Geometría proyectiva: Dynamic construction of a Tetrad of points [Geogebra]

Classically the constructions of geometry of the books show an image or drawing that we should explore carefully to determine the origin of the data and structures derived from them.

The interpretation of the sequence needed to obtain a certain construction is an added difficulty to the formative process of the different geometries.

Application “Geogebra” It allows you to develop dynamic constructions in which we can modify the position of the elements forming it, keeping the geometric constraints of these figures, allowing the invariants of the same show. This tool can be a valuable aid for students.

Professor Juan Alonso Alriols He has collaborated in the introduction of this tool in the teachings of “Graphic Expression” in the Superior technical school of engineering Aernonautica and space (ETSIAE) UPM (UPM), providing examples of high interest. You can see an example of his work in the “Dynamic construction of double reason for four points” accompanying this entry, that has added a driver text for use in our classes.

Thanks John for your contribution.

After seeing step by step the the process of obtaining the fourth item of a Tetrad known the value of its double reason, Let's see an example of dynamic geometry using the Geogebra program.

The problem provides the value of the Tetrad of points (ABXY)= m/n that point is unknown “And”. The exercise allows to vary at any time the value of the sliders on the left top (m and n), whose ratio is the value of the double reason sought. The same points can be moved “A” and “B” along the X axis to give rise to infinite variations of the initial data. By pressing the buttons in advance of the bottom of the screen, You can access the steps of construction:

1. Statement
2. The theoretical analysis of the problem becomes. To be invariant double reason projective, the Tetrad will be constant in the series of points resulting from projecting and disconnect them . So, You can determine the Tetrad (A1 B1 X 1 Y1)=(ABCD). If you are assessed that “X 1 = X” and point “Y1″ is in the infinite, It can develop the expression above until you get to (X 1 A1 B1)= m/n, whatever the position that “A1” occupy in the plane, as “A1” You can scroll with the mouse.
3. Determining the rays “to” and “b” that projected to “A“, “A1” and “B“, “B1” respectively the center of projection can be determined “V..“. You only need to project from “V..” the point of the infinite “Y1” to find the point “And” Searches.