# In theory, this should be easy

In theory, this should be easy....

... During the space of 12 hours, how many times will the big hand of a clock form a right angle with the little hand?

23 times ?

Close but no cigar.

It depend upon when you start counting the 12 hours.

59x2--each click on the clock can have a right hand right angle and left handed one...

It depend upon when you start counting the 12 hours.
Yes. If you start counting at 3:00 or 9:00 for 12:00+epsilon hours, with epsilon positive as small as you want, it seems it makes 23 times. If epsilon is negative, that probably makes only 22. Does it have an hour, minute as well as second hand? Or only hours and minutes?

I was thinking second hand before

Completely ignore the seconds hand, we are only considering the minutes and hours hand. For argument's sake, start at the current time as I post this here in my timezone, which is 1.05.

But it seems to me that humanino has probably got it. :)

But it seems to me that humanino has probably got it. :)
Well, I got the first (3:00) yesterday before going to bed, and I stupidily was happy about myself, so I forgot the other one (9:00) Well, I got the first (3:00) yesterday before going to bed, and I stupidily was happy about myself, so I forgot the other one (9:00) There are others as well.

Yes, so difficult the problem

Redbelly98
Staff Emeritus
Homework Helper
I have used a "brute force" method, simply writing down the (approximate) times at which right-angles occur. Highlight the text below for the solution.

Also ... somebody may want to check if I missed any There are 22 times where right-angles occur.

Note: times are approximate.

12:15
12:50
1:20
1:55
2:25
3:00
3:30
4:05
4:35
5:10
5:40
6:15
6:45
7:20
7:55
8:25
9:00
9:30
10:05
10:40
11:10
11:45

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There is an analytic way to approach the problem--

The position vector of the big hand is given by (where t=0 is noon)

$$\mathbf{x_b}(t) = \sin \omega_b t \mathbf{i} + \cos \omega_b t \mathbf{j}$$

while the position vector of the little hand is given by

$$\mathbf{x_l}(t) = \sin \omega_l t \mathbf{i} + \cos \omega_l t \mathbf{j}$$

and you require that they are perpendicular to each other so that

$$\mathbf{x_b}(t)\cdot\mathbf{x_l}(t) = 0 \Rightarrow$$

$$\cos\omega_b t \cos\omega_l t + \sin \omega_b t \sin \omega_l t = 0 \Rightarrow$$

$$\cos (\omega_l - \omega_b)t = 0 \Rightarrow$$

$$(\omega_l - \omega_b)t_n = \pi n \Rightarrow$$

$$t_n = \frac{\pi n}{\omega_l - \omega_b}$$

but

$$\omega_b = 2\pi/T_b, \omega_l = 2\pi/T_l$$ where the T's are the periods.

So we have

$$t_n = \frac{n}{2} \frac{T_b T_l}{T_b - T_l}$$
where
$$T_b = 12 h, T_l = 1 h$$

Kurdt
Staff Emeritus
Gold Member
Well its fairly easy to work out. angular velocity of each hand is simple.

Angular velocity of minute hand: $\omega_m=\frac{2\pi}{3600}$
Angular velocity of hour hand: $\omega_h = \frac{2\pi}{43200}$

Goverened by the rotational dynamics equations:

$$\theta = \theta_0 + \omega t$$

We want:

$$\theta_m - \theta_h = (2n+1) \frac{\pi}{2}$$

and we get:

$$t = \frac{(2n+1) \pi}{2(\omega_m - \omega_h)}$$

First couple of answers are (assuming you start from 12:00):

12:16:21.82
12:49:05.45
..etc

EDIT: I was wondering why my tex had come out as vectors for a second.

Oh yeah woops I made a rookie mistake-- should have been (2n+1)/2 \pi as Kurt has it and not \pi n. Duh!

I have used a "brute force" method, simply writing down the (approximate) times at which right-angles occur. Highlight the text below for the solution.

Also ... somebody may want to check if I missed any There are 22 times where right-angles occur.

Note: times are approximate.

12:15
12:50
1:20
1:55
2:25
3:00
3:30
4:05
4:35
5:10
5:40
6:15
6:45
7:20
7:55
8:25
9:00
9:30
10:05
10:40
11:10
11:45
You have the right number. Well done.

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wow someone brought maths into it.
i'd say for the first hour
when the minute hand is 15mins past the hour hand
for the last hour
when the minute hand is 15mins befor the hour hand

for the 10 hours in between
at 15mins before and after the hour hand

10*2 + 2*1 =22?

Are we counting every second, minute or hour? LOL, J/K.

NateTG
Homework Helper
This is pretty easy stuff:

The big hand makes 11 laps relative to the small hand. There are 2 times per lap that it forms a right angle, so the number is 2*11=22.

What's with making the text invisible? It doesn't even work, the text is still visible.

Kurdt
Staff Emeritus