Category: theorem

Triangulating a circle. How to do a drawing li…

Triangulating a circle. 

How to do a drawing like this? Start with a circle and divide it into n equal parts. I choose n=100 for this drawing. There is an easy formula for this, the i-th point is (radius*cos(i*2*PI/n),

radius*sin(i*2*PI/n)). Then for each i connect the i-th point to the point  (radius*cos(PI-2*i*2*PI/n), radius*sin(PI-2*i*2*PI/n)). This might not be one of the original points, but it is on the circle. And that is it 🙂    

Why is this mathematically interesting?  This way we get a set of nonparallel  lines such that there are a lot of triple intersections between them.  

The chromatic number of the plane is at least …

image

A very famous open problem is the following.

How many colors are needed to color the plane so that no two points at
distance exactly 1 from each other are the same color? (This number is known as the chromatic number of the plane)

For example 2 colors are not enough. Consider 3 points that are the vertices of an equilateral triangle with side length 1. Clearly no two of these points can have the same color, so we need 3 colors just for these.  

Until this week our best knowledge was that there is a possible coloring with 7 colors, and 3 colors are not enough but we didn’t know if the best answer is 4, 5, 6 or 7.  It took more than 50 years to get something more, a couple of days ago  AUBREY D.N.J. DE GREY presented a paper on arxiv, proving that 4 colors are not enough! See here: https://arxiv.org/pdf/1804.02385.pdf

Lets see the old results.

Why is 7 color enough? Consider the following coloring:

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Construct the hexagons such that the two opposite vertex of a hexagon are slightly closer to each other than 1. This way you cannot take two points from the same hexagon that are at distance one. If you take two points from different hexagons of the same color their distance will be at least bigger than 1. So this coloring is correct.

Why is 3 color not enough?  Consider the following 7 points. I connected the ones that are at distance 1. 

image

Can we color these points with 3 colors? Well, lets try! A,G and F must be all different, since they are connected. Also G, F and E must have three different colors. So if we color with 3 colors, A and E must have the same color. The same way we can see that A and D must have the same color. But this would mean that E and D have the same color which is not allowed! So 3 colors are not enough. 

Booth of these were very easy to see, but until now we couldn’t do better. 

So what is new?

AUBREY D.N.J. DE GREY proved that 4 colors are not enough. The proof is similar to the one above, but much more complicated. Some parts even used computer verification. On the following picture there are 1567 yellow points, and every pair is connected if their distance is 1. It is a lot of points, so it is not really visible :\

image

He showed that 4 color is not enough for these points, we need at least 5. This result doesn’t settle the problem, we still don’t know if 5 colors are enough for the whole plane, but we are closer to the answer. 

There is a lot of interesting things connected to this problem maybe I will make some posts about it later.  

Reuleaux triangles. These shapes have a very n…

Reuleaux triangles. These shapes have a very nice property, their width is the same in any direction. 


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Morley stabilized. Same as the previous one, but I changed it…

Morley stabilized. 

Same as the previous one, but I changed it such that the inner triangle is fixed instead of the circle. 

Morley theorem 2. The dashed lines are angle trisectors, so they…

Morley theorem 2. 

The dashed lines are angle trisectors, so they divide the angles of the triangle into three equal parts. The blue triangle is equilateral, which means that each blue segment has the same length. 

Morley’s theorem. In any triangle, the three points of…

Morley’s theorem. 

In any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle

Interestingly I moved the triangle in such a way that the blue triangle doesn’t change its direction. 

Simson line theorem. The three blue points always lie on a…

Simson line theorem. The three blue points always lie on a straight line. The blue points are the closest points to the moving red point on the lines. In other words the blue points are the projections of the moving red point to the lines.     

acosineofmadness: szimmetria-airtemmizs: curiosamathematica: Today I learned about a rather…

acosineofmadness:

szimmetria-airtemmizs:

curiosamathematica:

Today I learned about a rather remarkable open problem in mathematics, which looks tantalizingly easy. The question was posed by Ron Graham.

Consider the following recursively defined sequence:

image

Innocent question: is this sequence unbounded? Surprisingly, the answer to this is unknown—at least according to the source article dating from 2000, Unbounded orbits and binary digits by M. Chamberland and M. Martelli.

(Source: http://www.math.grinnell.edu/~chamberl/papers/mario_digits.pdf)

Here is a very similar problem where we know the answer.

The question is the same, is this sequence bounded or not? In other words, is it possible that the sequence never goes above a certain number?

Try to solve it! It is not easy but you don’t need any kind of advanced tools. 

The sequence is monotonic increasing, since

image

Suppose the sequence converges to a finite number, c:

image

(Contradiction)

This means that the sequence does not converge:

image

An increasing sequence that does not converge, increases without bound.

Correct!

There are a lot of ways to prove it. Here is an other one: Suppose the sequence is bounded by c. Then the sequence increases by 1/c in every step. But if it always increased by 1/c then it will go above c in c*c steps. Contradiction.  

szimmetria-airtemmizs: Johnson’s Theorem.  Pick a point in the…

szimmetria-airtemmizs:

Johnson’s Theorem. 

Pick a point in the plane (the red one). Draw three circle through it with equal radius (the black circles). Take the three intersection of these circles (the black points). Draw a circle trough the three point (the red circle). Then, the fourth circle have equal size to the first three.  

Someone asked for the proof, so here it is. It is very simple, you only need to know about vectors. Let the radius of the black circle be 1. 

Let the red point be the origin. Let a, b and c denote the vectors that point from the origin to the center of the black circles. Then it is easy to see that the three black point is just a+b, b+c and c+a. 

Consider the the point a+b+c. How far is a+b+c is from a+b? They differ in the vector c which has length 1, so their distance is 1. Similarly the distance of a+b+c and b+c is 1 and also the distance of a+b+c and c+a is 1. Therefore a circle  around a+b+c with radius 1 goes trough the black points. This circle has to be the red one, so the theorem is proved.    

A bit of extra: The eight point on the picture can be connected to form a drawing of a cube! Can you give a simple rule, which ones to connect?

Pascal’s theorem.No matter how you choose the red points, the…

Pascal’s theorem.

No matter how you choose the red points, the three blue point will lie on a straight line.

The red line is known as the Pascal line of the hexagon. 

https://en.wikipedia.org/wiki/Pascal%27s_theorem