Mathematics 11 - Puzzle Prime

Strips in the Plane

Section: Brain Teasers Category: Mathematics

Difficulty: Hard

Prove that you can not cover the plane with infinite strips which have a total sum of their widths equal to 1.

SOLUTION

Take a circle with radius 1 in the plane. A strip with width X covers at most an area of 2X of the circle. Therefore all strips cover at most an area of 2, which is smaller than the total area of the circle (~3.14).

Unknown Author February 17, 2018

0 Comments

Consistent Polyhedron

Section: Brain Teasers Category: Mathematics

Difficulty: Advanced

Can you construct a convex polyhedron, such that no two of its faces have the same number of edges?

SOLUTION

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.

Puzzle Prime February 6, 2018

0 Comments

Pawns on the Chessboard

Section: Brain Teasers Category: Mathematics

Difficulty: Expert

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.

SOLUTION

Assign the following weights on the squares of the chessboard:

1 on the main diagonal a8 – h1
1/2 on the diagonal b8 – h2
1/4 on the diagonal c8 – h3
1/8 on the diagonal d8 – h4
1/16 on the diagonal e8 – h5
1/32 on the diagonal f8 – h6
1/64 on the diagonal g8 – h7

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.

Unknown Author January 31, 2018

0 Comments

Integers on a Chess Board

Section: Brain Teasers Category: Mathematics

Difficulty: Hard

You are given an 8×8 chess-board, and in each of its cells, there is written one integer. If the difference between any two adjacent numbers is -1, 0 or 1, prove that some number is repeated at least 8 times on the board.

SOLUTION

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.

Unknown Author January 22, 2018

0 Comments

Obtuse Angle

Section: Brain Teasers Category: Mathematics

Difficulty: Expert Tags: Geometry

Prove that among any 9 points in (3D) space, there are three which form an obtuse angle.

SOLUTION

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 A_i and A_j, P must be bound between the planes passing through A_i and A_j which are orthogonal to the segment A_iA_j. 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.

Puzzle Prime December 25, 2017

0 Comments

Lost Boarding Pass

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Probability

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 in his own seat?

SOLUTION

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 been any preference made by anyone at any time towards any of these two seats, the probability that either of them is left last is 1/2.

Puzzle Prime December 16, 2017

0 Comments

Mysterious Polynomial

Section: Brain Teasers Category: Mathematics

Difficulty: Advanced

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)?

SOLUTION

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.

Puzzle Prime December 13, 2017

0 Comments

Larger or Smaller

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Probability

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?

SOLUTION

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%.

Unknown Author November 16, 2017

0 Comments

Wizards with Hats

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Probability

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%?

SOLUTION

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.

Unknown Author November 4, 2017

0 Comments

NASA and the Meteor

Section: Brain Teasers Category: Mathematics

Difficulty: Advanced Tags: Geometry

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?

SOLUTION

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.

Unknown Author October 10, 2017

2 Comments