# The objectives of the inventor Samvel Movsisyan

The author hopes that these geometric patterns and mathematical scales can sozdovat many interesting tasks.

For interesting problems the author promises a cash reward or prize.

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The author hopes that these geometric patterns and mathematical scales can sozdovat many interesting tasks.

For interesting problems the author promises a cash reward or prize.

For creation of this model at first we fix those rods, which have no metal pins. We fix them against each other.

By means of separately taken circle and the diameterthereof is possible to construct an inscribed triangle, quadrangle.

**Convertible telescopic geometric teaching and instructional manual on plane geometry and stereometry geometry "Triangular Pyramid"**

From this model it is possible to obtain all types of a triangular pyramid. Before starting working with models, it is necessary to know that for manipulations and transformations it is necessary to work consistently only with each of the sides separately. The pyramid which shows the theorem of three perpendiculars is of special interest.

SB⊥(ABC)

AC⊥CS follows AC⊥BC and AC⊥BC follows AC⊥CS

With this type of model it is possible to solve numerous tasks. From this model it is possible to receive all types of quadrangles and a triangle for the construction of which it is necessary to make the following: all sides of the basis of a pyramid A, B and C should be pulled, and the edges SA, SB and SC should be decreased in length until they all fit in the plane ofthe triangleABC.

*Creation of sections*

The auxiliary rods are included in the kitfor sections construction, which allow to obtain any sections. It is easy to be convinced, if we notice that the rods can be fixed in any combinations. In a triangular pyramid by means of auxiliary rods, as shown in figure, we demonstrate that section which represents a quadrangle. By the way, it is possible to construct with their help the height, the median, the bisector, etc. Auxiliary rods can also serve for replacement of the broken rods models.

Solving of tasks

Before passing to the solution of tasks, it should be noted that models are irreplaceable for obtaining drawings of spatial bodieson the background ofthe board. Appropriately holding the modelat the board, it is possible to draw itself as it is visible. For the solution of the tasks concerning a triangular pyramid it is necessary to use a quadrangular pyramid which thanks to the fact that is being modified in various triangular pyramids, gives the chance to demonstrate solutions to specific tasks.

*Task 1.*

We have a triangular pyramid with equal edges which basis is the rectangular triangle.

Height of such pyramid passes through the center of a circumscribed circle.The center of a circle is in the middle of a hypotenuse.

* *

*Task 2.*

* We have the pyramid with ribs a, b, c, which are mutually perpendicular. We need to find the volume of a pyramid. The task gets a simple solution if the pyramid is inverted. As we can see, we obtained a pyramid whose base is a right triangle, where the length of basement legs (and the base is a right-angled triangle) and the length of the altitude are known. Thus, the volume of a pyramid is equal and bc/6.*

With the help of our models it is possible to excellently demonstrate the possibility of obtaining the drawings of spatial geometric bodies on the board.

*Task 3.*

*To prove that it is possible to inscribe the sphere into any pyramid. This is one of the most difficult problems in stereometry. We wire the cross section which passes through bisector dihedral angle BACD. They intersect in a straight line AH. If we wire the third cross section, then it will intersect with the second cross section in a straight line CM. The point of intersection CH and CM is considered as the center inscribed sphere, as it is equidistant from all sides.*

This model allows receivingof all types of the quadrangular pyramid and pentagon. We remind that the transformations are carried out with the help of change of each of the sides separately. Of particular interest is the case when the basement of ABDC quadrate and the lateral edge are perpendicular to the plane of the basement.

Then according to the theorem of three perpendiculars we obtain: AC ⊥CD, which implies AC⊥SC.

It is possible to receive a triangular pyramid from a quadrangular pyramidas follows. All sides of a triangle ABD we extend up to the end, then we hold the top C and shortening SC gradually we put it inside (i.e. top of C) of the triangular pyramid ABDS until SC becomes ⊥ (ABD) and C ∈ (ABD). There is also the second option of transformation.

The top of C ∈ (SAD). As a result we obtain a triangular pyramid ABDS (Fig. 4) in which it is possible to construct the height BC with the help of an auxiliary rod. In the same way it is possible to obtain a triangular pyramid from the quadrangularpyramid. Shortening AD, we do so that the sides of DB and DC made one line to the line BC. It is clear that turning this pyramid, different types of triangular pyramids are created, which are used in the solutions of tasks.

For implementation of each of the following transformations it is necessary to close consistently each of the sides of the model up to the stop. The quadrangular pyramid turns into a regular pentagon with its diagonals.

We take from the box this model and first of allwe straighten AC and A1C1 and we obtain a cube.

All types of a quadrangular prism are turned out of it, for example, the inclined plane. We evenly increase the sides ABCD; we obtain a truncated rectangular pyramid.

This pyramid can easily be turned into a quadrangular pyramid if to increase AA1 as far as that A1B1B, in order A1C1C and A1D1D became straight lines. The various sections are visible in this modification of a quadrangular pyramid.

Each model can be modified into the previous model. We leave receiving from a cube a triangular pyramid to you. We receive from the cube a regular hexagonal pyramid with the top A or with the top C1. We hold either from the top A or C1. We hold either from top A or C1 and we pull the edges emerging from these tops until the hexagonal pyramid is formed.

It is very interesting when a regular octagon is turned out from a cube.