# Clock Puzzle

In our last department meeting, one teacher presented a puzzle/problem for us to figure out.

At 3:00, the hour and minute hands on a clock form a right angle. What is the next time that happens? The presenting teacher had a pretty darn elegant solution. But I enjoyed working it out using brute force. (That’s pretty much my go-to.) I’m going to type my solution down below the jump.

I saw the hour hand and the minute hand act like vectors coming out of the origin. Their position changed in time. $\vec{h}(t)=<\sin(\frac{2\pi}{720}t),\cos(\frac{2\pi}{720}t)>$ and $\vec{m}(t)=<\sin(\frac{2\pi}{60}t),\cos(\frac{2\pi}{60}t)>$, where $t$ is measure in minutes.

Now when are two vectors perpendicular? When their dot product is 0. $\vec{m}(t)\cdot\vec{h}(t)=0$ And lo and behold, naturally occurring in this problem, we see a difference of angles formula for cosine! $\cos(\frac{2\pi}{60}t-\frac{2\pi}{720}t)=0$ $\cos((1-\frac{1}{12})\frac{2\pi}{60}t)=0$ $\cos((\frac{11}{12})\frac{2\pi}{60}t)=0$ $\cos((\frac{11\pi}{360}t)=0$

And then we just have to find all values that make this true. Well the cosine function has an output of 0 when the input of the cosine function is $\frac{(2n-1)\pi}{2}$ (where $n$ is an integer).

That just leaves us with: $\frac{11\pi}{360}t=\frac{(2n-1)\pi}{2}$

From this, we solve for $t$ and we get: $t=\frac{(2n-1)180}{11}$

Let’s see what happens for various values of $n$! And so here we have our answer… around 212 minutes (3:32 and 44 seconds).

Pretty neat, huh? Okay, okay, once you look at the final solution for the general form for all the times, you might suspect there is a much easier thing going on here. And that’s what the teacher presenting the problem shared with us after we finished. But I’m not going to share it with you because I LIKE MY SOLUTION and HECK IF I DIDN’T SEE AND USE A SUM OF ANGLES TRIG FORMULA IN THE WILD! Huzzah!

1. cheesemonkeysf says:

SUPER-LIKE!!!!!!!!

1. Amy Zimmer says:

Ditto!

2. Ned Diamond says:

Sent from my iPhone

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1. Maya Quinn says:

There is a technique for many clock problems (e.g., a similar one with the hands at 180 degrees) in which one is convinced that an event occurs 11 times in some period of time. As yet another example, the hands overlap 11 times in twelve hours (beginning with 12:00, then 1 something, 2 something, … 10 something, but not 11 anything!).

A similar approach here involves observing the times at which you clearly have 90 degree angles: 3:00 and 9:00, and then dividing the 6 hours = 360 minutes between them by 11 to find 32.7272… minutes. And that’s how much time elapses in between each of the 90 degree occurrences, whence the solution is immediate: 3:32 and 0.7272 of a minute (~43.6 seconds).

Here is an example of a page that broaches similar problems:
http://puzzles.nigelcoldwell.co.uk/thirtyfive.htm

2. temur says:

Here is a solution that is kind of elegant. First of all, let us make a convention: We measure an angle by the time it takes for the minute hand to cover the angle. For example, 180 degrees is 30 minutes, 60 degrees is 10 minutes, 15 degrees is 2.5 minutes and so on.

It is clear that the event will occur just after 3:30. So set the time at 3:30. At 3:30 though, the two hands would not be perpendicular, because the hour hand would have moved a bit. Where would that be? Of course, it would be halfway between 3 and 4, which is, if we measure angle by minutes, 2.5 minutes. Now in order to “catch this up”, the minute hand must move an additional 2.5 minutes, pointing at 32 minutes 30 seconds, or 32:30. But during this process of setting the time at 3:32:30, the hour hand would have moved slightly, “ruining” again the right angled position between the two hands. How far exactly would the hour hand move? This is the movement corresponding to the time duration of 2.5 min, and taking into account that the hour hand moves 12 times slower than the minute hand, this angle is (2.5/12) min. This means that we now need to set the time at (2.5 + 2.5/12) minutes after 3:30. Now, the movement of the minute hand by (2.5/12) min will cause the movement of the hour hand by the angle of (2.5/12^2) min. Hence we need to set the minute hand at (2.5 + 2.5/12 + 2.5/12^2) minutes after 3:30. We repeat this process, to conclude that the time we are looking for is

2.5 + 2.5/12 + 2.5/12^2 + 2.5/12^3 + …

minutes after 3:30. This infinite sum can be computed easily:

2.5 + 2.5/12 + 2.5/12^2 + 2.5/12^3 + … = 2.5 * (1 + 1/12 + 1/12^2 + … ) = 2.5 * 12/11 = 2.72727…

1. samjshah says:

Um. This. Is. Awesome. Thank you for sharing! I’m going to share this with the original colleague who posed the problem.

3. Ned Diamond says:

I wrote last night for you to send out the elegant solution but a colleague and I spent some time this morning thinking about it and saw the elegant solution. It’s so cool when it all makes sense…of course the angular speed should be a constant. What a great linear relationship in a circular problem.

Thanks for sharing. I have a little folder of my Sam Shah stuff and it keeps growing. I want to tell my dept head that I don’t need to go to a conference for professional development. I just need time to digest all that you send out. :) I may have to check out Math Twitter camp one of these summers.

On Thu, Nov 5, 2015 at 6:58 PM, Continuous Everywhere but Differentiab

1. samjshah says:

Yay!!! I’m glad you figured it out. My brain would NEVER have figured out the elegant solution. It took me forever to understand it when I first heard it. And also: YES TWITTER MATH CAMP! It is a magical place.

4. Scott says:

Is there something more elegant than what Maya Quinn shared? My feeble mind cannot handle this!

1. samjshah says:

I think that what she shared is the most elegant/simple!

1. Maya Quinn says:

As far as figuring out “the most elegant/simple solution”: I learned the underpinnings of this technique from a book written by Kobon Fujimura and edited by Martin Gardner, entitled: The Tokyo Puzzles. The original question posed there asked for the number of times the hour and minute hands overlap in a day; take this idea – then add several years, a few more clock problems, and a dash of luck, and the approach finally emerges…

5. Nyima Drayang says:

Also just by fractions…
let m=minutes past the hour. Then m/60 * 1/12 + 1/4 is the position of the hour hand (as a fraction of 12 hours. The “+ 1/4” is because the hour hand starts at 3 o’clock). That simplifies to m/720 + 1/4. When the hands are 90 degrees apart, then m/720 + 1/4 +1/4 = m/60 (the extra + 1/4 is the 90 degree difference between the hands). This simplifies to 360/11 = m which makes m = 32 8/12 minutes (or 32 mins and roughly 44 seconds).

I probably missed some intuitive points there, but I had fun anyway : D

I really love problems that are open to completely different angles of attack. Thanks for sharing this (and everything else!).

6. Mimi says:

I solved this using algebra as well. I first visualized that it’ll happen a little after 3:30, so I let x represent the number of minutes after 3:00 when this happens. At that point, the minute hand is x minutes ahead of the 12 o’clock position. The hour hand is (15 + x/12) minutes ahead of the 12 o’clock position. So, the distance between them is x – (15 + x/12) = 15 minutes. Solving this, I get x is 32.727272727… or it happens at 3:32:43.6.

Fun!