Thanks to Gerry F. in Tampa for raising this interesting question. Given: a perfect watch with three sweep hands (hour, minute and second). (By “sweep”, I mean that the hands move continuously and do not “jump” every hour, minute, or second.)

Find: a list of all times at which the three hands divide the circle of

the clock into three equal sections – that is to say, any pair of hands

will form a 120° angle, or, in radians, an angle of 2π/3. (For the rest of the post I will use radians.)

The remainder of this post is my proof of the surprising result. If you’d like to try solving the problem yourself, stop reading now…

OK. First, let’s define some terms.

Let *t* be the time. I’m going to make a rather strange choice for the units. I have the obvious choices of hours, minutes, or seconds – instead, I’m going to choose a unit of **half-days**. So 12:00 corresponds to *t* = 0 or 1; 4:00 corresponds to *t* = 1/3; 9:00 corresponds to *t* = 3/4, etc. Note the clock has periodic behavior with a period of 1 (in units of *t*.)

Let *H* be the directed angle between 12 and the hour hand, increasing from 0 (at 12:00, *t* = 0) through π/2 (at 3:00) and all the way ’round to 2π at 12:00 again. Note that at 9:00 *H* is defined to be 3π/2 and not merely π/2. Note that *H* is also periodic with period 1.

Let *M* be the directed angle between 12 and the minute hand similarly. Note *M* is periodic with period 1/12: the minute hand goes all the way ’round in an hour, which is 1/12 of a half-day.

Let *S* be the directed angle between 12 and the second hand. Note *S* is periodic with period 1/720: the second hand goes ’round 720 times in a half-day.

We can now write explicit formulas for *H*, *M*, and *S* in terms of *t*:

Define *θ _{xy}* as the directed angle between hand

*x*and hand

*y*. In particular:

The key point is that the *angles between the hands are periodic.* The angle between the minute and the hour hands cycles from 0 through 2π exactly 11 times every half-day; the angle between the minute and the hour hands, exactly 12*59 times; and the angle between the hour and the second hands, exactly (12*60 – 1) = 719 times.

Now that we have formulas for the three relevant angles, we can answer the question “when are these angles 2π/3 or 4π/3?” In particular, define {*t _{xy}*} as the set of times at which the

*x*and

*y*hands make either a 2π/3 angle or a 4π/3 angle.

Let’s consider the angle between the hour and minute hand first.

So the angle is 2π/3 (or 4π/3) at multiples of *t* = 1/33 of a half-day, or every 12*60/33 = 21 and 9/11 minutes… except that at every third multiple of 21 and 9/11 minutes, the hands coincide, so those don’t count.

But rather than think of it as “21 and 9/11 minutes”, I’d like to urge you to think of it as “1/(3*11) of a cycle.”

Now consider the angle between the minute and the second hand:

The angle between the minute and the second hand is 2π/3 or 4π/3 at multiples of 1/(3*12*59) of a half-day (except that, as before, at every third multiple of 1/2124 the hands coincide instead of making a suitable angle.)

The following is a demonstration that {*t _{HM}*} and {

*t*} have no elements in common. This implies that the hour/minute angle and the minute/second angle are never 2π/3 or 4π/3 simultaneously. As this is a necessary condition for the three hands to divide the clock evenly, the three hands

_{MS}*never*divide the clock evenly.

Write {*t _{HM}*} and {

*t*} in lowest terms. How many factors of 3 are there in the numerator of each fraction? In the denominator?

_{MS}The numerator is easy. The numerator is never a multiple of 3 to begin with, and cancelation of common terms in the numerator and denominator will never add a factor, so after cancelation there are (still) no factors of 3.

The denominator for each *t _{HM}* has exactly one factor of 3; the denominator for each

*t*has exactly two factors of 3.

_{MS}*Thus, the two sets have no elements in common, and the clock is never evenly divided.* **QED**.

To illustrate the importance of the numerator being a nonmultiple of 3, let’s calculate {*t _{HS}*}:

719 happens to be prime.

It is tempting to put forth the following fallacious argument:

Fallacious argument: “{

t} consists of a bunch of fractions whose denominator contains the large prime factor 719. Neither of the other sets {_{HS}t} or {_{HM}t} contain this large prime factor in their denominators… therefore {_{MS}t} contains no elements in common with either {_{HS}t} or {_{HM}t}, QED.”_{MS}

The problem with this argument is that it attempts to prove something that is *false.* True, the intersection of {*t _{HS}*} and {

*t*} is empty, but the intersection of {

_{MS}*t*} and {

_{HS}*t*} is

_{HM}*not*empty – in fact, it consists of precisely the two elements { 1/3, 2/3 } (which correspond to 4:00:00 and 8:00:00 respectively.)

What happened to the factor of 719 in the denominator? It got canceled out by a multiple of 719 in the numerator. In particular, 1/3 = (3*239 + 2)/(3*719), and 2/3 = (3*479 + 1)/(3*719).