## Friday, October 30, 2015

### Unizor - Geometry2D - Area of a Circle

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Area of a Circle

1. Definition

No matter how small our area measurement unit (a square with a side of 1) is, we will always have problems to measure exactly the area of a circle of a radius 1.
So, we have no choice, but to resort to a process that leads to an area of a circle as its limit.
This process might involve filling a circle with as many as possible squares of a side 1, then choose a smaller unit square of a side size, say, 1/10 and fill the empty area and continue this process getting more and more precise value with smaller and smaller measurement units.

In this lecture I'd like to introduce a different method that leads to a formula for the area of a circle more directly through another process.

Let's inscribe a regular polygon into a circle. We can start with a hexagon, similarly to a process we used to evaluate a circumference of a circle, or any other regular polygon. Next, we will double the number of edges of this polygon on every step by constructing perpendiculars to all edges and taking new vertices at points of intersection of these perpendiculars with a circle.

Now we state without rigorous proof that inasmuch as the perimeter of these polygons tends to a circumference of a circle, their area tends to some limit that can be used as a definition of the area of a circle.

Moreover, it can be proven that the limit of the area of polygons exists and is always the same, regardless of which polygon we start with and how we increase the number of its vertices, as long as the longest edge of polygons decreases to zero.

With this theorem we can state that the definition of an area of a circle is correct. It exists as a limit of areas of polygons transforming during a process described above and is unique.

2. Formula for Area of a Circle

Our next task is to evaluate the limit of the area of these polygons. That limit will be the area of a circle.

Let's examine a polygon with N sides obtained at some point during our process of increasing the number of vertices. Connect all vertices of our N-sided polygon with a center of a circle and consider the area of a polygon as a sum of areas of all N triangles with one vertex at the center of a circle and two other vertices being adjacent vertices of a polygon.
All these triangles are congruent and the area of each of them equals to
a·h/2
where a is the length of the edge of a polygon serving as an edge of a triangle and h - the altitude of each triangle dropped from the center of a circle onto an opposite edge.

Since we have N such triangles in a polygon, the total area of a polygon equals to
N·a·h/2
Now notice that N·a is a perimeter of our polygon p. So, we can say that the area of a polygon equals to
p·h/2

Now it's time to observe the trend of this area. As the number of vertices increases to infinity, the perimeter of inscribed polygons tends to a circumference of a circle that is equal to 2πR, where R is the radius of a circle.
At the same time the altitude h tends to be closer and closer to a radius of a circle.

So, as we approach the limit, the area of a circle will be equal to
2πR·R/2 = πR²

## Tuesday, October 27, 2015

### Unizor - Geometry2D - Circumference of a Circle

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Circumference of a Circle

1. Definition

The first question to discuss related to a circumference of a circle is "What is a circumference of a circle?"
Unit of measurement used for segments is not good for measuring length of curves. So, we have to come up with a different definition.

Typical approach to this lies in approximation. We will rely a lot on intuitively obvious statements with full understanding that proving them rigorously might present some problem.
It's tempting to define the circumference of a circle as follows:
a) start a process by inscribing a regular hexagon into a circle and setting the first approximation to a circumference of a circle as a perimeter of this hexagon;
b) on each step we draw perpendiculars to all sides of a polygon obtained on a previous step and take intersections of these perpendiculars with a circle as new vertices of a polygon with twice as many edges as on a previous step;
c) continue this process of doubling the number of edges to infinity, and the limit of the perimeters of our polygons is, by definition, a circumference of a circle.

There are a few problems with this definition. For example, it is intuitively obvious that, if we start with an inscribed square instead of a hexagon, we should also gradually approach some limit by doubling the number of edges of a polygon. But is this limit the same as if we start with a hexagon?
Issues of existence of a limit and its uniqueness can be rigorously addressed, but are rather complex, and we just point the result that, no matter how our process is arranged, as long as the maximum distance between neighboring points goes to zero, the limit of the perimeters of polygons exists and is the same. That justifies the definition of a circumference as such a limit.

2. Similarity of All Circles

Therefore, if a circumference of one circle is greater than its radius by some factor, the same factor is applied to any other circle.

Traditionally, the factor between a circumference and a diameter of a circle is designated a Greek letter π. It is a constant for all circles. It's an irrational real number and can be approximated to any precision. To four decimal places it's equal to 3.1416.

4. Iterations

If d[N] is the length of a polygon, inscribed into a circle, on Nth iteration of replacing each of its sides with a pair of equal but smaller ones, this recursive equality is held:
d[N+1] = √{2 − √[4 − (d[N])²]}

5. Conclusion

As you see, our first process started with hexagon and we got one set of formulas for a perimeter of polygons obtained by doubling the number of edges.
The second process started with a square and formulas were different.
But in both cases a few first steps of iterative process led to very close results. This is exactly how it should be, because, no matter how we organize the process of approximation of the circumference of a circle with perimeters of polygons, as long as the maximum edge length tends to zero, perimeter tends to the circumference of a circle.

## Tuesday, October 20, 2015

### Unizor - Geometry3D - Similarity of Cylinders, Cones, Spheres

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Similarity of Cylinders, Cones and Spheres

We have discussed the fact that scaling of a straight line transforms it into a straight line and scaling of a plane transforms it into a plane.
We have also discussed that angles between lines and dihedral angles between planes are preserved by scaling.

Based on this, we can say that any polyhedron is transformed by scaling into a polyhedron with the same number of vertices, edges and faces, with the same angles between edges and with the same dihedral angles between faces. So, the general shape of a polyhedron is preserved.

In this topic we will attempt to prove that the scaling of other geometrical objects - cylinders, cones and spheres - preserves their type.

Cylindrical Surface

Cylindrical surface is characterized by its directrix and generatrix. In particular, we draw a line parallel to generatrix through each point on a directrix.
Now let's scale each such line. A transformed image of this line by a scaling will be, as we know, a straight line parallel to an original. Therefore, no matter how a directrix is transformed, the new object will still be a cylindrical surface because it will consists of straight lines parallel to the same generatrix.

Right Circular Cylinder

Recall that all angles are preserved by scaling. In a right circular cylinder any line forming its side surface is parallel to a generatrix, which, in turn, is perpendicular to a plane that contains a circular directrix and both base planes. In the image of this right circular cylinder all such side lines will be still parallel to an original generatrix and an image of a plane that contains a circular directrix and images of both bases would be corresponding parallel planes. Therefore, perpendicularity between a generatrix of a cylinder and a plane where its directrix lies is preserved. Therefore, scaling transforms a right circular cylinder into a right circular cylinder.

Right Circular Cone
A right circular cone is the one whose altitude (a perpendicular form an apex onto a circular base) is falling into a base's center.
Let point S be an apex of an source right circular cone, plane β - its base plane, point O - a center of its circular base and points A and B - two points on a base circle.
Since this is a right circular cone, SO⊥β, SO⊥OA and SO⊥OB.
Let S', β', O', A' and B' be images of corresponding points after scaling.
Since all angles are preserved, S'O'⊥O'A' and S'O'⊥O'B'. Therefore, S'O'⊥β', that is, S'O' is an altitude of a cone's image. Hence, an image of a right circular cone is a right circular cone.

Sphere

Since equality of the lengths of two segments is preserved by scaling, images of two points on a sphere, equidistant from its center by definition of a sphere, will be equidistant from an image of a center. Therefore, an image of a sphere after scaling is a sphere.

## Monday, October 19, 2015

### Unizor - Geometry3D - Cylinders - Not So Easy Problems

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Cylinders - Not So Easy Problems

The problems presented here are steps towards the last problem and their combined solutions are supposed to constitute a complete solution to that last problem.

The Last Problem

There is a cylinder inscribed into a cube ABCDA'B'C'D' as follows.
The main axis of a cylinder, connecting centers of its circular bases, lies along a main diagonal of the cube AC'.
One base of a cylinder, a circle, is tangential to three faces of a cube, ABCD, AA'B'B and AA'D'D, that share vertex A, one end of a cube's main diagonal.
An opposite base of a cylinder, a circle, is tangential to three other faces of a cube, C'D'A'B', C'CBB' and C'CDD', that share the opposite end of the main diagonal, vertex C'.
A section of a cylinder along its main axis is a square (that is, a diameter of its base equals to its height).
Find the ratio of a volume of a cylinder to a volume of a cube.

Analysis:

First of all, let's think about the problem in general. There is no information about the size of a cube, which implies that the ratio of a volume of a cylinder to a volume of a cube is independent of a size. Is it true?
All the cubes are similar to each other. All the cylinders with a diameter of a base equal to a height also are similar to each other. So, it looks like changing the size of a cube might proportionally change the size of an inscribed into it cylinder, so the ratio of volumes is constant and the problem does make sense. These are logical considerations rather than proof.
The real solution would be to assign to a cube some dimensions and calculate the dimensions of an inscribed into it cylinder. Then we can calculate the ratio of the volumes and the result should be independent of the initial size of a cube.

Let's assume that the cube has all edges of a length d. Hence, its volume is d³. This is the simple part of the calculations.

With a cylinder it's not as simple. The bases of a cylinder are perpendicular to the main diagonal of a cube. From considerations of symmetry, the plane perpendicular to the main diagonal of a cube should intersect its edges on equal distance from the vertex that is the end of a diagonal. So, if the plane of the base of a cylinder that is closer to vertex A, which we will call δ, intersects edge AB at point P, edge AD at point Q and edge AA' at point R, segments AP, AQ and AR should be of the same length. To prove it is the subject of Problem 1 below.

Assume, this is proven, and continue our analysis.

Next step is to understand that a circular base tangential to three faces of a cube that share vertex A (and lying within a plane δ defined by triangle ΔPQR) is a circle inscribed into this triangle PQR. This follows from the fact that our circular base completely belongs to plane δ, so its common points with any one of three faces of a cube should belong to intersection of that face and plane δ. But this intersection is a line - a side of a triangle ΔPQR, so a circle has only one common point with a side of a triangle, that is tangential to it.

Let's assign a size x to three segments AP, AQ and AR. They completely define plane δ where a base of a cylinder lies. Using this length, we can calculate the radius of a base circle tangential to three faces that share vertex A as a radius of a circle inscribed into triangle ΔPQR - see Problem 2 below.

Another plane, the one where the other base of this cylinder (closer to vertex C') is located, intersects three other edges of a cube, C'B', C'D' and C'C at, correspondingly, points P', Q' and R'.
It should be obvious from the formula connecting the radius of an inscribed into triangle ΔPQR circle to x that C'P'=C'Q'=C'R'=x. This simple conclusion is the consequence of the fact that both circles at the bases of a cylinder are congruent.

The final component is the height of a cylinder. This might be calculated based on the length of a main diagonal of a cube and the height of a pyramid with vertex A and base triangle ΔPQR - see Problem 3 below. This height should be subtracted twice from the length of a diagonal (once for a pyramid with apex A, another - for a pyramid with apex C') to get the height of a cylinder.
Equating this height of a cylinder to a diameter of its base should give us the value of x in terms of the length of a cube's side d (see Problem 4).

With all this calculated, we can determine the volume of a cylinder in terms of d and compare it with a volume of a cube to get the ratio of a volume of a cylinder to a volume of a cube.
The variable d should be reducible from a formula.

## Friday, October 9, 2015

### Unizor - Geometry3D - Cylinders - Easy Problems

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Cylinders - Easy Problems

Problem 1

Prove that an intersection of a cylinder with any plane, that is parallel to its bases and is located in between them, is a circle.

Problem 2

Prove that an intersection of a cylinder with any plane that contains a generatrix at its two positions around the base is a rectangle.

Problem 3

A cube with an edge length of d is inscribed into a cylinder such that one face of a cube (a square) is inscribed into one base of a cylinder (a circle) and an opposite face of a cube is inscribed into an opposite base of a cylinder.
What is the total area and a volume of a cylinder?

Area = 2πd²(1+√2)/2
Volume = πd³/2

Problem 4

Prove that the line connecting two centers of bases of a cylinder (center line) is its axis of symmetry.

Problem 5
Given a cylinder of a radius R and height H. A plane is drawn parallel to its center line on a distance d from it (less than a radius).
What is the area of a section of a cylinder cut by this plane?

2H√R²−d²

Problem 6

A regular tetrahedron with all edges equal to d is inscribed into a cylinder such that one of its faces (an equilateral triangle) is inscribed into one base of a cylinder (a circle) and an opposite vertex of a tetrahedron coincides with a center of an opposite base of a cylinder.
What is the total area and a volume of a cylinder?

Area = 2πd²(1+√2)/3
Volume = πd³√6/9

### Unizor - Geometry3D - Cylinders - Area and Volume

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Area and Volume of a Cylinder

For the definition of a cylinder and corresponding terminology please refer to topic "Elements of Solid Geometry" in this part of a course.
In short, a cylinder is formed by
(1) a generatrix (straight line) moving along a circular directrix perpendicularly to a plane where this circular directrix is located, thus making a side surface and
(2) two planes parallel to a plane that holds a directrix - top and bottom bases of a cylinder, they bound the cylinder from both ends..

Let's define two important parameters that fully characterize a cylinder.
They are:
(a) radius of a base circle, which we will refer to as radius of a cylinder,
(b) altitude or height of a cylinder (the distance between the top and bottom circular bases).

There are different approaches to defining an area of a cylinder. More rigorous approach involves full force of the theory of limits, but we would suggest here a different approach.

First of all, consider the side surface of a cylinder.
Since this side surface is formed by a straight line (generatrix) moving along a circular directrix always perpendicularly to the same plane where this directrix is located and, therefore, parallel to itself at different positions, it is intuitively obvious that, if we cut the side surface of a cylinder along one of these straight lines, we will be able to "flatten" it on a plane without stretching or squeezing, that is without any change to its area.

As a result of this transformation, we will obtain a rectangle with the width equal in length to a circumference of a circular base of a cylinder and the height equal to a height of a cylinder - both are known variables for any given cylinder.
Therefore, the area of a side surface of a cylinder is equal to the area of a rectangle and can be easily calculated.

The circumference of a circular base of a cylinder of a radius R equals to 2πR. If the height of a cylinder equals to H, the area of the side surface would be equal to 2πR·H.
The full area of a cylinder should include two areas of circular bases of radius R, each equal to πR².
That makes the full area of a cylinder of a radius R and height H equal to
2πR·H+2πR² = 2πR(R+H)

The situation with volume of a cylinder is less obvious and we will not be able to escape considerations based on the limit theory.

Let's inscribe into a circular base of a cylinder a regular N-sided polygon. Then construct a right prism with this polygon being a base and the height equal to a height of a cylinder. We obtain a prism inscribed into a cylinder.

Without rigorous proof, it is intuitively obvious that, as we increase the number of vertices N, the regular polygon inscribed into a circular base of a cylinder becomes closer and closer to a circle itself, and the prism, based on this polygon inscribed into a circular base of a cylinder, becomes closer and closer to a cylinder. So, the volume of a cylinder is a limit of the volumes of inscribed in this manner prisms as N→∞.

Since a volume of a prism is a product of an area of its base by height and, as N→∞, the area of the N-sided polygon inscribed into a circle of a radius R tends to the area of a circle itself, that is πR², while the height H remains constant, we conclude that the volume of a prism tends to πR²·H

## Monday, October 5, 2015

### Unizor - Geometry3D - Pyramids - Problem 6

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Pyramids - Problems 6

To effectively learn from problem solving, try to solve these problems just by yourself, then listen to a lecture and then try to solve them again by yourself.

Problem A

Given a triangular pyramid SABC with apex S and side edges having sizes:
SA = a,
SB = b,
SC = c.
Inside its base ΔABC we choose any point P.
From this point P we draw three lines parallel to three side edges of a pyramid until lines intersect the opposite faces:
PA' ∥ SA where A'∈ΔSBC,
PB' ∥ SB where B'∈ΔSAC,
PC' ∥ SC where C'∈ΔSAB.
Let the length of these new segments are
PA' = a',
PB' = b',
PC' = c'.
Prove that
a'/a+b'/b+c'/c = 1

### Unizor - Geometry3D - Pyramids - Problem 5

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Pyramids - Problems 5

To effectively learn from problem solving, try to solve these problems just by yourself, then listen to a lecture and then try to solve them again by yourself.

Problem A

Given a regular tetrahedron SABC.
Draw a median AM within its triangular base ABC from vertex A to opposite side BC (M∈BC).
From its apex S draw a plane γ perpendicular to median of the base AM.
Assume that the area of a section of a pyramid formed by this plane equals to s.
Pick a midpoint P of median AM and construct another plane δ that is parallel to plane γ. It also cuts some section inside a pyramid.
What is the area of this section?

9·s /16

### Unizor - Geometry3D - Pyramids - Problems 4

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Pyramids - Problems 4

To effectively learn from problem solving, try to solve these problems just by yourself, then listen to a lecture and then try to solve them again by yourself.

Problem A

Given a right rectangular prism ABCDA'B'C'D' with three edges with common vertex A having length x, y and z.
Let's connect with straight lines vertices
A and C,
A and B',
A and D',
C and B',
C and D',
B' and D'.
What is the volume of tetrahedron ACB'D' formed by these vertices?

x·y·z /3

Problem B

Given a triangular pyramid SABC with pairs of opposite edges equal to each other as follows:
SA = BC = a,
SB = AC = b,
SC = AB = c.
What is the volume of this pyramid?