The 2 o'clock problem

 A year ago, I took a math exam for a job application. I passed the exam, made it to the interviews, and actually got qualified for the training. Eventually, I decided not to take that opportunity, and I chose to pursue the job that I have right now.

But this article is not about that missed job or any jobs. This is about a particular math problem in the math exam. It was so interesting that I told myself I would write about it. And after 1 year, here it is.

The problem went like this:

It is 2:00. What time will the hour hand and the minute hand make a straight line together?

Before this exam, I had never solved a problem like this. I had no idea where to begin (let alone solve it). I could skip this problem and not give a fuck about this 1 point. But I did not. I wanted to solve it not for the point, but for the sake of solving it. I wanted to prove to myself that my knack for geometry from high school was still sitting behind my brain, awaiting to be tapped.

I so began thinking.

The first thing that was certain was that the earliest time the hour hand and the minute hand make a straight line would take less than 1 hour. Therefore, the hour hand would be between 2 and 3. If that's the area where the hour hand would be, then the minute hand should be between 8 and 9 to make a straight line with the hour hand.

I had narrowed down the possible answer. The time should be between 2:40 and 2:45.

Now, the math begins. To get the would-be position of the hands using geometry, I needed to look at the movement of the hands in terms of angles, not hours or minutes. Conveniently, the 360-degree complete rotation in a geometric circle is well divisible by 60 minutes--the complete rotation of the minute hand (but I need not bother about that in the meantime).

If I got the angles of the hour hand and the minute hand relative to the horizontal or vertical line, then I would be able to convert them to minutes, and therefore I would know the time.

To get those angles, I needed to solve them using algebra.

First, I laid down the most obvious truth to extract a formula out of it:

In a straight line, the hour hand and the minute hand would be 180 degrees apart from each other.

I get back to this later.

Now, I needed to know the initial position of the hour and minute hands at 2:00 in terms of angles.

Conversion:

360° = 60 dots = 60 minutes  →  6° = 1 dot = 1 minute

I established 12 as the reference 0°.

At 2:00, the hour hand was at 2. The distance between 12 and 2 was 10 minutes or 60°. On the other hand, the minute hand was at 12, at position 0°.

Earlier, I established the range of the possible answer: 2:40 and 2:45

From this, the end position of the hour hand must be between 60° and 90°, while the end position of the minute hand must be within 240° to 270°.

Those who have sharp math minds would have spotted that the range of the hour hand could be narrowed down further by getting the ratio and proportion with the range of the minute hand, but I did not bother, since that was not important in getting the answer. But to tickle that brain, I did you a service, and I actually calculated it this time:

40:60 = X:15  →  X=10
10/15 * 30 = 20°
45:60 = Y:15  →  Y=11.25
11.25/15 * 30 = 22.5°
Range: 60°+20° to 60°+22.5°  →  80° to 82.5°

Anyway, as I said, that was not important in getting to the answer using the method that I used back then. It was a good brain exercise nonetheless, and this is a good path to solve this problem using a different method.

To continue, I derived formulas from the position where the hour hand and minute hand make a straight line (blue line).

The distance of the hour hand from the initial position 2 would be x. The distance of the minute hand from the initial position 12 would be y.

Since the minute hand moves 12 (or 60/5) times faster than the hour hand, it means that when the hour hand moves by x, the minute hand would have moved 12x. 

Therefore:

y = 12x    (equation 1)

To find these values, I needed one more equation containing these two variables.

In the blue straight line, the hour hand and minute hand would make 180°.

Therefore:

y - (60°+x) = 180°    (equation 2)

Using equation 1 and equation 2, the values of x and y were solved:

x = 21.818°    y = 261.818°

Converting them to time:

x = 21.818° * (1 minute / 6°) = 3.636 minutes = 3.636 dots

y = 261.818° * (1 minute / 6°) = 43.636 minutes = 43.636 dots

The value of x = 3.636 dots being less than 5 dots indicated that the hour hand stayed between 2 and 3.

If I calculated the position in terms of angles, it can be proven that the calculated range of the position of the hour hand earlier was correct:
60° + x = 81.818° (between 80° and 82.5°)

The value y = 43.636 minutes was the position of the minute hand, also proving that it was between 8 and 9.

With these calculations, it can be concluded that the time when the hour hand and the minute hand formed a straight line was 2:43.636

or

2h:43m:38.18s

PS.

That exam was really a memorable experience for me. It covered electrical and electronics engineering and mathematics. In electrical and electronics engineering, some questions were impossible for me to answer without knowledge of the necessary formulas. So giving up on those questions that I could not solve was not heavy on the heart.

But in mathematics, it was different. No formulas need be memorized. It was a test of thinking. I genuinely enjoyed solving them. And to be honest, I actually learned more from solving those math problems than giving out my own knowledge in the form of answers. It was amazing, and I wanted to keep a memento of that moment, so I kept the scratch paper I used in that exam.