Time flies

To simplify somehow the problem let us make 60 ticks for all hands. This said the position of a hand can be modeled by the group $\mathbb{R} / 60 \mathbb{Z}$ (or in other words we are looking at the reals numbers modulo 60; see Details below if you are not familiar with this). This is required in order to deal with continuously moving hands (the second hand could be halfway between two seconds).

The actual possible positions for the hands are given by the triplet $(t, \frac{t}{60}, \tfrac{t}{720} )$ . This is because the minute hand turns 60 times slower than the second hand, and the hour hand turns 12 times slower than the minute hand (so 720 times slower than the second hand). In order that the hands split the clock in three equal thirds, it must hold that the difference between any two coordinates is $\pm 20 \text{ mod } 60$ (because one third of the clock is 20 units). This means $\begin{array}{rrll} (*) & t - \dfrac{t}{60} & = \pm 20 & \text{mod } 60 \\ (**) & \dfrac{t}{60} - \dfrac{t}{720} & = \pm 20 & \text{mod } 60 \\ \end{array}$ However note that $\begin{array}{rrll} (**): & \dfrac{t}{60} - \dfrac{t}{720} & = \pm 20 & \text{mod } 60 \\ \implies & 720 \cdot \dfrac{t}{60} - 720 \cdot \tfrac{t}{720} & = 720 \cdot \pm 20 & \text{mod } 60 \\ \implies & 12 \cdot t - t & = 0 & \text{mod } 60 \\ \implies & 11 \cdot t & = 0 & \text{mod } 60 \\ \end{array}$ Call this last equation $(***)$ . Next $\begin{array}{rrll} (*): & t - \dfrac{t}{60} & = \pm 20 & \text{mod } 60 \\ \implies & 60 \cdot t - 60 \cdot \dfrac{t}{60} & = 60 \cdot \pm 20 & \text{mod } 60 \\ \implies & 60 \cdot t - t & = 0 & \text{mod } 60 \\ \implies & 59 \cdot t & = 0 & \text{mod } 60 \\ \end{array}$ Call this last equation $(\bigstar)$ . Now it's not too hard to show $(***)$ and $\bigstar$ lead to a contradiction. $\begin{array}{rrll} (\bigstar): & 59 \cdot t & = 0 & \text{mod } 60 \\ \implies & 11 \cdot t +11 \cdot t + 11 \cdot t+ 11 \cdot t+ 11 \cdot t + 4 \cdot t & = 0 & \text{mod } 60 \\ \overset{(***)}{\implies} & 0+0+0+0+0+ 4\cdot t & = 0 & \text{mod } 60 \\ \implies & 4 \cdot t & = 0 & \text{mod } 60 \\ \end{array}$ Now repeat the same trick $11 t = 4t + 4t + 3t = 0 \text{ mod } 60$ and hence $3 t = 0 \text{ mod } 60$ . But then $4 t = 3t + t = 0 \text{ mod } 60$ so $t = 0 \text{ mod } 60$ .

So $t$ is an integer(!) and a multiple of 60. We could go on with the equations, but the contradiction is clear. This means the second hand is at the origin (mark 12 on the clock). Then the minute hand must be either at 20 or 40 minutes exactly (marks 4 or 8; in order to make the proper angle). But in that case the hour hand cannot be at marks 4 or 8 (because the minute hand is not at the mark 12).

Hence $(*)$ and $(**)$ imply that the answer is $\fbox{No}$ .

For those who want to go on with equations

Once we have that $t = 0 \text{ mod } 60$ , $(**)$ becomes, upon multiplication by 720: $\dfrac{t}{12} = 0 \text{ mod } 60$ . This means that $t$ is an integer multiple of 720, say $t = 720n$ . But then $(*)$ reads $12 n = \pm 20 \text{ mod } 60$ ( where $n$ is an integer ). The values of $12 n\text{ mod } 60$ are $0,12,24,36$ and $48$ . So there are no solutions, and the answer is no.

Details on $\mathbb{R} / 60 \mathbb{Z}$

The arithmetic of $\mathbb{R} \text{ mod } 60$ is essentially the same as the usual modulo arithmetic. For the clock, you can picture this as follows: take a real $r$ and add or subtract 60 until it lies in the (half-open) interval $[0,60)$ . For example $-20 \text{ mod } 60$ is represented by $40$ . Addition works fine: $53+26 \text{ mod } 60 = 79 \text{ mod } 60 = 19 \text{ mod } 60$ . When manipulating equations, multiplication by integers is also unproblematic (because it's just repeated addition). However multiplication by reals which are not integers does not make sense when you are manipulating an equation. Note also that if $r$ is some real then $60 r \text{ mod } 60$ could be different from $0\text{ mod } 60$ (for example if $r = \tfrac{1}{60}$ then $60 r = 1$ .

Note

Note that all allowed positions for the hands form a 3-dimensional space. However the set of correct positions is only one-dimensional. Lastly, the set of positions where the hands split the clock in equal thirds is also one-dimensional. This means that it would have been a big coincidence if the answer would have been yes, as in general two one-dimensional space inside a 3-dimensional space do not intersect.

In each hour the hour and minute hand make 60 degrees angle twice (with seconds hand pointing at 12). Now, the only condition required is that the second hand also makes an angle of 60 degrees with each hand which is like

But the angle between the minute and hour hand is a bit more or less than 120 degrees. Therefore, the given condition is impossible.

Let the angular displacements starting at time $t=0$ at 00:00 of the hour, minute, and second hands be $\theta_h$ , $\theta_m$ , and $\theta_s$ respectively in degrees. For $0 \le t < 24$ in hours, we have $\theta_h (t) = 30 t \bmod 360$ , $\theta_m (t) = 360 t \bmod 360$ , and $\theta_s (t) = 21600 t \bmod 360$ . Let us first consider when the hour and minute hands are $120^\circ$ apart. This happens when:

$\begin{aligned} \theta_m (t) - \theta_h (t) & = 120\color{#3D99F6}n & \small \color{#3D99F6} \text{where }n \text{ is a positive integer not divisible by 3.} \\ 360t - 30 t & = 120n \\ t_n & = \frac {4n}{11} & \small \color{#3D99F6} \text{Since }t < 24, \ n \le 65 \\ \implies \theta_s (t_n) & = \frac {86400}{11}n \bmod 360 \\ & = \frac {3240}{11}n \bmod 360 \end{aligned}$

There are two cases.

Case 1: When $n \bmod 3 = 1$ , $\theta_m(t_n) - \theta_h (t_n) = 120$ , we are checking if $\theta_s (t_n) - \theta_h (t_n) = 240$ . Let's see $\theta_s (t_n) - \theta_h (t_n) = \frac {3240}{11}n - \frac {120}{11}n = \frac {3120}{11}n$ , which is only an integer when $n$ is a multiple of 11 and only two of these $22, 55 \bmod 3 = 1$ . But $3120 \times 2 \bmod 360 = 3120 \times 5 \bmod 360 = 120 \ne 240$ .

Case 2: When $n \bmod 3 = 2$ , $\theta_m(t_n) - \theta_h (t_n) = 240$ , we are checking if $\theta_s (t_n) - \theta_h (t_n) = 120$ . Let's see $\theta_s (t_n) - \theta_h (t_n) = \frac {3240}{11}n - \frac {120}{11}n = \frac {3120}{11}n$ , which is only an integer when $n$ is a multiple of 11 and only two of these $11, 44 \bmod 3 = 2$ . But $3120 \times 1 \bmod 360 = 3120 \times 4 \bmod 360 = 240 \ne 120$ .

No , there is never a time that the three hands of the clock divide the clock face into three equal areas.

Assume that it is possible.

Let the angle made by second, minute and hour hand to be $\alpha=\dfrac{2\pi}{60}t,\beta=\dfrac{2\pi}{3600}t,\gamma=\dfrac{2\pi}{43200}t$ , then we will have $\begin{cases}\cos(\alpha-\beta)=-\frac12\\\cos(\beta-\gamma)=-\frac12\end{cases}$ Also we have $\alpha-\beta\equiv\beta-\gamma\equiv\pm\frac{2\pi}{3}\pmod{2\pi}$ If the sign of $\alpha-\beta\pmod{2\pi}$ and $\beta-\gamma\pmod{2\pi}$ are opposite, then will leads to $\alpha\equiv\gamma\pmod{2\pi}$ which is contradiction. Then for some suitable integer $k,h$ we have $\begin{cases}t\bigg(\frac1{60}-\frac1{3600}\bigg)=k\pm\frac13\\t\bigg(\frac1{3600}-\frac1{43200}\bigg)=h\pm\frac13\end{cases}$ Eliminate $t$ , we retrieve this equation: $\frac{3600}{59}(k\pm\frac13)=\frac{43200}{11}(h\pm\frac13)$ Which can be simplify to $1320k\pm4400=849600h\pm283200$ But if we take modulo 3 we will see $\pm2\equiv0\pmod3$ so this is a contradiction, this situation can NOT occur.

Let's look at the time it takes for the second, minute, and hour hand to travel $30$ degrees on the clock, where $S$ is the time in seconds:

$S_{sec}=\dfrac{30}{5}$
$S_{min}=\dfrac{30}{300}$
$S_{hr}=\dfrac{30}{3600}$

The condition that the minute and hour hand are exactly $120$ degrees apart is

$\left( \dfrac{30}{300}-\dfrac{30}{3600} \right) S=120+360n$

where $n$ is an integer, so that

$S=\left( \dfrac{14400}{11} \right) (3n+1)$

We then check to see if for any $n=1$ to $11$ the difference between

$S\; mod\;60 - \dfrac{1}{60}(S \;mod \;3600)$

is exactly $20$ seconds. The actual results are

$(27.2727, -10.9091, 10.9091, -27.2727, -5.45455, -43.6364, -21.8182, \ 0., 21.8182, 43.6364, 5.45455)$

so that although some are close to $20$ , none of them are exactly $20$ , so that the answer is "no".

This solution assumes that the clock is analog and all hands move continuously.