Medium | Discrete Maths |

On a table you have a square made of 4 coins at the corner at distance 1. So, the square is of size 1×1. In a valid move, you can choose any two coin let’s call them mirror and jumper. Now, you move the jumper in a new position which is its mirror image with respect to mirror. That is, imagine that mirror is a centre of a circle and the jumper is on the periphery. You move the jumper to a diagonally opposite point on that circle. With any number of valid moves, can you form a square of size 2×2? If yes, how? If no, why not?

Hint

Invariance

Answer

No!

Solution

Source: Saurabh Joshi's Blog

Enable Like and Comment Hard | Discrete Maths |

An optimist and a pessimist are examining a sequence of real numbers. The optimist remarks, ‘Oh jolly! The sum of any eight consecutive terms is positive!’ But the pessimist interjects, ‘Not so fast, the sum of any five consecutive terms is negative.’ Can they both be right? Atmost how large can this sequence be?

Hint

One easy proof is to construct rows, starting from ai, and satisfying one of the sum properties - optimist or pessimist. The column sum can follow the other property. Try making a contradiction.

Answer

11

Solution

Suppose such a sequence with terms {ai} has length 12

a1+a2+...+a5<0

a2+a3+...+a6<0

..

a8+a9+...+a12<0

Note that rows sums are negative and column sums are positive. This is contradiction to the total sum! So max length of sequence is 11. This sequence can be constructed but we wont cover the instruction. Example: 1 at positions 1,3,4,6,8,9,11 and -1.6 at positions 2,5,7,10.

a1+a2+...+a5<0

a2+a3+...+a6<0

..

a8+a9+...+a12<0

Note that rows sums are negative and column sums are positive. This is contradiction to the total sum! So max length of sequence is 11. This sequence can be constructed but we wont cover the instruction. Example: 1 at positions 1,3,4,6,8,9,11 and -1.6 at positions 2,5,7,10.

Source: Arthur Engel

Enable Like and Comment Hard | Discrete Maths |

We want to construct a structure made as follows: imagine that two long cylindrical pillars each with radius 1 intersect at right angles and their centers also intersect. What is the volume of this intersection?

Answer

16/3

Solution

If you cut the intersection by a horizontal plane at distance z from center, the cut will be a square with side-length 2*sqrt( 1-z^2). Integrate to get volume 16/3.

Another way is to imagine the largest possible sphere inscribed at the center of intersection. The sphere should have a radius of 1. At each cut perpendicular to the z-axis, the circle from the sphere is inscribed in the square from the intersection as well, So Area of cut-circle = (Pi/4)*Area of cut-square. This is true for all z, hence Volume of sphere = (Pi/4)*Volume of Intersection, this also gives 16/3

Another way is to imagine the largest possible sphere inscribed at the center of intersection. The sphere should have a radius of 1. At each cut perpendicular to the z-axis, the circle from the sphere is inscribed in the square from the intersection as well, So Area of cut-circle = (Pi/4)*Area of cut-square. This is true for all z, hence Volume of sphere = (Pi/4)*Volume of Intersection, this also gives 16/3

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