[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 340

New comment
 
ihor >>:

в непосредственной близости — необязательно
но как получить палиндром
ВВАААВВ + АА

And this is something the 2nd player should have prevented on the previous move. By swapping A with the inside B of the last group.


 
vegetate >>:

А вот этого 2-й игрок должен был недопустить на предыдущем ходу. Поменяв А с внутренней В последней группы.


then we need an algorithm

why we should get AVBABABABABABA from ABBA+AB

and not BABABAB

although for

how do we add ABBABAA?


 
Mathemat писал(а) >>
Two players play the following game: the first player writes the letters A or B in a row as he wishes (from left to right, one after the other; one letter per move), and the second player, after each move of the first player, swaps any two of the written out letters or changes nothing (this also counts as a move). After both players have made 1999 moves each, the game is over.
Can the second player play in such a way that any move by the first player results in a palindrome (i.e. a word that reads the same from left to right and from right to left)?


He can.
The second player has to make sure that one of the letters stacks up in the centre and the other stacks up to the right and left of the first.

 
ihor писал(а) >>

It doesn't work that way.

Let's say it was

ABABA or BAAAB.

add BB first



ABABA + BB

BAAAB + AA → no


BAAAB + BB

BBAAABB + AA → no

BABABAB +BB

BBBAAABB +AA → no



Cannot be added like this, as 1 letter is added per move (s.m. conditions)

 
PapaYozh >>:


Нельзя так добавлять, т.к. за 1-н ход добавляется 1-на буква (с.м. условия)

I was discussing the solution.

"The permutations must be done with an odd number of letters and starting with the third, so that each permutation respects the palindrome."

and I wanted to show that I'm missing the rule for building a new palindrome.
If you put some letters in the centre and others on the edges, you're OK.


And the answer is yes it can.

But I have a different tactic...

 
Hello again :) (с)

Suppose there is such a strategy. Then on an odd move following this strategy we will always have a palindrome.
Let's represent it like this
(P)X(R) where R (reverse) is a mirror image of P (part) and X can be any letter -- A or B.
Strategy:
The player writes x1
1. If x1 == X, append it to the middle and then x2 regardless of its value also to the middle. The result is (P X) x2 (X R) -- a palindrome.
2. If x1 != = X, add to the middle, you get (P) X x1 (R).
2.1. x2 == X we write as follows: (P) X x1 X (R) == (P X) x1 (X R) -- palindrome.
2.2. x2 != X, i.e. x2 == x1 we write so: (P) x1 X x1 (R) == (P x1) X (x1 R) -- palindrome.

All that remains is to find the initial palindrome step. This always exists -- it is the first step when the palindrome consists of a single letter.
 
Score, TheXpert. Next:
A flat convex figure is bounded by segments AB and AC and arc BC of some circle.
Construct some line which divides in two:
a) the perimeter of this figure;
b) its area.

P.S. It is probably assumed that AB != AC.
 

Again I don't understand what the trick is. It seems obvious: the line passes through point A and divides the arc into 2 equal parts.
I can't insert an image for some reason :)))
See you tomorrow.

 
Or is there something wrong here? )
Or we draw two circles.
one centered at B
one centered at C
with the same radius
radius greater than BA and smaller than BC
we obtain two points at the intersection of the circles which belong to the line we are looking for
 
Richie >>:

Опять не пойму, в чём прикол задачи. Кажется всё очевидно: прямая проходит через точку А и делит дугу на 2 равные части.
Не могу вставить изображение почему-то :)))
До завтра.


it's not like we have anything to measure the angle with
1...333334335336337338339340341342343344345346347...628
New comment