Can we identify the time if we know every angle between three hands of a watch?

Question

Let $M, H, S$ be the minute hand, the hour hand, the second hand of a watch respectively. Also, let $A_{MH}, A_{MS}, A_{HS}$ be the angle between $M$ and $H$, $M$ and $S$, $H$ and $S$ respectively.

Then, here is my question.

Question : Can we identify the time of the watch if we know all of $A_{MH}, A_{MS}, A_{HS}$ of a watch? If we can, how can we identify the time?

Motivation : I came up with this question while I was repairing my watch:) The answer seems yes, but I'm facing difficulty. Can anyone help?

Answer 1

The answer is yes.

If all three hands move continuously, then the angle of the hour hand is enough to completely determine the time, and therefore the angle of the other two hands.

Let $X$ be the angle of the hour hand from the $12$ o'clock position. Let $Y$ be the angle between the hour and minute hand. Let $Z$ be the angle between the minute and second hand.

The hour hand moves at a rate of $30$ degrees per hour, while the minute hand moves at a rate of $360$ degrees per hour. It we use a different frame of reference, it is possible to see that the minute hand moves at a rate of $330$ degrees per hour relative to the hour hand. This means that the value of $Y$ is equivalent every $360/330=12/11$ hours.

The second hand moves at a rate of $21600$ degrees per hour. Relative to the $360$ degrees per hour of the minute hand, it is traveling $21240$ degrees per hour. The value of $Z$ repeats every $360/21240=1/59$ hours.

Now, we need to find a number that is the smallest multiple of $12/11$ and $1/59$.

$$11*59 = 649$$ $$12/11=708/649$$ $$6/59=66/649$$ $$\text{LCM}(708,66)=7788$$ $$7788/649 = 12$$

As we can see, the unique combination of angle $Y$ and $Z$ only repeats every twelve hours.

Answer 2

HINT: The time has the same angles for 12:20:40, 4:40:00, etc.

Answer 3

EDIT: I am apparently four hours late :(

Assuming that the watch hands move continuously, it seems the answer might be yes, you can. A long and incomplete explanation follows. I'm pretty tired and am not thinking properly anymore, but I think this will eventually lead you to a positive answer.

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Modelling the Scenario

If the angles are measured in degrees, then let $T=\frac{1}{360}A_{SH}$ and $S=\frac{1}{360}A_{SM}$ and $B=\frac{1}{360}A_{MH}$. The motivation for the notation, to keep you oriented, is that seconds are Tiny compared to hours, and Small compared to minutes. Then, relatively speaking, minutes are Big compared to hours. The factors of $360$ are simply to keep everything between $0$ and $1$.

We will also assume that the angles are oriented clockwise, which means that at $12$:$00$:$15$, we have $T\approx \frac14$, but at $12$:$00$:$45$ we have $T\approx \frac34$. I think you could do it with unoriented angles, but I can't see a way to do this that would not devolve into very ugly casework. So for simplicity, we will keep the orientation.

We will do everything in terms of revolutions $r$ of the hour hand. Note that we can assume that $0\leq r<1$. In one revolution of the hour hand, the minute hand makes 12 revolutions, and the second hand makes 720 revolutions. Therefore, we have the following system of three equations:

\begin{align*} 720r + S &= 12r + n & \text{for some }n\in\mathbb{Z}\\ 12r + B &= r + k & \text{for some }k\in\mathbb{Z}\\ 720r + T &= r + m & \text{for some }m\in\mathbb{Z} \end{align*}

The lower case letters are acting as the variables here. $r$ will uniquely determine the time (since it encodes exactly how many hours have passed since midnight, to arbitrary precision) and the other three are simply accounting for the fact that we might have to make the faster hand go around many times before it hits in the right place with the others. Since they are only allowed to be integers, we ensure that only full revolutions are "the same".

Your question has thus been reduced to the nominally shorter question "Is there exactly one $0\leq r<1$ that satisfies the above equations?"

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Simplifying the Equations

First we combine the $r$ terms: \begin{align*} 708r &= n - S & \text{for some }n\in\mathbb{Z}\\ 11r &= k - B & \text{for some }k\in\mathbb{Z}\\ 719r &= m - T & \text{for some }m\in\mathbb{Z} \end{align*}

We want to eliminate $r$ and write everything in terms of the other three. I'll do this by first putting everything in one equation: $$ r = \frac{n-S}{708} = \frac{k-B}{11} = \frac{m-T}{719}$$ and then splitting it up again (I skip a lot of steps here: for each equation choose two terms and then cross multiply, and rearrange to isolate the variables): \begin{align*} 708k-11n &= 708B - 11S \\ 719n-708m &= 719S - 708T \\ 11m-719k &= 11T - 719B \end{align*}

A small remark here: Note that the LHS is only integers, so the RHS can only be integers as well. So in fact, the angles these three hands make are not arbitrary; they have these hidden relationships. Well, if you work it out, you find that the relationships are basically that they must add up to $360^\circ$, which upon some thinking is not very mysterious.

We make one final simplification: since the RHS elements are just constants, I will ignore their details.

\begin{align*} 708k-11n &= \alpha & \text{for some }\alpha\in\mathbb{Z}\\ 719n-708m &= \beta & \text{for some }\beta\in\mathbb{Z}\\ 11m-719k &= \gamma & \text{for some }\gamma\in\mathbb{Z} \end{align*}

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Interpreting a Solution

WolframAlpha will solve these equations without much difficulty: \begin{align*} 0 &= 719\alpha + 11\beta + 708\gamma \\ n &= \frac{708k-\alpha}{11} \\ m &= -\frac{719\alpha}{7788}-\frac{\beta}{708}+\frac{719k}{11} \end{align*}

The first equation is irrelevant for us, because these are constants; we got them from the problem statement, so they don't provide any new information.

The other two equations tell us something interesting, which is that $k$ is a free variable. That is, if we remove the integer restrictions, then there is a solution for any $k$. We'll resubstitute the old $SBT$ variables, to get (skipping a good deal of algebra, just substitution, though)

\begin{align*} n &= \frac{708}{11}(k-B)+S \\ m &= \frac{719}{11}(k-B)+T \end{align*}

Well, this is simpler: it is just $m-n-k=T-S-B$. However, remember that we had some conditions from before, namely that $0\leq B,T,S<1$, and therefore, $-2<T-S-B<1$. However, since $m-n-k$ is an integer, this means that $T-S-B$ is either $0$ or $-1$.

In the first case, $m=n+k$. However, we can go all the way back to the original three equations to see that if this is true, then $720r+S-(720r+T) = 12r+n-(r+m)$ and so $S-T=11r-k$. Rearrange to $11r=S-T+k$. The LHS is not negative, so either $k\geq 1$, or $k=0$ and $S\geq T$.

In the second case, $m=n+k-1$ and the same analysis shows that $S-T=11r+1-k$, which give $S-T+k-1=11r$, so similarly either $k\geq 2$ or $k=1$ and $S\geq T$. But notice that $r$ is the same as the first case either way, so we can ignore this one.

Furthermore, $11r<11$, and so $k\leq 10$. This gives, in general, 11 possible solutions for $k$. (There are twelve possibilities if $S\geq T$, but there are always at least eleven.) Furthermore each of these gives distinct values of $r$.

Therefore, we have found that for any valid configuration of angles, there are at least $11$ different times which could be found for that configuration.

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Parting Thoughts

Some headway could be made using a similar analysis. By this I mean, I have shown how to find the possible values of $k$, and by substituting into other pairs of the original three equations you can find possible solutions for $n$ and $m$. From there, you have to test these possible solutions somehow. I'm not sure how you would do this, exactly.

Brute force with a computer could probably run in reasonable time. But as a first refinement, you know that $m-n-k$ is either $0$ or $-1$; this should reduce the number of cases to check from 5 million to around 8,000. The fourth set of equations could probably get it near 1,000.

On the other hand, it seems impossible that there could be no "respectable" number-theoretic solution to this problem. Unfortunately, I am not able to find it right now :(