest to 20 from the hour and minutes, that came out as 9 and 1/11 or 9.05.27 3/33 and 2 and 10/11 or 2.54.32 8/11. Then to get a better approximation, I compromised with the second hand to 25 5/11 and 34 6/11, assuming that 2 seconds wasn’t going to make much difference to the min and hour hands.
Then to get the exact answer for the 9.05.25 time (I used minutes rather than degrees):
t is time in seconds since 9.05.00 (position of second hand)
m is position of minute hand = (t/60) +5
h is position of hour hand = ((t/60 +5)/12) +45
The total discrepancy, d, from 20 seconds between the hands varies linearly with t, but you get discontinuities because you have to vary whether you take m and h from t or t from m and h as you go from 25 to 26 seconds in order to get positive values.
From 25 to about 25.42 seconds it decreases:
d= (h-m-40) + (h-t-20) + (m-t+20)
So m is eliminated and d= 2h -2t-40
But when m-s passes 20:
d= (h-m-40) + (h-t-20) +(t-m-20)
So this time t is eliminated and d= 2h-2m-80
d continues to decrease, but more slowly until h-t passes 20 at about 25.45 seconds.
Then d= (h-m-40) + (t-h+20) + (t-m-40)
So h is eliminated and d= 2t-2m-40
(I think I typed in all my signs correctly above, but if not you get the general drift.)
Then as t continues past that point d increases, so the minimum is at the intersection of d= 2h-2m-80 and d=2t-2m-40
So h-40 = t-20
Substituting h in terms of t from the expressions at the top:
(((t/60) +5)/12) +45-40 = t-20
This works out as t=18300/719 or 25 and 325/719 seconds
You can then do a similar procedure with the 2.54.34 time, and if it’s correct you get the mirror image time of 34 and 394/719 seconds.

34. hex | PUZZLE MASTER | Profile

Is 0:00:00 excluded (the hands are evenly spaced)?
Would the exact solution yield exactly 120 degrees?

35. diegote | Profile

My previous answer is wrong, I tried to say 11:06:37, but I guess it´s wrong either. My new calculations gives me 9:05:25,342. I´m dizzy…too many spreadsheets and formulas…waiting for the right answer.

36. RK | Founder | Profile

Hey there Hex- 0:00:00 is excluded.

by the way, you must have jedi reflexes-saw your time on cover the spot :shock:

37. hex | PUZZLE MASTER | Profile

My horrible 6 pages of calculations show that there is no result yielding 120 degrees, that is unless the assassin gives a helping hand. I never properly double checked them (too tiring!) so I could be wrong.

As for cover the spot, it was Yoda who did it :shock: (where is the list of smileys to pick?)

38. RK | Founder | Profile

So far it looks like there’s 2 ways to solve this puzzle. If you’re interested, just shoot an email to info@smart-kit.com and I’ll send over both Bilbao and Bizarette18’s answer and method.

Hex-http://codex.wordpress.org/Using_Smilies

39. hex | PUZZLE MASTER | Profile

Having seen Bilbao’s answer, I realize I have not handled the last step in the solution properly.
The last step should be the following:
a1 (angle hours-minutes) = -11/120 x s + 485/2 – 360
a2 (angle minutes-seconds) = -59/10 x s + 30
a3 (angle hours-seconds) = 719/120 x s – 545/2

Total error = |a1 + 120| + |a2 + 120| + |a3 + 120|
which indeed gives the exact value by checking boundary conditions as it is a linear function. It cannot be done by differentiation since the derivative is not continuous (though the answer I received mentions setting the derivative to zero; an explanation would be appreciated)

On the other hand, we can calculate the minimum square error:
Total square error = (a1 + 120)^2 + (a2 + 120)^2 + (a3 + 120)^2
This function has a minimum at s=25.438309

40. hex | PUZZLE MASTER | Profile

My last reply is for 9:05 solution. The same method applies to the second solution.

Well done Bizarette, and great puzzle Bilbao!

41. michaelc | Profile

Well, I’ve found a more exact method after MUCH thought.

I won’t try to write down all the stuff in my comment, but maybe just explain the method somewhat.

My method was to let the hour and minute be equal to 9 and 5, and for the other mirror image, 2 and 54, and leave the exact second time as a variable. Subtract the portion of the positions from each other for hour – second hand, second – minute hand, and hour -minute hand.

From those 3 equations, subtract 1/3 and square the equation. Add these equations up and divide by 3. Thus, basically giving a form of the standard deviation of the angles.

Taking the derivative of the polynomial formed when solving for the standard deviation, gives a minimum value.

For 9 hours, 5 minutes, we find the seconds =
51,810,000/2,036,692

the fraction isn’t simplified, but my brain is tired!

For 2 hours, 54 minutes, we find the seconds =
70,391,520/2,036,692

again, the fraction isn’t simplified, but alas…

This one took me some time to solve, but I feel I learned a few things about finding minimum values during the process.

Thanks for puzzle Bilbao, and kudos to Bizarette!

42. Someone | Profile

I got an answer of 3:38:59 for a total difference of .2 degrees. I just used guess and check (with a computer) until I got that as the closest integer solution.

9:05:25.44 seems to be a popular solution. I don’t know why, maybe there’s something wrong with my program and my math.

43. suineg | PUZZLE MASTER | Profile

This is a little glimpse of the problem not a solution yet, this puzzle is hard thats for sure especially because this are periodic functions:
x being seconds
s(x)=6x with a period of 60 seconds m(x)=1/10x with a period of 3600 seconds and h(x)=1/120x with a period of 43200 seconds(this does not matter because when this happens all the model reset again) example 12:00:00
You add a constant for the periodic rate cs,cm,ch
If you plot this with x being seconds and y being degrees you have an idea of the magnitude of this problem jajaja periodic is cool.
one way could be subtracting integrals that you get three areas then equals the adding of this areas to 360 degrees and then calculate the limit of this adding when x tends to infinity, that should give you the answer( I guess), but all this stuff is really complicated back in college times when I had like 6 billion of more neurons jaja, another more complicated but possible way is by circular functions to get the areas an then do the reverse function to get the hour, minutes, seconds, but this need to be a monster in trigonometry, what really bugs me is that I tried to model this problem in a much easier way, using the simplex method, that is an optimization method that allows you to maximizate or to minimizate a function with the restrictions given, this did not work and I was pretty sure it should work, so I contact a very very good math professor, who could not answer me why simplex did not work with this problem, but in his defense, he never has time jaja; anyone(bilbao and RK included) who can answer if the simplex cannot be used to solved this problem and why that happen, please do so, dont wanna know the answer to the problem just why not the simplex, man

44. suineg | PUZZLE MASTER | Profile

Ok another “eassier way” would be Laplace transform man, hey bilbao this was dangerously hard jajajaj cool.

45. michaelc | Profile

I got RK to e-mail me the solutions that Bilbao and Bizaratte18 arrived at. Very impressive work!

I think when the all the answers are revealed, there may be somewhat of a debate over which is a better answer.

How do you like you’re angles? 1 perfect, and 2 not so perfect, or would you rather have them none of them perfect, but as a whole closest to perfection as can be.

Maybe like saying, how do you like your eggs?

I can’t eat a runny yolk in my egg, but I know lots of folks that wouldn’t have them any other way!

I might can add more (or less) when all answers are revealed.

46. hex | PUZZLE MASTER | Profile

michaelc,
If you examine my reply just above yours (#37), you’ll note that this is one of two methods I used, and I got 9:05:25.438309 for the 1st time, which is exactly yours. I am glad somebody had the same line of thought in this puzzle

But the question is:
Does the square of the error follow/represent the linear error? It seems not, as this method gives a slightly different solution than the linear error. But it is quite interesting as it allows you to differentiate the formula to calculate the minimums/maximums whereas the linear one contains absolute values which complicate the calculations.

Finally, many thanks as well to RK for this great site.

47. michaelc | Profile

Hi Hex,

Yes, I noticed I was not alone in my thinking!

I’m more into statistics, so I’m more inclined to the standard deviation, just because I’m more familiar with it.

I would not go so far as to say the answer using that method is better than the other using the linear error, just equally valid (at least until someone gives me convincing arguement to the contrary, I can change perspectives!).

If you lined up these hands of the watch by either method, you couldn’t visually tell a difference. You have to have a precise spreadsheet or calculator to even calculate the difference!

Yet, as we have seen by this example they are different.

So how do you like your eggs?

48. bilbao | Profile

Hi Hex,
when I sent my solution to RK I deliverately omitted several plots and the derivative expression to enlighten the solution I sent.
Regarding the derivative, it is a 3 page long expression, which Maple 12 showed on the screen. The point is that the method to calculate the error may be linear (vs square), but the error expression itself, before differentiating, is not linear at all since it drags on the previous expressions for x1, x2, x3, y1,y2,y3, d1,d2,d3 which are full of sin, cos, arctan that can be differentiated.

49. bilbao | Profile

However, it is true that depending on the type of error you calculate (linear or square) you get a different solution (extremely close one to another).
I may show both solutions when everything is up.
I chose linear error because in my opinion (so I may be wrong) square error is an statistical delimitation of the error and not an exact absolute value of the error.

50. bilbao | Profile

I mean,

using linear error leads you to a solution like:
‘I assure the distances between hands are: 120+e1, 120+e2, 120+e3 which happens at time t’

using square error (se) leads you to a solution like:
‘I assure the distances between hands are delimited in the following range: 120+/-se and this square error happens at time t’

These were my thoughts in the process of solving this puzzle :-)

51. Jimmy Anders | PUZZLE MASTER | Profile

Oh, we are to minimize the absolute value of the errors? In that case my answer would be at:
02:54:34 + 4334/7909 s and 09:05:25 + 3575/7909 s.

This is achieved by following my work right until the formula to be minimized, which would now be:

abs(a_1 – 120) + abs(a_2 – 120) + abs(a_3 – 120).

This is achieved by writing the equation as a piecewise function and then plugging in the limit points of each piece. The one that yields the smallest answer is 14400/7909 s, which when added to my t_1 and subtracted from my t_2 gives me my two above answers.

52. markodiablo | Profile

OK let’s give this a go. The closest I can see is 09:05 where the angle between the hour and minute hands is just under 120°. So if:

09:05 = Time
AH = Angle the hour hand moves by from Time 09:05
AM = Angle the minute hand moves by from Time 09:05

then:
AH = 360*MINUTE(Time)/60/60 + 360*SECOND(Time)/60/60/60
AM = 360*MINUTE(Time)/60+360*SECOND(Time)/60/60

Looking at a clock, the exact angle between hour and minute hands (AE) is:

AE = 90° – AH + AM
= 90° – [360*MINUTE(Time)/60/60 + 360*SECOND(Time)/60/60/60] + [360*MINUTE(Time)/60+360*SECOND(Time)/60/60]

Going up in seconds between the times 09:00:00 and 09:00:10 (11 seconds inclusive), we get this table:

Time AH AM AE Diff to 120°
9:05:00 0.500 30.000 119.500 0.500
9:05:01 0.502 30.100 119.598 0.402
9:05:02 0.503 30.200 119.697 0.303
9:05:03 0.505 30.300 119.795 0.205
9:05:04 0.507 30.400 119.893 0.107
9:05:05 0.508 30.500 119.992 0.008
9:05:06 0.510 30.600 120.090 0.090
9:05:07 0.512 30.700 120.188 0.188
9:05:08 0.513 30.800 120.287 0.287
9:05:09 0.515 30.900 120.385 0.385
9:05:10 0.517 31.000 120.483 0.483

From the table we can see that the exact time nearest to 120° (i.e. with the least difference) is…

09:05:05

Phew, I’m glad that’s done, I think my brain is melting.

53. hex | PUZZLE MASTER | Profile

Hi Bilbao,
I agree that the linear error method is more logical than the square error one. However, in the linear error method, the total error is the sum of the absolute values of the 3 error components as outlined in your solution (hence it is +/- too). Due to the absolute functions (which your solution uses), maximums and minimums cannot be solved for by straight differentiation.
Assuming that h:m=9:05 and using the simple error method, I have got the following:
a1 (angle hours-minutes) = -11/120 x s + 485/2 – 360
a2 (angle minutes-seconds) = -59/10 x s + 30
a3 (angle hours-seconds) = 719/120 x s – 545/2

Noting that a1, a2 and a3 are negative,
Total error = |a1 + 120| + |a2 + 120| + |a3 + 120|
This function does not have a continuous derivative where ai+120 changes signs. Specifically, the minimum occurs at one of the discontinuities. I guess the same applies to your formula, and hence the question.

When should we expect the next math puzzle? :mrgreen:

54. RK | Founder | Profile

will put up Bilbao’s solution as well as Bizarette18 later today, so you can better discuss this (incredibly) hard puzzle!

55. bizarette18 | PUZZLE MASTER | Profile

I think Jimmy Anders also has the right answer now.

56. RK | Founder | Profile

http://www.smart-kit.com/wp-co.....lution.jpg

57. michaelc | Profile

The minimal linear error solution gives the angles below,

120, 120.1668985, and 119.8331015. Solution has one perfect angle, and the other 2 equally off from each other.

The minimal standard deviation solution gives these angles,

120.0821332, 120.0860218, and 119.8318450.

I’m not convinced that 1 is better than the other.

Standard deviation “penalizes” data that is farther away from the average more so than linear error. That’s about the extent of the difference between the 2.

Love to hear more feedback from others. Anyone else out there have persuading arguements for or against either one of these?

58. hex | PUZZLE MASTER | Profile

The more I think about the choice of linear vs square error method, the more clueless I am. It all depends on the definition of “closest to being evenly separated”. Michaelc has pointed out that farther values are penalized more than closer ones in the case of the square error.

59. KMESON | Profile

Great problem… My physics background stymied me. I assumed that the optimal angles would minimize the sum of the unit vectors pointing in the direction of each hand. Then it is easy to represent each hand with a phasor and write down the expression to minimize. Sadly it was transcendental. I then briefly tried minimizing square errors. Didn’t even think of trying absolute error. Congrats to the solvers!

60. suineg | PUZZLE MASTER | Profile

Congratulations to Bizarette18 and Bilbao, Bizarette18 answer was a much more intuitive answer, like a recursive process, kind of my aproach using the simplex method for the continuos part( I will try to do it today, see if my result match yours and post the comment of the process)
But Bilbao answer was what I have in mind it was to be done to solve the problem in an elegant way using complex numbers, also using Laplace transform could be a way to get to a continuos function that you derivate is equal to 0 in order to get the answer, but is far way more complicated, I did not fully get the details of the solution, especially the first arguments of creating the basics equations, but cool, very cool this problem it make me review Laplace, jaja.

61. markodiablo | Profile

Wowsers B and B that was some solution. Well done.

62. bilbao | Profile

Hi Hex,
now I fully understand what you meant. The fact that Maple showed the derivative of the error function made me thought it was continuous (true) and could be differentiated in every point(false).
So when I tried ‘fsolve’ in Maple and didn´t come up with a solution I just thought I was missing something in Maple (it is first time I use it). So I pinpointed the solution from the plot of the error function.
Eager to quickly share it with you all I guess I rushed.
Jimmy Anders might show us how this last step can be fulfilled in a more elegant way (piecewise,…

63. hex | PUZZLE MASTER | Profile

Hi Bilbao,
Glad you got what I meant concerning the error function. Note that it is a linear function that changes slopes. It is pretty straight forward to find out where the changes of slopes occur (by setting each of the 3 terms in the sum to zero), add to that the boundary conditions for the validity of the function (0<=h<=11, 0<=m<=59, 0<=s<60), and check the function value at each of these points. The minimum is bound to be at one of these points due to the function’s linearity.
Very nice puzzle; When’s the next one due?

64. oddy | Profile

If the second-hand of the clock moves in one second increments the time when they are closest (the amount the three hands deviate from 120 deg apart is 1.15 deg in total) is 5:49:09 – any other time results in larger gaps.

This assumes that the second hand moves in 1/60th, the minute hand moves in 1/3600th and the hour hand moves in 1/43200th of a full circle increments at each click of a second, and there are no in-betweens.

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