Perpendicular hour and minute hand

Question

Starting at $12:00$ on a $12$-hour clock, how many times will the hour and minute hands be perpendicular to each other in a 12-hour period?

Each hour there are two instances when the clock has a $90°$ angle, so (naturally) I thought that the answer would be $12•2=24$, which is wrong. I have searched some answers before this, and one of them says that "as the hours progress, the minute hand lines up 90 degrees later in the hour", but I don't see how that interferes with the $12•2$ reasoning... no matter what the hour, the two 90° cases happen within the hour, don't they?

Answer 1

Sometimes a graph explains more than a thousand word.

On the horizontal axis I put the hours from $0$ to $12$ the purple line is the angle that the hour hand forms with the $00:00$ position.

On the vertical axis there is the angle that the minute hand forms with the $00:00$ position and is periodic since every hour the minute hand does the $360°$ round and then starts again from $0$. I have found the moments when the hands overlay and then shifting up and down by $90°$, $15$ minutes, the moments when the hands are perpendicular.

Keep in mind that the system is not isometric. To make a clearer drawing I used a $1:4$ scale. In other words four units on the $y-$axis is one hour like one unit on the $x-$axis.

If you want to know at what time the hands are perpendicular for the first time you solve the system formed by the equations of the two lines: $y=x$ the line of the minute hand and $y=\dfrac{x}{12}+\dfrac{1}{4}$ the hour hand shifted of $90°$ (=fifteen minutes). We find $\left(\frac{3}{11},\;\frac{3}{11}\right)$ which means $3/11+3/11\times 60\approx 00:16:38.2$ that is noon sixteen minutes and $38.2$ seconds.

In a similar way can be found all the other moments, changing a little the equation

hope it helps

Answer 2

Oh, let's do this the annoying way.

The minute hand moves at $360\frac {\text{degrees}}{\text{revolutions}}\frac {1 \text{revolution}}{60 \text{minutes}}=6\frac{\text{degrees}}{\text{minutes}}$

The hour hand moves at $\frac{360 degrees}{1revolution}\frac{1revolution}{12hours}\frac{1hour}{60 minute}= \frac 12 \frac{degree}{minute}$.

If a minute hand is $x$ degrees away from the hour hand how long will it take for the minute hand to become $x + 180$ degrees away from the hour hand?

Well that's a matter of solving $x + 6t = x + \frac 12t + 180$ or $t= \frac {180}{\frac {11}{2}}=\frac {360}{11}$. (Roughly $32$ minutes.)

At $3:00$ the hour hand and the minute hand are perpendicular. The next time that will happen will be when the minute hand moves for $90$ degrees before the hour had to $90$ degrees after the hour. That will happen in about $32$ minutes. (At $3:32$). Every $\frac {360}{11}$ theminute hand will move 180 degrees further than the hour hand and the hands will be perpendicular.

In a $12$ hour (or a $60*12 = 720$ minute period) this will happen $\frac {720}{\frac {360}{11}} = 22$ times.

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Here's a better way, although it was easy to make annoying sign and logic errors if you aren't careful.

The minute hand travels $6$ degrees a minute. ($\frac {360}{60} = 6$). The hour hand travels $.5$ degrees a minute. ($\frac {360}{12*60} = .5$).

In a 12-hour period there are $12*60=720$ minutes.

Let's say $\theta_t = 6t$ is the angle of the minute hand after $t; 0 \le t < 720$ minutes. Note: it's very possible that $\theta(t) > 360$.

Let's say $\phi_t = .5t$ is the angle of the hour hand after $t$ minutes..

Obviously $\theta_t \ge \phi_t$.

$\theta_t$ and $\phi_t$ are perpendicular if $\theta_t = \phi_t + 90 + k*180$ for some non-negative integer $k$.

So we need the find out how many solutions there are to:

$6t = .5t + 90 + k*180; 0 \le t < 720$ there are.

So $t = \frac {90 + k*180}{5.5}$ will have one solution for each integer $k$.

So how many $k$ are there so that $0 \le t = \frac {90 + k*180}{5.5} < 720$ are there?

Multiply everything by $11$: $0 \le 2(90 + k*180)< 11*720$

Divide everything by $180$: $0 \le 1 + 2k < 11*4$

$-1 \le 2k < 11*4 -1$

$\frac 12 \le k < 2*11 - \frac 12$

And as $k$ is a non-negative integer $0 \le k < 22$. There are $22$ possible $k$s which give a solution.

This also gives us the times this occurs:

$t = \frac {90 + k*180}{5.5} = \frac {180}11 + k *\frac {360}{11} \approx 16.36 + 32.72k$ (or a little more than every half-hour):

So at 12:16, 12:49, 1:22, 1:54, 2:27, 3:00, etc.

Answer 3

In each time interval $[n, n+1]$ hours, the minute hand is perpendicular to the hour hand twice, and there are 12 of these intervals. However, since the hands are perpendicular at 3:00 and 9:00, which each belong to two of these intervals, we overcounted by 2. Thus, the hour and minute hands are perpendicular 22 times.