## Thursday, August 27, 2015

### Unizor - Geometry3D - Similarity - Cavalieri - Theorems 1

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

Mini Theorems 1

Let's prove the condition of Cavalieri's principle (see the previous
lecture explaining this principle) in a couple of very simple cases
using the similarity. This would justify our decision to present the
Cavalieri's principle among topics related to similarity.

Mini Theorem A

(two-dimensional case to introduce the approach to a proof)

Given a straight line d on a plane and two segments on it, AB and CD, that have equal lengths.

Let S be a point on this plane outside the line d.

Prove that the condition of Cavalieri's principle is held for triangles ΔSAB and ΔSCD.

If proven, we can state that the areas of these triangles, according to Cavalieri's principle, are the same.

Mini Theorem B

(three-dimensional case that will be used in calculation of the volume of a pyramid)

Given a plane δ and two triangles on it, ΔABC and ΔDEF, that have equal area.

Let S be a point in space outside this plane δ.

Prove that the condition of Cavalieri's principle is held for triangular pyramids SABC and SDEF.

If proven, we can state that the volumes of these pyramids, according to Cavalieri's principle, are the same.

Mini Theorem C

Given a plane δ and parallelogram ABCD on it with diagonal AC.

Let S be a point in space outside this plane δ.

Using the Cavalieri's principle, prove that the volumes of two pyramids, SABC and SACD, are equal.

## Wednesday, August 26, 2015

### Unizor - Geometry3D - Cavalieri's Principle

Unizor - Creative Minds through Art of Mathematics - Math4Teens

The
purpose of this lecture is to introduce the Cavalieri's principle that
will be used to derive the formula for a volume of a pyramid, the next
topic of this course.

The problem with a volume of a pyramid, as
with many other measures of geometric objects is that absolutely
rigorous derivation of the formulas for their area and volume can be
obtained only within a framework of higher levels of mathematics,
usually addressed in colleges. However, using the theory of limits on an
intuitive level, as presented in this course in a topic Algebra -
Limits is sufficient to derive these formulas for high school students.

The
Cavalieri's principle, actually, hides the complexity of this issue by
postulating certain property of geometric objects. This principle is not
an axiom mathematicians accept, it is a theorem that can be proven with
above mentioned higher levels of mathematics using integration.
However, accepting this principle as an axiom at this stage allows to
shortcut the derivation of formulas for area and volume of geometric
objects. The Cavalieri's principle sounds very natural and is
intuitively obvious. So, we will use it without any hesitation.

There
are two related to each other parts of the Cavalieri's principle - one
for area of a geometric object on a plane and another for volume of
solid objects in three-dimensional space.

Two-dimensional case

Assume we have two geometric objects X and Y on a plane.
Assume
also that there is a base line d on this plane such that any line h
parallel to this base line has the following property:
its
intersection with object X (a segment or a set of segments for
irregularly shaped object X) has the same linear measure as its
intersection with object Y.
Then the areas of objects X and Y are equal.

This
principle is a sufficient condition for equality of the areas of
objects X and Y, but by no means a necessary one. There are objects of
equal areas for which this principle is not true. For instance, take a
2x2 square and 1x4 rectangle. Their areas are equal, but you cannot find
a base line d such that any line parallel to it produces sections of X
and Y of the same linear measure.

Three-dimensional case

Assume we have two solid geometric objects X and Y in a three-dimensional space.
Assume
also that there is a base plane δ on this plane such that any plane γ
parallel to this base plane has the following property:
its intersection with object X (a flat object) has the same area as its intersection with object Y.
Then the volumes of objects X and Y are equal.

This
principle is a sufficient condition for equality of the volumes of
objects X and Y, but by no means a necessary one. There are objects of
equal volumes for which this principle is not true. For instance, take a
2x2x2 cube and 1x4x2 right rectangular prism. Their volumes are equal,
but you cannot find a base plane δ such that any plane parallel to it
produces sections of X and Y of the same area.

## Wednesday, August 19, 2015

### Unizor - Geometry3D - Problems on Similarity

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Problems on
Similarity in 3-D Space

1. Prove that all cubes are similar.

2. Prove that all regular tetrahedrons are similar.

3. Prove that all spheres are similar.

4. Consider a sphere inscribed into a cube of a unit size. It takes up certain volume in the cube, while certain part of a cube's volume remains free.
Now let's take this sphere out from a cube, take certain number of spheres with a diameter equaled to 1/Nth of the original sphere and fill the cube with these smaller spheres positioning these spheres in rows, columns and layers, one strictly above the other. They will also take up some space, while certain part of a cube's volume will remain free.
In which of these cases the free part of a cube will be larger by volume?

5. Consider two persons drinking martini from the same conical glasses. One takes the glass filled with martini to the brim, while another said to a barman that she wants only half a glass. Barman fills her glass to half its height. How much martini is in this glass relatively to a full glass of her partner?

6. Specify requirements for two right rectangular prisms that assure their similarity.

## Tuesday, August 18, 2015

### Unizor - Geometry3D - 3D Similarity

Unizor - Creative Minds through Art of Mathematics - Math4Teens

3-D Similarity

The definition of scaling or homothety in three-dimensional space is exactly the same as on a plane. The transformation of scaling requires the center of scaling (a point in a space) and the factor of scaling (non-zero real number) with exactly the same functionality as on a plane.

Assume, point X is a center of scaling and f is a factor of scaling. Any point M in a space not coinciding with the center X is transformed by a scaling using the following procedure (exactly as on a plane):
(a) connect points M (given) and X (center) by a segment;
(b) construct a new segment with the length equal to the length of XM multiplied by an absolute value of the factor |f|;
(c) starting at the center X along the line XM position this new segment in the direction from X to M for positive factor f or to an opposite direction for negative f;
(d) the end of this new segment is the result of transformation of point M by scaling with center X and factor f.
If point M coincides with center X, scaling with any factor does not change its position since the length of a segment XM is zero and will be zero after multiplication by any factor.

The most important property of scaling with factor f is the fact of changing of the length of any segment MN into another M'N' with the length that is equal to a product of the length of the original segment and factor |f|, regardless of a center of scaling X. This simple fact follows from the similarity on a plane of triangles ΔXMN and ΔXM'N'.

Now, with scaling transformation defined, we can define similarity of two geometrical objects ξ and η in three-dimensional space as the possibility to transform one into another using scaling, as defined above, optionally combined with congruent transformations of parallel shift (translation), rotation and symmetry. The word "possibility" means here the existence of a particular point X in space that can be used as a center of scaling and a particular non-zero factor of scaling f that, if used in the scaling of object ξ, transform this object into another, congruent to object η.

The three-dimensional expansion of this statement is that the scaling of a plane in three-dimensional space is a plane parallel to the original. To prove it, it's sufficient to choose any two intersecting lines on the original plane and compare them with their counterparts on the resulting plane. Since these lines are correspondingly parallel, the planes they belong to are parallel as well.

Dihedral angles are also preserved by scaling since they are measured by angles between two lines, correspondingly belonging to two planes and perpendicular to a line of their intersection, and the angles between lines are preserved.

It's easy to prove (admittedly, not absolutely rigorously since we do not use the full power of the theory of limits) that scaling of any polygon by a factor f effects its area by a factor f².

It can also be proven (admittedly, not absolutely rigorously since we did not use the full power of the theory of limits) that scaling of any polyhedron by a factor f effects its volume by a factor |f|³.

### Unizor - Geometry3D - Prisms - Problems

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Problem 1
Calculate a volume V and the total surface area S of a cube ABCDA'B'C'D', if the length of its main diagonal AC' equals to d.

V = d³·√3/9
S = 2d²

Problem 2
Calculate a volume V and the total surface area S of a right regular hexagonal prism with all edges of the same length d.

V = 3√3d³/2
S = 3d²(√3+2)

Problem 3
Prove that the area of a square with a side equal to a main diagonal of a right rectangular parallelepiped is greater than or equals to a half of this parallelepiped's total surface, and equality takes place only if this parallelepiped is a cube.

Hint:
Use inequality
x²+y² ≥ 2xy

Problem 4
Prove that the total area of all side faces of any prism (not necessarily the right one) equals to the length of a side edge multiplied by a perimeter of a polygon obtained by cutting the prism by a plane perpendicular to a side edge.

Hint:
Side face is a parallelogram, the area of which equals to a product of a side edge by its altitude.

Problem 5
Given a slanted triangular prism ABCA'B'C' with an isosceles triangle ΔABC as its base, AB=AC.
Side edge AA' forms equal acute angles with base edges AB and AC, that is
∠AA'B = ∠AA'C.
Prove that AA'⊥BC and, consequently, that side face BCC'B' is a rectangle.

Hint:
Prove first the following two simple theorems.
If projection p of line d onto plane γ is perpendicular to some line q on this plane, then the original line d is also perpendicular to line q.
Similarly, if original line d is perpendicular to line q lying on plane γ, then projection p of line d onto this plane γ is also perpendicular to line q.
Symbolically:
p=Projγ(d); q∈γ; p⊥q
⇒ d⊥q
p=Projγ(d); q∈γ; d⊥q
⇒ p⊥q

## Monday, August 3, 2015

### Unizor - Geometry3D - Prisms - Polygonal Prism

Unizor - Creative Minds through Art of Mathematics - Math4Teens

A prism with any polygon as its directrix is called a polygonal prism.
Triangular prisms and parallelepipeds are particular cases of polygonal prisms. So are pentagonal (pentagon as a directrix) and hexagonal (hexagon as a directrix) prisms.

According to a mini-theorem proven about all prisms, the upper and lower bases of a polygonal prism are congruent polygons and the side faces are parallelograms.

Obviously, we can talk about right polygonal prism if all the side edges are perpendicular to bases.

Of particular interest is the volume of a polygonal prism. To approach this problem, we will perform the following its transformation.

Assume we have a pentagonal prism with a lower base ABCDE and an upper base A'B'C'D'E' (any other polygonal prism can be addressed similarly). Let's break it into a combination of three triangular prisms by cutting it vertically with planes ACC'A' and ADD'A'. The three triangular prisms are: ABCA'B'C',
ACDA'C'D' and
All three triangular prisms have the same altitude H - the distance between the two base planes. The volume Vol of each of them equals to a product of the area of a base S and an altitude:
VolABC = SABC·H
VolACD = SACD·H

Summarizing the three equations above, we obtain the volume of an entire prism:
But the expression in parenthesis constitutes the area of a polygon of the base of this prism. Therefore, our formula for the volume of any polygonal prism is still the same - product of the area of the base and the altitude:
Vol = S·H

Let's do some counting now.
Assume we are dealing with a polygonal prism with an N-sided polygon as the directrix.
The number of vertices is:
V=N+N=2N.
The number of edges is:
E=N+N+N=3N.
The number of faces is:
F=N+2
Let's check the Euler's formula V+F-E=2 that we mentioned in one of the introductory lectures about polyhedrons:
2N+N+2−3N=2
So, the formula holds, as expected.

### Unizor - Geometry3D - Prisms - Triangular Prism

Unizor - Creative Minds through Art of Mathematics - Math4Teens

A prism with a triangle as its directrix is called a triangular prism.

According to a mini-theorem proven about all prisms, the upper and lower bases of a triangular prism are congruent triangles and the side faces are parallelograms.

Obviously, we can talk about right triangular prism if all the side edges are perpendicular to bases.

Of particular interest is the volume of a triangular prism. To approach this problem, we will perform the following transformation of a triangular prism.

Assume we have a triangular prism with a lower base ABC and an upper base A'B'C'. All side faces are parallelograms, including BCC'B'. In that parallelogram we choose a point of intersection of its diagonals
P=BC'∩B'C.

Now let's use this point P as a center of symmetry and reflect each vertex of our triangular prism relatively to this center. That means, we connect each vertex of a prism (say, A) with this center P by a segment (AP in this case) and extend that segment further behind a center P by a length of this segment, getting a centrally symmetrical point. In this case a new point D' will be symmetrical to point A and a new point D will be symmetrical to point A', while all other vertices will be symmetrical among themselves: B - to C' and C - to B'.

Prisms ABCA'B'C' and BCDB'C'D' are symmetrical and, therefore, congruent and have the same volume. But, considered together as an object ABCDA'B'C'D', they form a parallelepiped. It follows from the properties of central symmetry of transforming lines to lines, planes to planes, segments to segments of equal length and angles to angles of equal measure. We leave the details of this proof to students and will be happy to include it into this writing with all the credentials if submitted.

Now, since our original triangular prism is a half of parallelepiped ABCDA'B'C'D', its volume VTprism is also a half of the volume of this parallelepiped VPped, which is equal to a product of the area of its base SPped by its altitude H:
VPped = SPped·H
Therefore,
VTprism = (1/2)·SPped·H
But, since area of the base of our triangular prism equals to a half of the area of parallelogram ABCD,
STprism = (1/2)·SPped
and, therefore, we came to the same formula for a volume of a triangular prism, that is it's equal to a product of the area of the base by the altitude:
VTprism = STprism·H