CAA: Christian Anime Alliance

OK. I had a weird thought today. It's a math puzzle, and it struck me that some of you may want to help figure this one out.

At what time exactly (if any) will a normal analog clock be divided into equal parts by it's hands?

In other words, 4:00:40 isn't right, because at 40 seconds, the minute hand is 2/3 of the way to 01.

It's a three part problem.

1. Is there a "real" time when this will happen? (I.E. - seconds must be whole integers).

2. If we allow for real numbers, what time will it occur?

3. And for the true mathphiles out there... How many times a day will this occur?

Bonus question: Derive a solution from a function.

Anyway, I'm sure most people will run screaming, but I'm hoping some people will take this as a puzzle. Oh, and there's probably something like this out on the internet somewhere. DON'T POST SOMEONE ELSE'S WORK!!!


GOOD LUCK!!!

Maybe I'm just stupid and I've missed the point but wouldn't 00:00:30 work? Or 00:30:30, or 00:30:00?

I'm confused as to what's being asked....if you take any number on the clock face and draw a line from it to the number it's directy across from (ex: 2 would connect to 8) then that divides the clock exactly in half. So, would an answers like 8:10:00 or 2:40:00 be acceptable? 6:00:00 is the easiest to see. As for how many times this would happen a day (assuming "day" means a full 24 hour cycle)...then would that be 2(6*2) which is 24....er..yeah. There are 6 pairings times 2 to account for 8:10:00 and 2:40:00 like occurances. And then you double it to account for AM and for PM.

Right?

*peeks in*

@_@;;;

*runs out screaming*

She was smart...my head hurts.....@_@

..... school fries my brain enough.... I'm not even gonna try to think about it or I'll make myself frustrated *stays and watches to see if anyone gets it right*

Yay! math ... unfortunately I am taking a semester (or longer) break from real math, but if someone comes up with a formula post it because I miss calc and would love to derive it!

I've seen something like this before.

I would have to say since there is a second hand on many analog clocks (like my wristwatch here), that just 6:00:00 wouldn't work. Just a simple division of three all around, I've come up with 8:00:20 and 4:00:40. Twice a day for both, actually, based on a double 12-hr period as opposed to military time. Then it would be 24 numerals divided by three 'stead of 12. Also complicated would be getting between the hours (like 4:30 or 12:27) and finding a spot for the second hand so that the face is still divided equally into three parts. As for a solution from a function...I'm going to go take a nap.

GAHHH!!!!! I can't think of anything.... but my brain is now fixated on finding a formula to do this with....this is sad...

For part 1, is it somewhere around 12:23:43?

is there a certain number of equal parts there has to be? Can there be 2 or must there be 3? If there can be 2 equal parts it occures at 6:00:00 am and at 6:00:00 pm. But that's if the second hand is aloud to be on the big hand. If there is suppost to be 3 equal parts then i will continue to try and figure this out. BUt i dont' think it's 12:43:43 b/c there is a big chunk about 75% uneveness between 1 and 43. I'm a math minor and i can't figure this one out...
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there may be no answer/solution. I dont' think a clock can divide into 3 equal parts given the second hand. Perhaps it's a trick question or I'm missing something.
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*walks i

Actually, the answer to all math problems is Pi/2.

I actually am very busy this week, so I can't work on this problem just yet. When I get some free time, I will try to solve it, as this does sound like a fun one...

You killed me, by making myself think to hard... *Dies* -_-;;

o_O

Brain... overload....

*falls over in fetal position*

OK. I was going for 3 equal parts. Remember, you can look for fractions of a second if you have to.

As for my homework, that's funny. I haven't had math homework in... Huh. Let's just not go there, OK?

SH: Did you mean 12:23:43?

OK, well, as a short answer... no, it never occurs that the hour hand, minute hand, and second hand split the clock face into three equal parts. It does come very close, though.

Let's just deal with the hour and minute hand first. Let h be the location of the hour hand (in degrees) on the clock face, with 0 degrees at 12 O'Clock.

Let m be the location of the minute hand.

The hour hand traces out 360 degrees in 720 minutes, and so advances 0.5 degrees per minute. The minute hand traces out 360 degrees in 60 minutes, so advances 6 degrees per minute.

We're interested in all circumstances where h and m are 120 degrees apart. We'll start at 12 PM and go with advancing 120 degrees, so our equation is:

h + 120 = m

Substituting, we find

0.5*t + 120 = 6*t, where t is the amount of time that has passed since the start of the hour.

So, we find the answer to our first question is

5.5*t = 120, or

the hands are 120 degrees apart at 12:21.81.

Now, we know exactly where the second hand is here -- at about 48.6 seconds. Problem is that it needs to be at 41.8 seconds to exactly be one-third the way around, so it doesn't fly.

We need to make adjustments to the equation in order to take into account the "starting" time of the hour hand - so, for during the one o'clock hour, we adjust the equation to read

5.5*t = 150, since the hour hand moves 30 degrees per hour. Again, we find that the hour and minute hands are at the correct angle at 1:27.27, but the second hand is sitting at 16.2 seconds, rather than where we want it (about 47 seconds), so that's no good either.

And so it goes. In the interest of completeness, here are all the times when the hour and minute hand are plus/minus 120 degrees apart:

12:21.82
12:43.64
1:27.27
1:49.09
2:32.72
2:54.54 *
3:38.18
4:00.00
4:43.63
5:05.45
5:49.09 *
6:10.91 *
6:54.55
7:16.36
8:00.00
8:21.81
9:05.45 *
9:27.27
10:10.91
10:32.72
11:16.36
11:38.18

I've noted four times when the clocks are exceptionally close to being divided equally. For example, at 2:54.54, the second hand is only 2 seconds away from actually splitting the clock evenly. So I guess it depends how picky you want to be (or how thick the clock hands are).

I'm not going to try an equation (although one is certainly possible).

[EDIT: Started out doing this for two, not three hands! Also had to correct the table.]

Okay, you said from a function, so, in PHP:
[php]// clockexperiment.php by Kaligraphic
// file header omitted


// yes, it's a stupid bubble sort. And yes, it could be done easier
// with the array functions. but this works.
sort($one, $two, $three)
{
if ($one < $two)
{
$first = $one;
$second = $two;
} else {
$first = $two;
$second = $one;
}

if ($three > $first)
{
$sortedarray[0] = $three;
$sortedarray[1] = $first;
$sortedarray[2] = $second;
} elseif ($three > $second) {
$sortedarray[0] = $first;
$sortedarray[1] = $three;
$sortedarray[2] = $three;
} else {
$sortedarray[0] = $first;
$sortedarray[1] = $second;
$sortedarray[3] = $three;
}

return sortedarray;
}

// note that to account for possible slight imprecision due to uncertain data types,
// we will be using approximations. We will display all outcomes that are "good enough".
// where "good enough" is defined as two degrees variance between hand intervals.
// adjust to taste.
oldphilosopher()
{
$maxinterval = 2; // maximum interval in degrees - change here
for ($i = 0; $i <= 86400000; $i++)
{
$hour = ($i * (0.5 / 60000)); //based on the per minute values in milliseconds
$minute = ($1 * (6 / 60000));
$second = ($i * (360 / 60000));
$positionarray = sort($hour, $minute, $second);

if ((abs(($positionarray[0] - $positionarray[1]) - ($positionarray[1] - $positionarray[2])) <= $maxinterval) && (abs(($positionarray[0] - $positionarray[1]) - (($positionarray[2] + 360) - $positionarray[0])) <= $maxinterval) && (abs(($positionarray[1] - $positionarray[2]) - (($positionarray[2] + 360) - $positionarray[0])) <= $maxinterval))
{
echo "At $i milliseconds, the hour hand is at $hour degrees, the minute hand is at $minute degrees, and the second hand is at $second degrees. This is within two degrees of equality.";
}
}
}[/php]
This code will show you all the times (in milliseconds) where the hand intervals are within two degrees. You can adjust the cutoff by changing the value of $maxinterval.

oldphilosopher wrote:SH: Did you mean 12:23:43?

Me, yes. *goes and edits even though she might be completely off on whether it can be solved*

As long as I am here, if you like math, CAA has a math club in the Goof off forum. It's called Equating X: The Society for the Support of Numbers/ CAA's Math Club. We usually do something like this.

It's too bad the silence is gone. I'd have her ask her lil' brother to code it in perl.

Well done you guys. A PHP function, huh? I didn't expect THAT. heh heh. OK. I guess I'll see if I can come up with the other kind of function for this one on my own then.

I think ClosetOtakuOH has gotten it. that makes sinces. I wonder why i didn't figure that one out. I forgot all about degree's. I was hoping for an easy solution I'm good at math but not so good at puzzles :'(.
-----------------------l

I basically just used ClosetOtaku's ideas and put them into PHP. It ought to work, though. (actually, upon thinking about this, you could just check for 120 degree separation (or multiples thereof) and weed out times where two hands are in the same place instead of the calculations I put in to check for equidistance. That would also remove the need for a sorting function. (the one I did is semi-scalable - that is, it's a generic form that could easily be adjusted to an arbitrary number of hands - I was originally going to write it to use the array more, but I wanted to make sure that the code's purpose was obvious to those less familiar with the language.)

It is quite difficult to develop an anlytical expression for this problem, so as has been demonstrated approximate solutions may be arrived at through simulation. The reason for the difficulty is that the angular position of each hand is a function that is modulo 2*pi (on a 24 hr clock), which is not an easy situation to deal with. You could also represent the position of each hand as a sawtooth function, but that's not really any help. In any event, both expressions result in a periodic, discontinuous function that is not easy to manipulate.

OK. Here's a starting point to work it with fractions. Can someone point out where (if anywhere) this function is off?

Time = approx 08:20:03.9005736

Think like this for a minute... I started by assuming that one time this might work is around 8:20, which (without min/hr hand advance) puts all 3 hands on the numbers. Now, we have to count for the fact that the hour hand is part of the way to 9, the min hand is just past the four, and the second hand is what done it to 'em.

...Prepair for major jump in logic...

So our function has to account for the number of seconds in a 12 hour period (60 * 60 * 12 = 43200). X is the number of seconds past 8:20 we are...

Our function, therefore, looks like this:

43200 = x + 60(20 + X) + 3600(8 + x)
43200 = x + 120 + 60x + 28800 + 3600x
14280 = 3661x
x = 14280/3661
x = 3.09005736 seconds.

Convinced?




Incidental Note: I'm very evil, so I may very well have done this just to show you "logic errors" people make.

NO! the thing is, if the minute hand was on 30, then the our hand would be in BETWEEN two numbers! therefore the minutes ins't "20" or "40" or so... hmm intresting

I have a solution based on a different approach than what I tried earlier. Here I don't try to measure everything using a common unit. I haven't tried oldphil's logic yet, but here's my own.

In this case, I have assumed continuous motion of the hands on a 12-hour clock (although a 24 hour clock will work just as well).

The angular frequency of each hand is:
w1=2*pi/12 radians/ half-day
w2=2*pi/60 radians/hour
w3=2*pi/60 radians/min

This means that when a 12 hr/60 minute, etc period has passed, the hand will have completed one revolution (through 2*pi radians)

We know that the angular difference between each hand must be 120 degrees or ~2.094 rad. So, for one configuration,

w1*t1-w2*t2=2.094
w2*t2-w3*t3=2.094
w3*t3-w1*t1=2.094

This is a system of 3 linear equations and 3 unknowns. However, the resulting matrix is rank 2, meaning that we will have one free parameter which we shall allow to be t3 (seconds), which is restricted to the interval [0,60).

This gives us:

t3 is in [0,60)
t2=(w3*t3-2*2.094)/w2
t1=(w3*t3-2.094)/w1

Acceptable solutions occur when t2 is in [0,60) and t1 is in [0,12). One such solution exists at:
t1~=5 (hours)
t2~=5 (minutes)
t3=45 (seconds)

Neither t1 nor t2 are constrained to be integers, but for this solution t3 was (although it need not be). Please note that there is some associated approximation error, since I truncated the radian measure of 120 degrees to 2.094. *edit* I seem to have made a mistake here in decoupling the times. There may be no way of avoiding modulo 2*pi functions after all.

wow, you're very smart!