(inspired from a good Russian algebra problem)

sometimes you just think too much …

PROBLEM

Let and be (monic) polynomials with real coefficients (the first coefficient being equal to ), and . Prove that if the equation has no real solutions, then   has a real solution.

a beetle trapped … (for Katerina :-))

(Infinite pigeonhole principle) Given an infinite set of objects, if they are arranged in a finite number of places, there is at least one place with an infinite number of objects.

PROBLEM:

A 100 × 100 board is divided into unit squares. In every square there is an arrow that points up, down, left or right. The board square is surrounded by a wall, except for the right side of the top right corner square. A beetle is placed in one of the squares. Each second, the insect moves one unit in the direction of the arrow in its square. When the beetle moves, the arrow of the square it was in moves 90 degrees clockwise. If the indicated movement cannot be done, the beetle does not move that second, but the arrow in its squares does move. Is it possible that the beetle never leaves the board?

this is nice ” yeah, yeah, yeah” as the Beatles sing …

Problem:

small insects move inside a unit square. Prove that it is always possible to find of them that are inside a circle of radius

 

another nice easy one from combinatorics geometry. If any reader likes a clue I would be glad to give one…

find the invariant

It’s a bit tricky such and a  beauty too:

Choose 4 real numbers . Write down:

.

In the following line write the new set: where:

Continuing this procedure prove that after some steps we will take:

Note: This sequence is called ducci.

Coloring problem’s

I recently came across Sperner’s Lemma. Again. Sperner’s lemma is a combinatorial result which can prove some pretty strong facts, as Brouwer fixed point theorem. A fast description goes like this: Take a triangle RGB (R red, G green, B blue) labelled counterclockwise, and subdivide it into smaller triangles in whatever way you like (see the picture below). Then label all the new vertices as follows:

• vertices along RG may be labelled either R or G, but not B
• vertices along BC may be labelled either G or B, but not R
• vertices along CA may be labelled either B or R, but not G
• vertices inside triangle ABC may be labelled R or G or B.

Now shade in every small triangle in the subdivision that has three different labels.
Use two different shadings to distinguish the triangles which have been labelled counterclockwise (i.e. in the same sense as triangle RGB) from the triangles which have been labelled clockwise (i.e. in the sense opposite to that of as triangle RGB).

Then there will be exactly one more counterclockwise triangle than clockwise triangles. In particular, the number of shaded triangles will be odd.

For a nice proof and more details take a look here

So two problems I tried that have to do with coloring and combinatorics:

Problem 1: (All Russian Math Olympiad 1980) Some unit squares in an infinite sheet of squared paper are colored red so that every 2 x 3 and 3 x 2 rectangle contains exactly two red squares. How many red squares are there in a 9 x 11 rectangle?

Problem 2: (JBO 2004)  Consider a convex polygon having vertices, We arbitrarily decompose the polygon into triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is also a side of the polygon and in white those triangles that have no sides that are sides of the polygon.

Prove that there are two more black triangles that white ones.