Real MATHEMATICS – Geometry #19

How Much Chocolate?

It is midnight and my stomach is talking to me. I hope to find something to eat in the kitchen and I see a chocolate bar:

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Immediately made myself a cup of coffee and broke a piece of the chocolate bar:

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After I “killed” the broken piece I started having second thoughts about my decision: Oh God; Did I eat too much chocolate?

I placed the leftover on a grid. This way I found where both whole and the broken piece lies on the grid:

çiko1

The broken piece is shaped like a simple polygon. My goal is to calculate the area of that piece. There are several ways I could calculate the area. Although, the first thing comes to my mind is a theorem called “Gauss’ shoelace theorem”.

Gauss’ Shoelace Theorem

The shoelace theorem can only be applied to simple polygons. In order to use the theorem, I have to find where the edges of the simple polygon lie on the grid:

çiko2

Theorem uses these points just like shoelaces. But first we have to define the edges and make a list of them:

çiko3

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Do not forget to add the first edge to the bottom of the list.

Now you can multiply the numbers diagonally; from right to left and left to right. Then add left to right and subtract it from right to left ones:

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{(0*0) + (5*1) + (4*3) + (5*3) + (0*0)} – {(0*5) + (0*4) + (1*5) + (3*0) + (3*0)}

{32} – {5}

27.

The area of that simple polygon can be found by dividing the result above:

27/2

13,5

M. Serkan Kalaycıoğlu

Real Mathematics – Killer Numbers #5

Equilateral Triangle and Irrational Number

In the previous article I asked you to prove whether it is possible or not to draw and equilateral triangle on the system of lattice points. Here is one of the possible proofs for that.

Proof by Contradiction

Let’s assume that we have an equilateral triangle that has sides of length 2 units:

akuie2

Here, we will make a critical assumption: Corners of this triangle sits on the system of lattice points. Because of that the triangle must have an area that is rational. Why?

Because Pick’s theorem says that whenever we are inside the system of lattice points the numbers of points and area of the polygon are directly related with each other. And since we can’t count irrational number of points (eg. we can’t have √3 points, can we?!), area of the polygon must be rational too.

euake3

We already know that we can find a triangle’s area: ½(height x base). Then let me draw the height of the base and find its length using Pythagoras’ theorem:

h2 + 12 = 22

h2 = 4 – 1

h2 = 3

h = √3.

We just found the height of our equilateral triangle an irrational number. From here we will find the area

1/2(2*√3) = √3 units.

CONTRADICTION

This result is a contradiction. Despite what Pick’s theorem says (polygons inside the system of lattice points must have rational areas) this result shows an irrational number. Then we can conclude that this or any equilateral triangle can never be drawn on the system of lattice points.

One wonders…

Can you find another polygon which you can’t draw on the system of lattice points?

M. Serkan Kalaycıoğlu

Real Mathematics – Killer Numbers #4

Lattice Points

Imagine a page from a squared notebook. Take all the edges of little squares and leave the vertices. If you assume that horizontally and vertically distance between two consecutive points is exactly 1 unit. I name this plane as the system of lattice points.

In the system of lattice points every point is represented by a number duo:

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Pick’s Theorem: An alternative method in order to find the area of a polygon.

In the system of lattice points one can construct as many polygons as wanted with joining points together. Pick’s theorem is handy for finding the areas of these polygons.

According to the theorem only two things are needed to find the area of a given polygon: Number of points polygon has on its edges (let’s call it e) and the number of points staying inside the polygon (let’s call it i). Pick’s theorem gives the following formula for the area of a polygon sitting on the system of lattice points:

Area = i + (e/2) – 1

Example 1: Triangle.

üçge

Assume that there is a triangle as shown above. Number of points on its edges e is equal to 4 as the number of points inside the triangle i is equal to 0. Hence Pick’s theorem says the area of this triangle is:

Area = 0 + (4/2) – 1

          = 0 + 2 – 1

          = 1 unit.

We know from basic geometry that the area of a triangle is the half of height times base: (2*1)/2 = 1 unit.

Example 2: Square.

karre

In the picture we can see that e=12 and i=4. Thus the area is:

4 + (12/2) – 1 = 4 + 6 – 1 = 9 units.

We know that area of a square is the square of the lenght of its edge which makes 3*3=9 units.

Square and triangle are easy examples and perhaps you are thinking that Pick’s theorem is redundant. Then let’s continue drawing a more complex polygon and find its area.

Example 3: Polygon.

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Now Pick’s theorem shows its strength. Without the theorem we’d have to divide this polygon into various polygons and then calculate areas one by one. But with Pick’s theorem it is just counting points:

e=12 and i=72. Hence the area of the polygon can be found with:

Area = 72 + (12/2) – 1 = 72 + 6 – 1 = 77 units.

Equilateral Triangle

Up to this point it looks like we are dealing with geometry but the headlines said that it is an article about numbers. Let me change the course of the article with a question.

Q: Is it possible to draw an equilateral triangle on our points system while the corners of the triangle sits on the lattice points?

For instance let me draw the base of an equilateral triangle that has 2 units of length:

ekui
Third corner (Q) sits between lattice points.

As seen above third corner of the triangle won’t be on a lattice point.

One wonders…

Can you prove that this is the case for each and every equilateral triangle on the system of lattice points?

To be continued…

M. Serkan Kalaycıoğlu

Real Mathematics – Geometry #12

Cogito Ergo Sum*

*I think, therefore I am. You probably heard these words before. They belong to Rene Descartes. Descartes is believed to be the person who invented modern philosophy. Actually he was a groundbreaker for mathematics.

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Rene Descartes (1596 – 1650)

One of his best discoveries for mathematics has a mesmerizing story. Even though it is unknown whether this story is true or not, I find it amusing.

It is a known fact that Descartes had a rough childhood. He was constantly battling with various illnesses which were major setbacks for his early school days. Every day he was able to attend to his school around noon. This has become a habit for Descartes. For the rest of his life he spent his mornings in bed. (Except final days of his life in Sweden)

Story of Descartes’ discovery is based on this fact. Allegedly one morning as he was lying down, he saw a fly on his ceiling. He started thinking about fly’s position. There was one question in his mind: “How can I describe this fly’s position to someone who hasn’t been with me in this room?”

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Descartes began his answer with assumption. If we assign a corner of the ceiling as the starting point, it is possible to reach the fly with only two directions for movement from that point: To the width and to the length.

This is the story of how Cartesian coordinate system was first discovered. Analytic geometry was born with Descartes’ publication on the issue.

Analytic Geometry = Cartesian Geometry = Coordinate Geometry

Cartesian Coordinate System

Let’s say that left bottom corner of Descartes’ ceiling is the starting point. There are two available paths for us: Up or right.

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According to Descartes one can reach any point of the ceiling with two specific movements from the starting point: X amount to the right, Y amount to the up.

In that case the point (or the object) would have a position which could be expressed with two different numbers. If X represents amount of movement to the right, and Y to the left, then position can be shown as (X, Y).

Example: Assume that unit of movement is cm, and fly sits on the ceiling as shown:

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If we sit at the starting point and make 4 cm to the right & 3 cm to the left, then it is safe to call the position of the fly (4,3).

Catching the Thief

Materials:

  • Cartesian coordinate system that has size 6×6.
  • Dice.
  • Pencil/pen and paper.
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Coordinate system needed for the game.

Catching the thief is a multiplayer game. One player gets to be the detective as the other one gets to be the thief. Detective’s mission is the catch the thief as soon as possible.

Thief rolls the dice twice and determines its position: (First dice, Second dice).

Detective makes its first guess. If he/she is right, then the game is done.

If detective is wrong in his/her first guess, he/she receives a text message from the thief. Message reads a number. This number represents the total of the differences between these positions.

Let’s assume that the thief is at the position (2,2) and detective makes his/her guess as (4,1). Differences of the positions are: 4-2=2 and 2-1=1. Add them together and we’ll get 2+1=3. Hence thief sends the number 3 in his/her message.

Detective makes his second guess according to this information and this process continues until thief is caught.

Example

Thief rolls the dice. First try reads 3, second reads 4. Hence thief’s position is determined: (3,4).

Detective makes his/her first guess with (2,2) and misses.

Thief calculates the differences of the positions: 3-2=1 and 4-2=2. Thief adds them together: 1+2=3.

Detective receives the message: “3”. Now detective is in a slightly better situation.

Possibility 1: Detective adds 3 to the x. Guess: (5,2).

Possibility 2: Detective adds 2 to the x, and 1 to the y. Guess: (4,3).

Possibility 3: Detective adds 1 to the x, and 2 to the y. Guess: (3,4).

Possibility 4: Detective adds 3 to the y. Guess: (2,5).

Possibility 5: Detective subtracts 2 from the x, and 1 from the y. Guess: (0,1).

Possibility 6: Detective subtracts 1 from the x, and 2 from the y. Guess: (1,0).

Possibility 7: Detective subtracts 2 from the x, and adds 1 to the y. Guess: (0,3).

Possibility 8: Detective subtracts 1 from the x, and adds 2 to the y. Guess: (1,4).

Possibility 9: Detective adds 2 to the x, and subtracts 1 from the y. Guess: (4,1).

Possibility 10: Detective adds 1 to the x, and subtracts 2 from the y. Guess: (3,0).

After the message from the thief, detective knows for sure that the thief is at one of these 10 positions. Detective continues to make his guesses and uses the same strategy until he/she catches the thief.

One wonders…

  1. Determine how many possibilities can the detective have in case he/she guesses (3,0)?
  2. I am thief and you are the detective. Your first guess is (3,3) and I send you the text “2”. Where am I? Leave your answer to the comments.

M. Serkan Kalaycıoğlu