Polar Fox: I have a problem with this one!

Red Fox: What's wrong with it?

Polar Fox: Nothing moves

*continuously*?Red Fox: Come again?!

Polar Fox: The universe is discrete! THERE IS NO CONTINUUM.

Red Fox: Then how do the things move, flow, or slide?

Polar Fox: Nothing moves! Absolutely nothing moves!!

Red Fox: But the time ticks away, no? Clock arms DO advance.

Polar Fox: No, they don't! But they die in one instance and resurrect in the next one. In between no measurable time passes. Matter oscillates between existence and non-existence continuously.

Red Fox:

*Continuously*? Isn't that ironic?Polar Fox: Between any two existence, there is nothing but emptiness.

Red Fox: I am having the feeling that your intelligence fall into non-existing state just now.

Polar Fox: Hold on. I think I am about to jump back into the reality. Wait a sec.

Red Fox: Take your time dude. Just take your time!

So we have,

ReplyDeletex + y + z = 360 and sinx*siny*sinz

-> f(x,y) = sin(x)*sin(y)*sin(360-x-y)

-> f(x,y) = sin(x)*sin(y)*-sin(x+y)

Now to find the critical points of f(x,y), we just need to find the partial derivative with respect to x and y and solve for 0.

f_x(x,y) =

sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)

solve for 0.

sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)=0

-> tan(x) = -tan(x+y)

Since the function is symmetric, we should get the same partial derivative for y.

-> tan(y) = -tan(y+x)

-> tan(x)=tan(y)

-> x = y or they are opposites. However, if they are opposites, the original function just becomes 0. Thus, x = y.

Now substitute in x for y in the original equation and find its critical points.

Eventually, you will get sin(3x)=0

x = 120 degrees

y = 120 degrees

z = 120 degrees

Therefore, your answer is D.

Flawless execution! It's nice to solve this analytically.

ReplyDeleteThank you six.

ah ah! may I respectfully object?

ReplyDeletethe 3 clock arms move "continuously" but not "independently". Unless otherwise stated the clock is not broken, so the 3 angles should also correspond to a correct time of day, which isn't the case for the optimum solution.

from quick informal search, the biggest figure I could get is at 06:09:50.002858 (+/-1 microsecond) where angles are 121.983, 118.983 and 119.034. This is 99,879% from optimum. But this may not be the "absolute" max.

Maybe somone could come up with an analytic solution for this max?

bleaug

http://www.wolframalpha.com/input/?i=max%28sinx-siny-sin%282pi-x-y%29%29

ReplyDeleteBleaug,

ReplyDeleteAre you telling us that you, too, are believing in the nothing-moves theory as stated by Polar Fox. If you are assuming that Second-hand advances 6 degrees every second DISCRETELY (as most clocks do), then you may be right. Otherwise, I can't see why 120-120-120 perfect split is not possible.

Let's try this way. Suppose the clock is at 00:00:00. About 21 minutes later (say around 00:21:00) alpha is very close to 120...

Wait a second.

Wait a second.

I am thinking.

Please wait a second.

Wooow..

I think you are right Bleaug! There is a chance that when the Second-hand reaches 120 degrees (beta=120), then alpha might have already advanced beyond 120 degrees. Actually there is a chance that 120-120-120 may never be possible!

Woow, are we saying that 3 clock arms (discrete or continuous) can NEVER be equally separated???

my informal search above was based on wrong calculations. However the conclusion still holds.

ReplyDeleteIf t is the time of day expressed in seconds, the maximum we look for must be an extremum of the function:

f(t)=sin(11.pi.t./21600).sin(708.pi.t/21600).sin(719.pi.t/21600)

some properties:

- this function has a period of 43200 = 12h which makes sense

- if T is an extremum, 43200-T is also an extremum

- extremums can be split in two sets: those which approach a 120-120-120 angle separation (max) and those which approach a 60-60-240 angle separation (min)

Based on this, the maximum is achieved for the two following timestamps:

a) 2:54:34.56169

b) 9:05:25.43831 = 12:00 - 2:54:34.56169

modulo 12:00, i.e. AM/PM of course.

This max is 99,999139% from theoretical max = 3.sqrt(3)/8

bleaug

@8foxes

ReplyDeletewell, looks like another version of Zeno's paradox (http://en.wikipedia.org/wiki/Zeno%27s_paradoxes)

bleaug

PS: you can call me "Polar Fox" ;-)

Yes, Bleaug is absolutely correct. Thank you for pointing that out.

ReplyDeleteIf we have

11xmod12 (minute hand)

719xmod12 (second hand)

This should give us the distances from the hour hand, where the hour hand is just xmod12. So if we want a clock with perfect 60 degree angles between each clock, then these set of equations should be satisfied.

11x = 4mod12

719x = 8mod12

x can't be an irrational number, for then the product would be irrational and not 4mod12 or 8mod12. However, we also see that since 11 and 719 are prime, the only rational numbers that can work for this expression are n/11 and t/719, respectively. However, 719 and 11 are coprime, therefore x must be an integer.

i.e. Denominator must be 719*11, but say, divide 719 by 719*11, and you will see that x must be a multiple of 11 in order for the product to be an integer, vice-versa for 11 divided by 719*11.

Since x must be an integer, there are no solutions to this system of equations.

The general solution for the first is 8 + 12n, and the general solution for the second is

4 + 12k.

Ah, and I forgot to check the case where the second hand is 4 units away from the hour hand. But the results should be the same.

ReplyDeleteSo here we are:

ReplyDeleteBleaug and his alter-ego Polar Fox, rightfully objected the claim and proved their case. But the amount of work in this effort should not be wasted. Here's what I am suggesting:

1. update the options for this fox (so that the truth shall be spoken all over the land)

2. construct a new fox for which six's good analytic solution remains valid, and

3. construct a new fox to coin the claim that "Clock arms can never be equally-separated".

These should be completed within this week.

Any objections? Please ring the bell, otherwise, thank you for the great work!