Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 75

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MetaDriver:

It's nice. But it's complicated. The menu is funnier. Any one-color arc should be decorated with three dots, two at the edges and one in the middle. Connect them with straight lines. You get an isosceles triangle.)

// Don't tell me all the arcs are infinitesimal, I'll split them all in half anyway. ;-)

Moderators put it this way: a circle can be coloured by some Dirichlet function (well, whatever). A Dirichlet function is 1 (red) if the number is rational, and 0 (blue) if it is irrational. I.e. there is no question of any continuity.

Neither are we talking about "infinitely close to each other points", because in the case of real numbers it does not exist in principle. The case is most general.

2 alsu: this is something original, I haven't seen a square in the basis of construction yet. I'll show my proof later, but I'll try to figure out yours.

P.S. I figured it out. It is quite correct.

But think about the border between the states.

 
Mathemat:

The moderators put it this way: a circle can be coloured by some Dirichlet function (well, whatever). A Dirichlet function is 1 (red) if the number is rational, and 0 (blue) if it is irrational. I.e. there is no question of any continuity.

Neither are we talking about "infinitely close to each other points", because in the case of real numbers it does not exist in principle. The case is most general.

// They're assholes out there, your moderators.... Tell them so. :-)

// Always suspected that the desire to confuse interlocutors (apparently it's their "purpose") leads to mental inferiority.

Brother Dirichlet is on my side in this battle. I only need two adjacent (having a common point) segments to construct an isosceles triangle, and a set of rational numbers will provide me with plenty of them. But I'm sure your moderators will start wriggling out again and put you into a helpless trance with some more bookish nonsense.

So I had to take irresistible measures and invent such a construction:

Let's inscribe a regular pentagon into a circle. Now let us make the assertion: there is no way to paint the points which are vertices of this pentagon such that it is impossible to construct an isosceles triangle on any three of its points.

For example, arranging the dots as in the picture allows you to construct an inescapable isosceles triangle depicted in blue.

No amount of changing the colours of the dots will save your moderators from defeating the corresponding construction.

Let them stay awake now.

// In fact, any regular N-gon, where N > 4. N=5 is just a minimal case.

 

Yeah, they'll have to wave the white flag. You even disarmed Dirichlet himself :)

// Уроды они там, эти твои модераторы.... Так им и передай.  :-)

Yeah, well, sometimes it seems that way.

 
MetaDriver:

// In fact, any regular N-gon is suitable, where N > 4. N=5 is just a minimal case.

Actually, it's even simpler - the arithmetic mean of two rational numbers is always a rational number, so for any red points there are two suitable isosceles triangles.
 
MetaDriver:

Refutation:

Let's draw two arcs of length Pi/3 of radian from any point on the circle "coloured" by this "method" and at the same time let's construct an isosceles triangle on these points (its two sides lengths will be equal to R). :)

Obviously, only one corner of it is in the shaded point (the reverse contradicted the statement about irrationality of Pi). So, as it turned out, there are at least twice as many holes on this circle as there are shaded points. :))

// What's in inverted commas - read with a snide tone.

The sum of irrational numbers can be a rational number. Example: 1+sqrt(2) and 1-sqrt(2)

Your example should rather use the transcendence of pi, but that in no way prevents me from constructing segments that are irrational, but not transcendent with respect to pi.

 
Actually, I can suggest another construction: We mark an arbitrary diameter in red, divide the resulting arcs in the relation, for example 41/59 or in any irrational relation (not in half, so as not to get isosceles triangles at once), colour in blue, repeat to infinity. In the limit we have a coloring that has no arcs, but, nevertheless, for it the construction as I have above is valid. In general, one can think of as many such colourings as one wants, the main thing is to obtain a set of continuum power, but of dimension less than 1, a kind of fractal.
 
MetaDriver:
///
Volodya, don't you have anything else to occupy your mind with? You already have a task, but you're in no hurry to solve it. If you don't want to do it, just say so.
 

In the comments I found a verse solution to the problem of 23 people who need to be divided into a judge and two teams:

King Saltan and Black Sea<br / translate="no">.
1. CONDITION

In the palace of Tsar Saltan,
Intriguer and tyrant,
In the tsar's warlike retinue
served faithfully.
Twenty-three mighty men
With Chernomor in command.

With his armour shined to the shine,
Chernomor goes to his chambers:
"King Saltan, your soldiers
Waiting for their due wages.
A property of such cunning:
Firing any man to the reserve -
I could divide my retinue
Into two equal halves
So that the amount of money
In each platoon the same.
And all served in the same way
Some better, some worse.
"You must admit, it's not right
to pay all with one coin."

The tsar thought for a while,
And then... Follow strictly
King Saltan's logic,
What did he say to the ataman?


2. DECISION

We shall prove for the answer
The impossibility of this estimate.
Be the wage packets,
As the Blacksmith wished,
The payout would be even
Or, on the contrary, odd,
or else, believe my word,
They'll put such a man in the reserve,
So that the other revenues
Shall not be divided by two platoons.

If such a set be found,
Then the Saltan will deduct
The lowest wages,
And cheap soldiers
In the will of the sultan's evil will
They'll be with zero pay.

If even the best
Won't get a penny,
It means that in the beginning
Everyone got an equal amount.

Or else the tsar will do a bad deed
He'll do a bad deed:
He'll give half as much to all,
As long as everything can be divided.

As a result, in the new estimate
all the zeros will be in place,
And those who have been honored
Will be paid in odd amounts.

Isn't it time to call it a day?
We've proved above
The basic property of an estimate,
But we have no such thing.


3. ANSWER

The king thought a little,
And then, forgetting God,
to the Chernomor, howling,
He said: "If this is not the case,
You and all your soldiers
"You and all the soldiers are out of pay!"

* * *

Businessmen in different countries,
"Sons" of King Saltan,
♪ There's a lot of them now ♪
They're called "scammers."
 
there are two boxes standing on a smooth surface, connected by a spring
  ---------- | | | --------- | | | | | M |--/\/\/\/\/--| m | | | | | | | | | ---------- --------- ========================================
The masses of the boxes are M and m,(M > m) the coefficient of friction is K.
A constant force F acts on one of the boxes.

What is the minimum force F required, and on which box to apply it, to make both boxes move.

_________________________________________________________________________________________

Please do not google, and do not write answers and reasoning.

 
TheXpert:
On a smooth surface there are two boxes connected by a spring
.

The masses of the boxes are M and m,(M > m) the coefficient of friction is K.
A constant force F acts on one of the boxes.

What is the minimum force F, and on which box to apply it, to make both boxes move.

F[M]=M*K*g

F[m]=m*K*g

F[M+m]=K*g*(M+m)

You can apply a force to any box and in any direction - eventually both of them will start moving.

PS This is a simple problem.

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