Can anyone help me with this problem?
Suppose a sphere is colored in two colors: 10% of the surface is white, and the remaining part is black. Prove that there is a cube inscribed in the sphere such that all vertices are black
thanks
~matt
There is a famous problem called the tower of hanoi in which you have n disks of increasing diameter, and three pegs. The object is to move them from one peg to another in such a way that no larger disk is on top of a smaller disk.
the number of moves is defined by
T(n)=2^{n}-1
A far...
so... the problem that you are proposing is,
given probability functions based on the distance from one player to another, what is the ideal initial setup to allow for an equal chance of each player living?
someone had said...
Player 3 is the most likely to win followed by player 2. Player 1 is the least likely to win.
this all has to do with the values of x,y, and z...
imagine that player one is a great shot, and hardly ever misses, and that player 2 and 3 can't hit the broad side of a...
you have all heard of a duel... but what about a truel...
well... a truel is a duel with three people...
in this specific truel, there are three gentlemen players who each wait their turn before they shoot.
player 1 will hit the person he shoots at with probability x
player 2 will...
I actually think that i found a solution for player 1 to always tie...
if player one chooses the middle square first... and then mirrors all of player 2's choices (i.e. if player 2 chooses top right, then player 1 then chooses bottom left), it works out to a tie every time...
Im doing a project on the data encryption standard (DES), and was wondering if anyone had any ideas on where to find papers that discuss the mathematics behind the system.
thanks
first of all, i just realized that i have the bounds wrong...
they should be from 0 -> pi... sorry about that, but
to get the \frac{\pi}{2} part
take the equality... and subtract one from the other...
\int_{0}^{\pi} xf(\sin{x})dx = \frac{\pi}{2} \int_{0}^{\pi} f(\sin{x})dx...
i found the solution of it...
the hint that I got was--
\int_{0}^{1} xf(\sin{x})dx = \frac{\pi}{2} \int_{0}^{1} f(\sin{x})dx
from there, it isn't to bad...
The game of reverse Tic-Tac-Toe, called Eot-Cat-Cit, has the same rules as Tic-Tac-Toe, except that the first player with 3 markers in a straight line looses. Can the player with first move avoid being beaten?
(note: a tie counts as a win for player 1)
so, more or less
\sum_{n=1}^{[\frac{n}{2}]} \frac{1}{2n}
(that is, if the weight of the ant = 0)...
do you remember how you came up with that?
thanks...
no, no weights or glue can be used...
and I thought that it would be 1 as well (well, i thought that the placement for the cards would be related to 1/2^n... but i tried it out... and can definitely get more than 1 card length over the edge...
any ideas?
first of all I would like to say that this is a question to which I do not have the answer...
imagine that you are building a set of stairs for an ant out of a deck of n cards... what is the greatest horizontal distance from a table out over an abyss that can be covered over by the cards...