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Can you construct a convex polyhedron, such that no two of its faces have the same number of edges?
No, you can not construct such a polyhedron. Assume the opposite and consider its face, which has the largest number of sides, say k. Then the polyhedron contains at least k more faces with different numbers of sides, all less than m. However, this is clearly impossible.
Six pawns are placed in the middle squares of the main diagonal of a chess board – b7, c6, d5, e4, f3, g2. You are allowed to take any pawn on the chessboard and replace it with two pawns – one on the square above it and one on the square on its right, in case there are empty squares there. If after several moves there are no more pawns on the main diagonal, show that all the squares above it except for h8 are covered by pawns.
Assign the following weights on the squares of the chessboard:
Every time you make the splitting move, the total sum of the numbers of the squares covered by pawns remains a constant. At the beginning that sum is 6. Since 7/2 + 6/4 + 5/8 + 4/16+ 3/32 + 2/64 = 6, all 27 squares above the main diagonal, except the top-right corner (on which you can not place a pawn in any way), must be covered by pawns at the end.
You are given an 8×8 chess-board, and in each of its cells
Consider the intervals spanned by the numbers in the first row, second row, third row, etc. If all of these intervals intersect each other, then there is a number, which appears in all of them. If not, there are two intervals, which are disjoint, and a number between them, which does not appear in the two rows. Now it is easy to see that this particular number will appear in every column.
Prove that among any 9 points in (3D) space, there are three which form an obtuse angle.
Let the points be labeled A1, A2, … , A9, and P be their convex hull. If we assume that all angles among the points are not obtuse, then the interiors of the bodies P + A1, P + A2, … , P + A9 should be all disjoint. That is because, for every Ai and Aj, P must be bound between the planes passing through Ai and Aj which are orthogonal to the segment AiAj. However, all of these 9 bodies have the same volume and are contained in the body P + P, which has 8 times larger volume. This is a contradiction, and therefore our assumption is wrong.
There are 100 passengers which are trying to get on a plane. The first passenger, however, has lost his boarding pass, so decides to sit on an arbitrary seat. Each successive passenger either sits on his own seat if it is empty or on an arbitrary other if it is taken. What is the chance that the last person will sit
The chance is 50%. Indeed, the last passenger will either have to sit in his own seat or the one which belongs to the first passenger. Since there hasn’t
You are given a polynomial P(x) of unknown degree with coefficients which are non-negative integers. You don’t know any of the coefficients, but you are allowed to evaluate the polynomial at any point you choose. What is the smallest number of evaluations you need to use, so that can find the degree and the coefficients of P(x)?
Just 2 evaluations are enough. First, get P(1). This will give you an upper bound on the number of digits of the largest coefficient of the polynomial, let’s say that is N. Now evaluate the polynomial at the point M = 10 to the power of N. This will make all of the coefficients of P appear in the number P(M), separated by strings of zeros.
Alice secretly picks two different integers by an unknown process and puts them in two envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss) and shows you the number in that envelope. Now you must guess whether the number in the other, closed envelope is larger or smaller than the one you have seen.
Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?
Choose any strictly decreasing function F on the set of all integers which takes values between 0 and 1. Now, if you see the number X in Bob’s envelope, guess with probability F(X) that this number is smaller. If the two numbers in the envelopes are A and B, then your probability of guessing correctly is equal to:
F(A) * 0.5 + (1 – F(B)) * 0.5 = 0.5 + 0.5 * (F(A) – F(B)) > 50%.
There are 2 wizards and each of them has infinitely many hats on his head. Every hat has 50-50 chance to be white or black, and the wizards can see the hats of the other person, but not their own. Each wizard is asked to identify a black hat on his head without looking, and they win if both succeed to guess correctly. If the wizards are allowed to devise a strategy in advance, can they increase their chance of winning to more than 25%?
Each wizard guesses the position of the lowest black hat on the head of the other wizard. Then the chance of winning becomes 1/4 + 1/16 + 1/64 + … = 1/3. It can be shown that this is an optimal strategy as well.
NASA locates a meteor in outer space and concludes that it has either a cubical or spherical shape. In order to determine the exact shape, NASA lands a spacecraft on the meteor and lets a rover travel from the spacecraft to the opposite point on the planet. By measuring the relative position of the rover with respect to the spacecraft throughout its travel on the planet (in 3D coordinates), can NASA always determine the shape, no matter the route taken by the rover?
The answer is NO.
The question is equivalent to analyzing the intersection of a cube and a sphere which share a common center. Thus the question gets reduced to figuring out whether such intersection, which is a curve, can connect two opposite points on the sphere/cube.
Let the edge of the cube has length 1. If you pick the radius of the sphere equal to √2/2, the intersection will consist of 6 circles inscribed in the sides of the cube. Then the rover can just move along these circles from one point to its opposite and NASA won’t be able to figure out the exact shape.
Remark: It is not hard to see that 2:√2 is the only edge-radius ratio, for which NASA can’t figure out the shape.
A town consists of 3 horizontal and 3 vertical roads, separated by 4 square blocks. A policeman and a thief are running along the roads with speeds of 21km/h and 10km/h respectively. Show that the policeman has a strategy ensuring he will eventually see the thief.
Remark: The policeman can see the thief if they are on the same road at some moment. He has no idea about his position at any time.
A working strategy for the policeman would be to go to the center and to start encompassing the four blocks clock-wise one by one, in a clockwise manner.
Since the policeman is twice as fast as the thief if the thief is in the center of the town at some point, then there exists a moment in which the policeman is in the center, and the thief is not on the boundary, i.e. he gets shot.
Now, assuming the thief never visits the center, his angle with respect to the coordinate system defined by the two middle roads changes continuously. The angle of the policeman with respect to the same coordinate system can be defined to change continuously as well. Since the policeman needs less time to increase his angle with 360 degrees than the thief, there will be a moment when the two have the same angle. However, this implies that the policeman will be able to see the thief and shoot him.
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