## Tuesday, July 21, 2015

### Unizor - Geometry3D - Symmetry - Congruence

Unizor - Creative Minds through Art of Mathematics - Math4Teens

The purpose of this lecture is to prove that symmetrical objects in three-dimensional space are congruent.

Mini-theorem 1

A centrally symmetrical counterpart to a segment is a segment of the same length.

Mini-theorem 2

A counterpart to a segment reflected relatively to a plane is a segment of the same length.

Mini-theorem 3

A centrally symmetrical counterpart to a triangle is a triangle congruent to the original.

Mini-theorem 4

A counterpart to a triangle reflected relatively to a plane is a triangle congruent to the original.

Mini-theorem 5

A centrally symmetrical counterpart to an angle is an angle congruent to the original.

Mini-theorem 6

A counterpart to an angle reflected relatively to a plane is an angle congruent to the original.

Mini-theorem 7

A centrally symmetrical counterpart to a dihedral angle is a dihedral angle congruent to the original.

Mini-theorem 8

A counterpart to a dihedral angle reflected relatively to a plane is a dihedral angle congruent to the original.

Let me add more philosophical statement that is based on the above theorems.

Since

most of complex three-dimensional objects contain elements discussed

above (like pentagon consists of five segments and five angles,

parallelepiped consists of six parallelograms and eight dihedral angles

etc.), these complex objects are transformed into congruent ones by a

transformation of symmetry (central or relative to a reflection plane).

## Monday, July 20, 2015

### Unizor - Geometry3D - Plane Reflection

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Plane reflection in three-dimensional space assumes the existence of a fixed plane of reflection.

To construct a point A' reflectively symmetrical to a given point A relatively to a plane of reflection γ, we have to drop a perpendicular from point A onto plane γ and extend it beyond the base point by the same length.

Based on this construction, it's easily observed that the process is reversible. If we start from point A', drop a perpendicular to the same plane γ and extend it by the same length, we will arrive to the original point A. We can say, therefore, that the process of reflection is reflexive, that is, if point A' is a reflective image of point A relative to plane γ, then point A is a reflective image of point A' relative to the same plane γ.

Theorem 1

A reflectively symmetrical counterpart of a straight line is a straight line.

Theorem 2

A reflectively symmetrical counterpart of a plane is a plane.

## Friday, July 17, 2015

### Unizor - Geometry3D - Central Symmetry

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Recall that on a plane we had two types of symmetry - symmetry relative to a point (central symmetry) and symmetry relative to a straight line (symmetry about an axis or reflection).

Here we will discuss the central symmetry.

On a plane point A', centrally symmetrical to point A relative to a center of symmetry P, can be constructed by connecting A and P by a segment and extending this segment beyond point P (center of symmetry) by the same length, thus obtaining point A'.

Central symmetry in three-dimensional space requires analogous construction.

If we are given point A and a center of symmetry P, then point A', symmetrical to point A relatively to center P, is located on a continuation of a segment AP beyond point P by the same distance as the length of segment AP.

The construction of a symmetrical point is, obviously, reversible. If we start from point A', connect it to the center of symmetry P and extend by the same length, we will be at point A. So, if point A'

is centrally symmetrical to point A relatively to center P, point A is symmetrical to point A' relatively to the same center of symmetry P.

Theorem 1

A centrally symmetrical counterpart of a straight line is a straight line parallel to the original one. Original line and its symmetrical counterpart are equidistant from the center of symmetry.

Theorem 2

A centrally symmetrical counterpart of a plane is a plane parallel to the original one. Original plane and its symmetrical counterpart are equidistant from the center of symmetry.

## Monday, July 13, 2015

### Unizor - Geometry3D - Prisms - Mini Theorems 1

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Mini Theorem A

If a directrix of a cylindrical surface is a straight line non-parallel to a generatrix, this surface is a plane.

Mini Theorem B

If a directrix of a cylindrical surface is a polygon ABCDE lying within a plane γ non-parallel to a generatrix, then any other plane δ, parallel to plane γ, intersects a given cylindrical surface along a polygon A'B'C'D'E' congruent to polygon-directrix ABCDE.

Corollary 1

Immediate consequence of the above theorem is that two polygons that are the bases of a prism are congruent with correspondingly parallel and equal edges and correspondingly equal angles.

Corollary 2

Each side face of a prism is a parallelogram.

## Thursday, July 9, 2015

### Unizor - Geometry3D - Prisms - Volume of Parallelepiped

Unizor - Creative Minds through Art of Mathematics - Math4Teens

The volume of a parallelepiped (as a volume of any solid body) is an additive measure that quantifies the space occupied by it.

The unit of volume is a cube with the length of each edge equaled to a linear unit. So, if linear unit is 1 meter, the unit of volume is 1 cubic meter (1m³).

Consider a rectangular parallelepiped with three edges sharing a vertex A having linear lengths of a, b (along the sides of a base) and c (height).

For now, consider these three lengths as integer numbers. Obviously, the number of unit cubes that fill the area of the base of this parallelogram in one single layer is S=a·b.

There are c such layers along the height of the parallelepiped, so the total number of unit cubes that fill this rectangular parallelepiped equals to V=S·c=a·b·c.

Next step is to consider the lengths of edges to be rational numbers. We can always assume them to have the same common denominator, so

a=p/n; b=q/n; c=r/n

A unit cube can be filled with smaller cubes with edges of 1/n. The number of these small cubes is, obviously, n³. So, it's natural to assume that the volume of a cube with the edge of 1/n is V=1/n³.

Since volume is an additive measure, we can count the number N of small cubes with the edge of 1/n that fit into our parallelepiped with above mentioned lengths of edges. Then the volume of a parallelepiped would be this number N multiplied by a volume of a small cube 1/n³. Obviously, N=pqr.Therefore, the volume of a parallelepiped is

V = (pqr)/n³ =

= (p/n)·(q/n)·(r/n) = abc

Again, as we see, the volume of a rectangular parallelepiped equals to a product of lengths of three edges sharing one vertex.

Extending without a rigorous proof this result to irrational lengths, we can state that, in general, the volume of a rectangular parallelepiped with lengths of three edges sharing the same vertex equaled to a, b and c is equal to

V=abc

Let's consider a more general case of a right parallelepiped. The difference between right and rectangular parallelepipeds is that the base of the right one is a general parallelogram, while the base of the rectangular one is a rectangle. In both cases the side edges are perpendicular to bases.

Assume the bottom base of our right parallelepiped is a parallelogram ABCD and the top base is, correspondingly, A'B'C'D'.

Within the bottom base draw perpendiculars BM and CN from vertices B and C onto side AD (or its continuation). Assume for definitiveness that point M lies in between points A and D, while point N is outside this segment.

Obviously, triangles ΔABM and ΔDCN are congruent - a trivial statement from plane geometry.

Now draw a plane through edge BB' and line BM on the bottom base. It cuts from our parallelepiped a right prism with a triangles ΔABM and ΔA'B'M' as bases and side edges AA', BB' and MM'.

Draw another plane through edge CC' and line CN. Consider a right prism with bases ΔDCN and ΔD'C'N' and side edges DD', CC' and NN'.

Two right prisms with triangles as bases are congruent since their bases and edges are congruent - an easy to prove statement. Therefore, their volumes are equal. What's interesting is that if we cut off the first prism from our original right parallelepiped and add the second, we convert our right parallelepiped into a rectangular one. This new rectangular parallelepiped has rectangles BCNM and B'C'N'M' as bases and segments BB', CC', NN' and MM' as side edges.

Since we formed this new rectangular parallelepiped by cutting one triangular prism from the original right parallelepiped and adding another, equal in volume, the volumes of the original right parallelepiped and a new rectangular one are the same.

The volume of the latter equals to a product of an area of the base S = BM · MN by the height BB'. But the area of the base (rectangle) BCNM equals to the area of the original base (parallelogram) ABCD. Therefore, the volume of the right parallelepiped with base ABCD also equals to a product of the area of the base (parallelogram) ABCD, that is S = AD · BM, and the height of the parallelepiped AA'.

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