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MO Proof May 1, 2015

MO’ Proof

(Gerry’s Modified KZ’s Proof)

The Famous Watch Story

This proof is a variation on KZ’s proof.  The main difference is that it is “less restrictive” in terms of how one might decide to “measure” the angles between any two hands of the clock.  Also, I think that this rendition is more clear and easier to follow.

Again, we restate the basic problem. What times are there (and what are they) such that the angle between the hour, minute, and second hands are the same. That is, the angle between h and m, h and s, s and m are 120o or 240o       (2/3 * Pi or 4/3 * Pi). Or put another way, the hour, minute, and second hands divide the face of the clock into 3 equal pieces/areas. The location of the hands must represent a legitimate time. That is, they can not be simply placed to arbitrary positions such as 12, 4, and 8. Also, the clock doesn’t “tick” but rather it runs smoothly (more on this a little later).

Again, to solve the problem, we start the clock at 12:00:00 and let it run. That is, all hands start at “zero degrees”.

The position of all of the hands, in degrees or radians, is described by:

h = h
m = 12*h
s = 720*h

 

That is, the second hand travels at a rate of 60 times the minute hand, and the minute hand travels at a rate of 12 times the hour hand. The clock runs “smoothly” and doesn’t “tick” so h, m, and s can be any (positive) value… not just integers!

For this proof, we’ll need to know what the positions are, relative to zero, for each hand. Using the “12” position as “zero”, we’ll need to know the “relative zero” angle/position of each hand regardless how many degrees the hands have actually traveled. For example, if a hand had traversed 450o then its position relative to zero (12 o’clock) would be 90o.

We’ll designate:

 
h0 as the “relative zero” position for the hour hand.

 
m0 as the “relative zero” position for the minute hand.

 
s0 as the relative zero position for the second hand.

 

At any instant in time, it’s always, and only, the “relative zero” positions that we see. Remember… we are looking for the time(s) such that the current position of the hands of the clock “divide” it into 3 equal pieces. That is, where the “relative zero” positions of the hands, at that instant, “divide” the face of the clock into 3 equal pieces.

For such a case to occur then the angle between any two hands must be one of: 120o, -120o, 240o, or -240o.

For example, let’s say the hour hand is exactly at the “4” position. Then in order to divide the face of the clock into 3 equal pieces the second hand would have to be at “12” (-120/+240)   or at “8” (+120). The same would hold true for the minute hand!

If you buy into this then two things become apparent:

 
(s0h0 ) must be one of: 120o, -120o, 240o, or -240o

(s0m0) must be one of: 120o, -120o, 240o, or -240o

In other words,

The angle between s0 and h0 is +-120o or +-240
The angle between s0 and mis also +-120o or +-240o

 

Now, here come some key points:

     h0 = h
     m= m – 360*nm
     s0 = s – 360*ns

Some notes on the above:

nm and ns are simply positive integers, or zero. In other words, for any hand, the “relative zero” position is simply the remainder of the total angle traversed divided by 360. For the hour hand, we don’t really need a relative zero position because after 12 hours all of the hands start back at zero again anyway!  So far everything should make sense and be very straightforward.

Now we want to calculate the angles between the “relative zero” positions of the second and hour hands, and between the second and minute hands. That is,

We will calculate (s0 h0) and (s0 m0).

 First we calculate  s0 h0

 
s0 – h0 = s     – 360ns – h
s0 – h0 = 720h  – 360ns – h
s0 – h0 = 719h  – 360ns

 

Since s0 – h0 must equal 120, or -120, or 240, or -240 then we have:

 

719h – 360ns = 120*(+- 1 or 2)

 

h    =   120*(+- 1 or 2)   +   360ns
         ~~~~~~~~~~~~~~~~      ~~~~~~
             719                719

 

 

Now we will calculate s0 – m0  (the “relative zero” angle between the Second and Minute hands).

s0m0 = s – 360ns – (m – 360nm)

s0m0 = 720h – 360ns – 12h + 360nm

s0m0 = 708h + 360(nmns)

 

For simplicity, let the integer nms = (nmns).

Again, s0m0 must be 120, or -120, or 240, or -240 so we have:

708h + 360nms = 120*(+- 1 or 2)

708h = 120*(+- 1 or 2) – 360nms

 

h = 120*(+- 1 or 2) – 360nms
    ~~~~~~~~~~~~~     ~~~~~~~
         708            708

 

 

       On the prior page we calculated s0 – h0 and s0 – m0 and then for both cases we solved for h. So, now we can equate the two and see what happens…
       

 

 s0 – h0                                 s0 – m0
<<<<<<<<<<<<<<<<<<<<<<<<<<<<<        >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>


120*(+-1 or 2) + 360ns          120*(+-1 or 2) – 360nms
~~~~~~~~~~~~~~~ ~~~~~~      =   ~~~~~~~~~~~~~~   ~~~~~~~~ 
      719        719               708              708

 

120*708(+- 1 or 2) + 708*360ns =    120*719(+- 1 or 2)  –  719*360nms

 

360(708ns + 719nms)            =    120(719(+- 1 or 2)  –  708(+- 1 or 2))

 

708ns + 719nms                    120 * (719(+- 1 or 2)  –  708(+- 1 or 2))
                                     ~~~
                                     360
 

 

INTEGER        =  NON-INTEGER

 

On the left side of the above equation we have an integer. On the right side of the equation we have 1/3 times the following 16 possible values (below) none of which are evenly divisible by 3:

 

 719 – 708             = 11

719 – (-708)           = 1427

719 – 708*2            = -697

719 – (-708*2)         = 2135

 

-719 – 708             = -1427

-719 – (-708)          = -11

-719 – 708*2           = -1416 – 719 = -2135

-719 – (-708*2)        = 1416 – 719 = 697

 

719*2 – 708            = 1438 – 708 = 730

719*2 – (-708)         = 1438 + 708 = 2146

719*2 – 708*2          = 1438 – 1416 = 22

719*2 – (-708*2)       = 1438 + 1416 = 2854

 

-719*2 – 708             = -2146

-719*2 – (-708)          = 708 – 1438 = -730

-719*2 – 708*2           = -1438 – 1416 = -2854

-719*2 – (-708*2)        = -1438 + 1416 = -22

 

In all of the 16 possible cases the right hand side of the equation results in a non-integer. Since an integer can’t equal a non-integer we have a contradiction and therefore there are not any times such that the hands of a clock cut the clock into 3 equal pieces!

*************** end of MO’ Proof *************

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