Shollenbarger's Clock

expectedvalue

Shollenbarger's Clock:

Excluding 12:00, Is there ever a time on a clock where the second, minute and hour hands are equidistant from each other? (In other words, the three hands are all 120° apart? If yes, give the EXACT time(s) that this occurs. If no, prove that no such time exists. Good Luck!

DM

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If you include 12:00
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The second hand could be on 12:00, the minute hand on 4:00 and the hour hand on 8:00

Time would be 8:20

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If you exclude 12:00
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I personally don't think it's possible.

If any four hour marks distance stand for 120° (each hour mark is 30°), then 12:00 must be included.

Thus no such time exists!

Parlyne

But, at 8:20:00, the hour hand of a clock doesn't point directly at the 8. In fact, it will be one third of the way from 8 to 9. Thus, at 8:20:00 the hands will not all be [tex]120^\circ[/tex] apart.

I think he was saying to exclude the time 12:00:00 from consideration, rather than the direction of pointing at the 12.

expectedvalue

Parlyne is correct. I wanted to exclude 12:00:00 from consideration as an answer, because all three hands are on top of each other. But one hand can be pointing at 12.

8:20:00 is not a valid solution:

From second to minute: 120°
From minute to hour: 130°
From hour to second: 110°

DM

What about:

hour hand on 4:00, minute hand on 12:00 and second hand on 8:00

Time = 4:00

expectedvalue

This time is: 4:00:40. This is not a valid solution:

From minute to hour: 119 2/3°
From hour to second: 119 2/3°
From second to minute: 120 2/3°

The hour hand cannot be at 4:00 or the minute hand at 12:00 when the second hand is at 8:00, because it would now be 40 seconds after 4:00, thus the hour hand and the second would have moved slightly.

Keep trying. (Fractions of a second are allowed as valid answers, if needed.)

DM

My last attempt of the day is:

Hour hand on 3:00

Minute hand on 0:59

Second hand on 0:00:39

Time = 3:59:39

expectedvalue

No, unfortunately 3:59:39 is not a valid answer:

From minute to hour: 121.925°
From hour to second: 114.175°
From second to minute: 123.9°

geniusprahar_21

its not possible.......one way of proving it is assume the time as xhrs, yhrs, and z secs........and put appropriate conditions on them... then find angles and solve..........

DM

This is a very messy explanation. "Put appropriate conditions on them" doesn't give us the reason.

What conditions?

gnomedt

I too believe it's impossible. I haven't finished the proof, but the idea is that I make the hour, minutes and seconds hands of the clock into unit vectors which are a function of time, then take dot products and set them equal (because the dot product of any two of the aforementioned vectors must be |1|*|1|*cos[120 deg], i.e. 4 hour-marks apart) and finally show that there's no way that that's possible. There might be an easier way, but I like vectors.

Now here's another brainteaser, if this one turns out to be impossible: what time on Shollenbarger's Clock will make the hands the closest to 120 degrees?

DaveC426913

Correct me of I'm wrong, but no one said they have to be at exact integer times, right?

Don't these three hands line up at 120 degrees from each hour approximately 11 times in every 12 hour period? (I'll leave it to others to determine the exact times.)

Parlyne

No. The hour and minute hands line up 11 times in each 12 hour period. The second hand will only be in line with these one of those 11 times.

Parlyne

I'm going to go out on a limb and say that there is no such time. To demonstrate, let's assume there is a time when the hands are all [tex]120^\circ[/tex] apart. Call the hour x, the minute y and the second z. Because of the format times are cited in, x and y must be integers.

Using these, we can express the angle at which each hand sits. The hour hand sits at [tex]\alpha = \left (\frac{x}{12} + \frac{y}{720} + \frac{z}{43200} \right)*360^\circ = 30x + \frac{y}{2} + \frac{z}{120}[/tex]. The minute hand sits at [tex]\beta = \left ( \frac{y}{60} + \frac{z}{3600} \right )*360^\circ = 6y + \frac{z}{10}[/tex]. And, the second hand sits at [tex]\gamma = 6z[/tex].

We set the second and minute hands [tex]120^\circ[/tex] apart by setting [tex]\beta = \gamma \pm 120[/tex]. This gives the condition [tex]y = \frac{59}{60} z \pm 20[/tex]. Since y must be an integer, z must be an integral multiple of 60/59. Thus, we can write [tex]z = \frac{60}{59}n[/tex] and [tex]y = n \pm 20[/tex], where n is an integer.

Finally, to put the hour hand in the correct position, we set [tex]\alpha = \gamma \pm 240[/tex]. This reduces to the condition that [tex]x = \frac{11}{59} n \pm \frac{23}{2}[/tex].

We can see that, for x to be an integer, n must be valued such that [tex]\frac{11}{59} n[/tex] is an odd half integer. But, since 59 is odd and 11 and 59 are relatively prime, the only way for [tex]\frac{11}{59} n[/tex] to be an odd half integer is if n is an odd half integer multiple of 59 - [tex]n = \frac{2m+1}{2} 59[/tex]. Since all factors in the numerator are odd, this can't be an integer. Thus, if n is an integer, x is not.

Thus, there are no x, y, and z satisfying the relationships above such that x and y are both integers.

geniusprahar_21

conditions like:
x,y,z should be integers...........or atleast x and y should be integers(if u want to accept decimal seconds). further 0<= x <12....... 0<= y <60 and 0 <= z < 60........with these conditions u'll never get an appropriate answer

ceptimus

It gets close enough for practical purposes at

02:54:34.576

expectedvalue

I tried to clarify this earlier...

Decimal seconds are allowed. The seconds do not have to be integers. But give your answers with fractions.

Unfortunately, this is a brain teaser and not a practical problem. So, 'close enough' is not equidistant.