Crazy Hand and Torus-Clock

Victor Gutenmacher

A Clock. Linkage between hands.

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A dozen problems:

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The following problems are not very difficult,

except for the problem 12, but solving the problem 12 is

our goal.

0. Compose (make a picture) funny clocks, watches and puzzles.

For instance:

A clock is a plate, the hour hand is a knife,

the minute hand is a fork, the numbers are fruits.

A necklace-watch has a tie-pendulum.

A clock has the shape of a hand with five fingers.

A clock has a cuckoo (a bird, a cat, a dog).

A clock has an alarm with dynamite.

A sun-clock with Roman numerals has a cover.

A clock with half face.

A mirror-clock with opposite direction, but

you can watch in the mirror.

(people use the word "clock-wise"

for the orientation)

Some funny metronomes.

What else ?

We will also invent in the end of this paper some funny things:

a clock with a Crazy-Hand and a Torus-Clock.

Remark:

The following puzzle needs a picture. Let do it yourself:

A puzzle: on the picture there are several clocks.

Which of them are wrong or stopped ?

1. What time is it ?

There are a few pictures of a clock with two hands,

which of them make sense?

For instance, the clock shows 6 o'clock. Let us exchange (switch)

the minute and hour hands. Does this clock show any time,

or the linkage between hands is broken.

2. Can we figure out the time when our clock has only one

hour hand ?

We can show on a picture a position of the hour hand

and ask to point approximately the position where the minute hand

has to be.

3. How many times faster is the second hand than

the minute hand ?

4. How many times faster is the minute hand of a clock than

the hour hand ?

In one hour there are 60 minutes and also

in one minute there are 60 seconds,

but the answers for problems 3 and 4 are different.

5. Sometimes the minute and hour hands coincide to each other.

It happens when the minute hand catches up to the hour hand.

How many such positions are there ?

Experiment:

Certainly we can watch a clock for a long time but

maybe you can rotate them yourself.

There is a linkage between hands of the clock.

If you will get the answer: 12 positions, know that this is wrong.

6. How many times does the minute hand point perpendicularly to

He also created all sorts of problems and this is one of his favorites:

"Let us pretend that we have two clocks.

One of them has stopped completely.

The other one gains half a minute every twenty-four hours.

Which of the two clocks tells the exact time more often ?"

8. What is the approximate time when the minute hand catches up

for the first time to the hour hand after 12:00?

9. What is the exact time when the minute hand catches up

for the first time to the hour hand after 12:00?

10. Can we always figure out the time by a clock with

two hands of equal size ?

The answer: Certainly, yes, because the movement of the

minute hand shows that this one is the minute hand.

11. Is there a moment when we can switch (exchange) the minute

and hour hands and get some other reasonable time ?

Certainly, if they are coincide to each other we can exchange

them, but maybe there exist another positions.

Try to watch them on a clock.

This is the end, because it is too difficult to move further.

12. How many positions are there when we can switch (exchange)

hands and get a position which makes sense?

Experiment: We can watch a clock (or, better, rotate its hands) and

find out that such position appears approximately every 5 minutes.

Now, we will give solutions which don't use

algebra. They can be represented in computer in very nice way

we are optimists.

Crazy-Hand Solution of the problem 12.

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Let us tell a story.

We had a guess ( a hypothesis):

Let a clock be with three hands - the hour hand,

the minute hand and the second hand.

if the second hand catches up to the hour hand,

we can exchange the minute hand with the hour hand

and get a position which also make sense.

But when we watched a clock, we saw that this is wrong,

but after every five rotation (5 minutes) of the second

hand it was almost right.

So, we came to the idea, that, if we will invent another

hand instead of the second hand, it will be true.

We were lucky and could also solve the problem 12.

What is a crazy hand ?

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Let us pretend that there exists a "crazy" hand which

goes around the clock 12 times faster than minute hand.

(It is like the second hand, but the second hand

is 5 times faster than this crazy hand).

It turns out that all such "switch positions" happen

when the crazy hand catches up to the hour hand.

(we will prove this a little bit later but you can check yourself).

It happens 144-1 times (the crazy hand 12 times faster

than minute hand and the minute hand 12 times faster than

hour hand, 12x12=144).

The answer to the problem 12

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Therefore the answer to problem 12 is that

there are 144-1=143 such positions.

We subtracted 1 from 144, because initial and last positions

are coincident.

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Among these positions, as you know (if you solved the problem 5)

there are 11 position when the minute

and hour hands are coincident (see the problem 5).

If we subtract this number 11 from 143,

we get the number

143 - 11 = 132.

These 132 positions appear by pairs, because they are

switch positions,

And we can also say that there exist 132:2 = 66

switch pairs.

What we have to do - only to prove that crazy hand

works in such way which we say.

Let us begin our proof in the way which mathematicians like:

the definitions - what we mean by some names,

the theorem - what we want to prove,

the lemma - what can help us.

But, if you would like, we can construct a clock with crazy

hand in a computer and check it with experiment.

Let's pretend that when the crazy hand catches up to

the hour hand we hear a bell.

Definitions

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Let' s name a hand which goes faster 12 times faster

than the minute hand a crazy-hand.

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Let's name switch-position the layout of two hands on the clock

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when switching these hands gives a reasonable time

(and, certainly, another switch-position).

Let's name crazy-watch a watch with two hands : the shorter one

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is a minute hand, the other one is a "crazy hand".

We can propose the following theorem.

The theorem

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Let a regular clock be given a third hand - the crazy hand.

A switch-position happens if and only if (iff)

the crazy hand coincides with hour hand.

(the crazy hand runs 12 times faster than the minute hand)

The proof

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Let's prove at the beginning the following lemma.

Lemma

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The switch-positions on a crazy-watch with minute and crazy

hands are the same as the switch-positions as on a regular watch.

Proof of Lemma

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The linkage between the hands on a regular

watch in on a crazy watch the same.

Indeed, the only difference between a crazy-watch and

a regular watch is that the crazy-watch runs 12 times

faster. Thus, the switch positions on both watches

are the same, and

the lemma is proved.

Now let us prove the theorem:

Let's pretend, for convenience, that there are two watches -

a regular watch and a crazy-watch and these watches

lie near each other and their minute hands are

moving parallel to each other.

On either watch, the position of the slower hand determines

uniquely the position of the faster hand, and if the slower

hand is pointing in one of the two directions of a switch

position, then the faster hand is pointing in the other.

Suppose the regular watch has just attained a switch-position.

Then the minute hand of the crazy watch (being parallel to

the minute hand of the regular watch) is pointing in one

direction of a switch position, and the crazy hand is

pointing in the other. Since, by the lemma, the switch

positions of the regular and crazy watches are the same,

the hour hand on the regular watch and the crazy hand on the

crazy watch must be parallel. This

means that in this moment of switch-position

on a watch with three hands, both hands -- the hour

and the crazy hand -- are coincident.

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Torus-Clock and a new solution of the problem 12.

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Let us invent a more interesting clock:

One small round clock with one hand

moves around another big round clock

that the plane of the first one all the time

is perpendicular to the plane of the other clock.

Sweeping of the small clock looks like a bagel.

The surface of the bagel is called "torus".

The different positions of the small clock are

called meridians of the torus, the sections of the torus

which are parallel to the big clock are called

its latitudes or just parallels.

The more advanced solution of the problem 12

is the following (we are thinking about Configuration Space).

Let us introduce the clock coordinates on a meridian

and a latitude circles of the torus. Let the minute hand

move around the meridian circle and hour hand moves around

the latitude circle.

Let a point P of the torus have the coordinates:

the end of hour hand and the end of the minute hand.

The trajectory of the point P is a spring on the torus

which rotates one time around the latitude and 12 times

around the meridian.

Now, let us switch the minute and hour hands on the torus.

Then, the trajectory of the point P will be the spring

which rotates 12 times around the altitude and 1 time

around the meridian. The intersections of these two

springs correspond to the switch positions.

These two springs intersect each other in 144-1 points.

Indeed, the first spring intersects every latitude 12 times,

but the second spring has 12 latitudes.

We subtracted 1 because these springs go out and come back

to the same point.

(we draw such torus-clock and spirals in computer).

The next problem (this is a joke):

Again:

0. What time is it ?.

1p.m, 2p.m, 3p.m .....

Algebra for a clock.

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Let us try to solve the problem:

9. What is the exact time when the minute hand catches up

for the first time to the hour hand after 12:00 p.m.?

It is clear that minute hand hand catches up the hour

hand for the first time only after 1 p.m. o'clock.

Let us pretend that we stretch the circle of a clock into

a straight line - a trail and

there are 2 runners (or cars) M and H

on the such trail (it marked as a clock):

M--> H->

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| | | | | | | | | | | | | |

12 1 2

In 1 p.m. o'clock M and H start and run. We know that M runs

12 times faster that H. The speed V of M is equal 1 mark per minute,

the speed v of H is equal 1/12 of one mark per minute.

Suppose the t is the time when M catches up H.

For this time M has ran the distance 1*t, but H has ran the distance

1/12 * t. The first distance, on 5 marks more than

the second one. Thus,

t = 1/12t + 5

or 11/12t = 5.

>From this equation we find out that t=12/11*5 or t = 5+1/11 minutes.

The answer: the minute hand catches up the hour hand at

1p.m and 5+1/11 minutes.

If we would like to know when the minute hand catches the hour

hand the next time , we can write the new equation:

t = 1/12t + 10.

If we would like to know when the minute hand catches the hour

hand the third time , we can write the next equation:

t = 1/12t + 15.

What the exact time the minute hand catches the hour hand

last time before 12 a.m. ?

Advanced PreCalculus.

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The previous solutions can be rewritten in the following

manner:

if x is the coordinate of the hour hand and y is the

coordinate of the minute hand, then

y - 12{x} = 0, 0 =< x < 12, 0 =< y < 12;

where {x} is the fractional part of the number x.

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If we exchange these hands we will get the other condition:

x - 12{y} = 0, 0 =< x < 12, 0 =< y < 12;

(one can draw these graphs by different color).

The switch-positions correspond to 144-1 points (x,y)

of the intersections of two graphs of the functions:

y = 12{x} and x = 12{y}.

Appendix

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The e-mail-solution which I've got from one professor:

Let q denote the angle (say, in the clock-wise direction) between the

hour and minute hands of a clock. Thus, using Victor's terminology,

every switch- position corresponds to a q such that both q and 2np-q

make sense (p is "= pi").

For each q, the number of times q occurs in a 12-hour cycle corresponds

to a number of switch-positions associated to q.

0. Fact: the minute-hand move 12 times faster than the hour-hand.

1. By fact 0, every q that makes sense happens 12 times in a 12-hour

cycle.

2. Let q be such that q and 2np-q both make sense. By fact 0, the

angle

(i.e. 2np-q, n > 0) subscribed by the minute-hand from a switch

position associated to q to its corresponding counter-position is equal

to 12 times the angle (i.e. q) subscribed by the hour-hand from the

same switch-position to its

corresponding counter-position. Thus

12q =3D (2np - q)

13q =3D 2np

q =3D 2np/13.

Possible solutions for q modulo 12-hour are thus given by

n =3D 1, 2,...,12.

3. By 1. And 2., there are therefore 12 times 12 =3D 144 switch

positions.

My remark: it has to be 143 switch positions

Cube-clock

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I've visited my friend Leonid Levin

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he is the professor at BU and

one of founders of Computer Science.

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My gift to his 3 gifted children was the list problems with clocks.

After sunset on Saturday we sat round the big table

and were solving these problems. It was a fun.

They've solved all problems. All children suggested

ideas about new odd clocks.

Leonid himself also suggested an idea about new clock.

His idea rotating Cube-Clock. A cube has 6 faces, 12 edges and

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its group of transformations is a group with 24 elements:

this group is the group of permutations of its 4 big diagonals (4!=24).

The number of even permutations is equal to 12.

Also, the group of permutations of 5 elements consists

of 5!=120 elements, and respectively the number of even

permutations is equal to 60 and can be illustrated by the other

polyhedron.

This is a good fabulous for introduction of Modern Algebra.