The Modelling Process of a Paper Folding Problem in Geogebra 3d 1

—In this research; a problem situation, which requires the ability of thinking in three dimensions, was developed by the researchers. As the purpose of this paper is producing a modeling task suggestion, the problem was visualized and analyzed in GeoGebra3D environment. Then visual solution was also been supported by algebraic approach. So, the capability of creating the relationship between geometric and algebraic representations in GeoGebra was also presented in 3D sense.


INTRODUCTION
There are several studies on modeling the real life situations in GeoGebra [2,3,4,5].In this research, a problem situation has been suggested and modeling process in GeoGebra has been explained.Zbiek and Conner pointed out, modeling contributes to understand the fore known mathematical concepts thoroughly by demonstrating the applicability of mathematical thoughts to real life, to learn new mathematical concepts, to establish inter disciplinary relations and to both conceptual and operational development of the students studying in modeling processes [6].Furthermore, the algebraic and geometric representations are needed to be connected in two ways [1,7].That is, the modeling should present the advantage of understanding how algebraic facts effect the observed situations.The problem can be described as follows according to the figure 1   intersect?" in the research (Figure 2).The mathematical concepts related to the solution can be summarized as algebraic approach, line equations, slope, point and vector in three dimensional space and scalar triple product.As a short result, it can be stated that at least one of the points A and B must be in the midpoint of the segments on which they are located after visualization in the environment of GeoGebra 5.0.It is expected that this kind of real situated activities are attractive for the students.Both of the lines' equations can be constructed by using a point which belongs to the line and its tangent.
The equation of the line DA d can be calculated as follows; After the revision of equation, following form can be obtained; The equation of the line . After the revision of equation, following form can be obtained; The equation of the line By finding the common solution of the equation ( 4) and (5) the coordinates of d K can be obtained as follows; Finding the coordinates of the point d L : The intersection point of the line BC d and the line Both of the lines' equations can be constructed by using a point which belongs to the line and its tangent.
The equation of the line BC d can be calculated as follows; www.ijacsa.thesai.org After the revision of equation, following form can be obtained; The equation of the line After the revision of equation, following form can be obtained; By finding the common solution of the equation ( 7) and (8) the coordinates of d L can be obtained as follows; After the revision of equation, following form can be obtained; By finding the common solution of the equation ( 10) and (11) the coordinates of d M can be obtained as follows When the triangles ,, DOA AKB BLC and CMD are folded perpendicular to the floor on the sides DA, AB, BC and DC respectively, the corner points O, K, L and M will be in their new places.Let's call these revised points as t O , t K , t L and t M respectively.These points will be in three dimensional space and the coordinates of ,,    Now, we have the points t O , t K , t L and t M in space.We can think as the mathematical question in our main problem is "what is the condition for the line segments tt OLand tt KM are intersected".The basic answer will be the vectors tt OL and tt KM must be coplanar and not parallel.Since the position of the points t O , t K , t L ve t M in space, we can be sure that these vectors are not parallel and has intersection points when they are reflected on the floor.Let's check the vectors tt OL and tt KM coplanar when they are intersected in space.
For this check, the fact of "three vectors' scalar triple product must be zero to be coplanar" can be used.We have two vectors for this operation.By choosing the vector tt OK additionally, we can calculate the scalar triple product.
The components of the vector tt OL; and figure 2. Let's call the intersection point of the segment [] KL and the line passing through the point D and parallel to segment [] OM as B and the intersection point of the segment [] DK and the line passing through the point C and parallel to the segment [] KL as A. So, the rectangle DABC and the triangles DOA  , such that the points D and A can be moved dynamically on the segments on which they are located. 11 A short summary of this article has been submitted to The International GeoGebra Institute Conference 2012 on 21-23 September 2012 and supported by Ahi Evran University.

Figure 1 :
Figure 1: Statement of pre-folding The problem was stated as "for which locations of the dynamic points A and B, the segments tt OL   and tt KM  

Figure 2 :
Figure 2: Statement of post-folding II.UNDERSTANDING THE MODEL VISUALLY The basic structure of the problem can be constructed as follows where the values of a, b, 0 x and 0 y defined as a slider tool.The points (0,0) O , ( ,0) Ka , ( , ) L a b , (0, ) Mb are the corners of the rectangle.The points 0 ( ,0) Ax ,

Figure 3 :
Figure 3: Problem statement III.ANALYZING THE MODEL ALGEBRAICALLY After the triangles ,, DOA AKB BLC and CMD are folded on the sides DA, AB, BC and DC respectively, the coordinates of the points ,, d d d O K L and finding the common solution of the equation (1) and (2) the coordinates of d O can be obtained as follows; the coordinates of the point d K :The intersection point of the line AB d and the line d KK d will provide us to determine the coordinates of the point d K .the fact of the product of the tangents of two perpendicular lines, it can be obtained that of the lines' equations can be constructed by using a point which belongs to the line and its tangent.The equation of the line AB d can be calculated as follows; the fact of the product of the tangents of two perpendicular lines, it can be obtained that 9) Finding the coordinates of the point d M : The intersection point of the line DC d and the line d MM d will provide us to determine the coordinates of the point d M .the fact of the product of the tangents of two perpendicular lines, it can be obtained that of the lines' equations can be constructed by using a point which belongs to the line and its tangent.The equation of the line DC d can be calculated as follows; and d M will be the first two components of them.The third components of each point will be the lengths of d OO ,

Figure 4 :
Figure 4: Comparing two and three dimensional position of folded paper.These lengths can be calculated by using the points O, K, L and M and the equations of the lines DA d , AB d , BC d and DC d .After the proper calculations the lengths are found as follows;