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

lak1

 

 

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 #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:

latis3

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 #13

Baking Cookies

You started craving for some chocolate-chipped cookies and you are trying to convince your mother to bake some… or maybe more than some. Eventually you and her met halfway: Your mom will cook only one oven tray of cookies and your task is to place them on the tray. Obviously all cookies must have identical shapes. They can’t be placed randomly and they should not overlap on the tray. In the end you have to come up with a system that will give you the most cookies possible.

 

Dimensions of the oven tray: 46,5 cm x 37,5 cm.

Diameter of a single cookie: 5 cm.

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Q: How many cookies can you fit on this oven tray?

Honeycomb

First we must analyze how honey is produced in order to solve the cookie problem. Honey bees use fantastic mathematics and their masterpiece architecture is honeycomb.

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Have you ever thought why do honey bees construct honeycomb as they are (which is in a hexagonal shape)?

Honey bees are in need of storage units for the honey they produce. For now try to imagine those storage units in two-dimensional plane. This way honey bee problem turns into a problem of finding the optimal regular geometrical shape in order to tessellate a paper.

Responsibility of the Honey Bee: “I must tessellate a paper with a regular geometrical shape and I shall do that with creating the maximum area with minimum circumference.”

(Maximum area: More honey storage. Minimum circumference: Less time spent on constructing the storage.)

Three Regular Polygons

There are only three regular geometrical shapes to fill a paper without a gap: Equilateral triangle, square and hexagon.

bad tessellates
Here is how 7, 8 and 9-sided regular polygons look like on a paper. You could try a polygon and see the result on your own.

Let’s start with a square that has side length of 2 units. It means the area of one square is 4 units and its circumference is 8 units.

 

Now I’ll consider constructing an equilateral triangle that has area of 4 units. Its circumference would make a little over 9 units. This means that equilateral triangle tessellations will take more time for the honey bee.

 

Next one is the hexagon. Whenever a hexagon has an area of 4 units, its circumference will be around 7,45 units. This result shows that a hexagon requires less time to be built than a square does. This is why our honey bee chooses hexagonal tessellations in order to construct its honeycomb.

 

Circle to Hexagon

We still don’t know how honeycombs are being constructed. Scientists consider two possible scenarios for it: Bees use either circles or hexagons in the construction of honeycombs. One theory suggests that bees build circle storages and those storages turn into hexagonal shapes within time.

Following picture illustrates how circles and hexagons are related with one another:

 

One wonders…

Solution of the cookie problem is hidden inside the relationship between circle and hexagon. Then please find how many cookies you can place on the oven tray.

M. Serkan Kalaycıoğlu

Real Mathematics – Geometry #10&11

Q: How many regular polygons are there?

In the previous articles I have shown how to construct an equilateral triangle using Euclid’s method which means using a compass and an unmarked ruler. An equilateral triangle is a regular polygon as it consists of equal sides and angles. We can even conclude that an equilateral triangle is the smallest regular polygon.

Then let me continue with increasing the number of sides to construct more polygons. A regular polygon with four equal sides and angles… Hmmm… A square!

A regular polygon with five sides: Pentagon.

A regular polygon with six sides: Hexagon.

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A regular polygon with fifty sides: Pentacontagon.

There are no limits for the number of sides of regular polygons we can create. Although after certain number of sides it is almost impossible to distinguish a regular polygon from a circle.

Pentacontagon (left) and its comparison with a regular polygon that has 200 sides.

Regular polygons exist in two dimensional worlds. What if we try to construct regular objects in three dimensions?

Q: How many regular polyhedrons are there?

There are two specific properties for regular polyhedrons: Each of their faces is the same regular polygons and there are same numbers of regular polygons meeting at each corner.

Let me give you the answer right away: There are five different regular polyhedrons. Exactly five!

The very first time I heard about it, I thought it was bizarre to have only five different regular polyhedrons while there are infinite number of regular polygons. How can coming to three dimensions from two dimensions changes so much?

School of Broad Shouldered

Around 427 BC a baby named Aristocles was born in Athens. When he grew up, Aristocles had wide shoulders which resulted with him adopting the nickname “Plato” that meant “broad” in Greek. (This story is from C.J. Rowe’s Plato.)

plato
Plato (427 BC. – 347 BC.)

Majority of the people have no idea that Plato was an integral individual for mathematics’ development. His persistence for clearer explanations for proof and hypothesizes evolved mathematics completely. Although his biggest contribution, not just to mathematics but to all branches of science, was the school he founded.

In 387 BC he attempted to build a school in Athens, on the land of a guy named Academas. He gave the school his name: Academy. For over 900 years Plato’s Academy was the home of countless philosophers. (Just for comparison to its worth: University of Bologna was built 1400 years after Academy was founded.)

geometri-bilmeyen-giremez
Above Academy’s door: “Let no man ignorant of geometry enter here.”

Platonic Objects

Polyhedrons are also called Platonic objects because he was the first to explain that there are exactly five of them.

  1. Tetrahedron: An object formed by four equilateral triangles.
  2. Cube: An object formed by six squares. I guess I didn’t need to explain it.
  3. Octahedron: An object formed by eight equilateral triangles.
  4. Dodecahedron: An object formed by twelve pentagons.
  5. Icosahedron: An object formed by twenty equilateral triangles.

platonic_solids

Why Five?

When examined carefully one might see the properties that are exclusive to the regular polyhedrons. These properties both explain and prove why there are exactly five of them.

I’ve already mentioned that regular polyhedrons are three dimensional objects. If you study a regular polyhedron, you’d see that at least three regular polygons meet at a point (corner). Let’s try it with the smallest regular polyhedron.

Take any corner on a tetrahedron. You’ll see that three equilateral triangles meet at that corner. When it is reduced to two dimensions, it would look like as follows:

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The corner has 360 degree around itself. One equilateral triangle has 60 degrees of internal angle on this specific point which means there are

360 – (60+60+60) = 180

degrees of empty space around the corner. By means of this empty space it is possible to construct a three dimensional object. If there were no space, then this shape would not have the flexibility and thus it would not be turned into a three dimensional object.

We are getting close. Let’s use induction and go from tetrahedron to general.

In order to construct a polyhedron we must follow these rules:

  • Draw a point on a paper.
  • Draw three regular polygons which meet at that point.
  • Add the internal angles of the regular polygons. If it doesn’t exceed 360 degrees, then it is possible to construct a regular polyhedron with these regular polygons.
  • Also one can continue adding same regular polygons as long as summation never exceeds 360 degrees.
  • If the summation of the internal angles of the regular polygons is 360 degrees or more, then this shape can’t be converted into three dimensions. This means a regular polyhedron can’t be constructed with such regular polygons.

Example 1: Tetrahedron.

Take a point and draw three equilateral triangles around it. One internal degree of equilateral triangle is 60, which make 60*3=180 degrees around the point. It is less than 360, that is why we know it is possible to construct a Platonic solid with them: Tetrahedron.

Example 2: Tetrahedron + 1 more equilateral triangle.

If we add one more equilateral triangle, internal angles around the point will make 240 degrees, which is still less than 360. Hence it is possible to construct a Platonic solid with four equilateral triangle: Octahedron.

Example 3: Tetrahedron + 2 more equilateral triangles.

We continue to add more equilateral triangles. Now we have five of them around a point, which gives 300 degrees. We are still under 360 degrees, thus we can construct a Platonic solid with them. This is our third Platonic solid and we constructed them with equilateral triangles: Icosahedron.

Example 4: Tetrahedron + 3 more equilateral triangles.

We add the sixth triangle. There is 60*6=360 degrees around the point. It is impossible to construct a Platonic solid with these triangles. Actually this shape stays only in two dimensions.

Example 5: Cube and adding a square to it.

We are done with equilateral triangles, hence we move to the next regular polygon: Square. Internal degree of a square is 90, which gives 90*3=270 degrees around the point. It is less than 360 degrees, and that is why it is possible to construct Platonic solid with three squares. This is the fourth Platonic solid: Cube.

If we add one more square, then internal angles around the point will add up to 90*4=360 degrees. It is impossible to construct a Platonic solid in this case.

Example 6: Dodecahedron and adding one more pentagon.

Since we are done with squares, we move to the next regular polygon: Pentagon. A pentagon has internal degrees 108 at its corners. Three pentagons around a point will give 108*3=324 degrees which is less than 360 degrees. So it is possible to construct a Platonic solid with three pentagons meeting at a point. This is our fifth Platonic solid.

Although if we a forth regular pentagon, internal degrees will be 108*4=432 degrees. It is more than 360, which makes the pentagons overlap. Hence it is impossible to construct a Platonic solid with these.

One wonders…

Continue to the next polygon: Hexagon. Examine why it is impossible to construct a Platonic solid with them.

M. Serkan Kalaycıoğlu