Generic dynamic constructions

Create and observe families of dynamic constructions

A generic construction is a family of constructionally isomorphic constructions called examples. All examples in a generic construction generate the same kind of objects in the same sequential order, but may differ in the operations used. All examples of a generic construction in a given construction steps, for example, create a line, but one example creates a line through two given points, another creates the bisector of the line segment through the same two points, etc. Generic constructions are a generalisation of common constructions, where in some steps instead of an operation that generates an object there is a rule, i.e. a"union" of equivalent operations all yielding the same kind of objects. A generic construction can contain dozens, hundreds, or even thousands of examples.

Examples of a generic constructions are ordinary dynamic constructions. Generic constructions can be recognised by the red Gen button in the tool bar on the right side of the display. The same button also invokes the form for managing the examples contained in a generic construction.

With OK Geometry you can create, observe and analyse generic constructions.
  • Each of the example constructions of the family is immediately available, can be displayed, observed, analysed, exported, etc.. just like an ordinary dynamic construction.
  • It is possible to search the family for examples with certain properties. For example, you can search for all constructions in a family where three given points are collinear. Searching for specific properties in a family of constructions is very fast.
  • It is possible to search the family for examples, where a property occurs anywhere. For example, it is possible to search a family for all examples in which  any two triangles with the same area occur.
  • It is also possible to perform Triangle analysis and Formulae observation simultaneously for all or part of the examples in a family.
You create a generic construction in the same way you create an ordinary dynamic construction. In the steps where you want a "union" of operations instead of one operation, you use a rule from the Generic menu of the Sketch Editor. 

For details about generic constructions, related commands, and managing examples, see the OK Geometry Plus Reference manual. 

Example 1

The left figure below is a standard exercise in elementary geometry.  

ABCD is an arbitrary quadrilateral.
E is the midpoint of AB, the points F, G, H are defined cyclically.
Prove: EFGH is a parallelogram.

The exercise has obvious generalisations and analogies that can be presented as a generic construction. Instead of the midpoint operation, we consider a rule that contains several operations to determine points on the sides of the quadrilateral (e.g., midpoint, point at 1/3 of the side, point on the golden section, etc.). And instead of a random quadrilateral, we consider a rule consisting of several types of quadrilaterals built on a segment. We look for examples in  the generic construction, where points E,F,G,H are cocyclic and examples where lines EG and FH are orthogonal. 
In our case, we have created a generic construction that contains 28 kinds of quadrilaterals and 12 operations to create points E, F, G, H, so that the generic construction contains altoghether 336 examples. We have also declared the parameter Perpe_EGFH, which checks the ortogonality condition, and the parameter Cocyc_EFGH that checks whether EFGH is a cyclic quadrilateral.

The Gen button opens the form for managing examples of the generic construction. Using the form, we scan through the examples. Among the examples with cocyclic points EFGH (many of them are trivial to prove) we present the following two: 
       
ABCD is a quadrilateral.
|AB|2+|CD|2 = |BC|2 + |DA|2  (Pythagorean quadrilateral)
E, F, G, H are midpoints of the sides of ABCD.
Prove: EFGH is a cyclic quadrilateral.

ABCD is a quadrilateral.
AC is orthogonal to BD.
E, F, G, H are midpoints of the sides of ABCD.
Prove: EFGH is a cyclic quadrilateral.



Among the examples with orthogonal lines EF and GH (many of them are trivial to prove) we present the following four:
     
ABCD is a quadrilateral.
AC is orthogonal to BD.
E, F, G, H are midpoints of the sides of ABCD.
Prove: EG is orthogonal to FH.

ABCD is a rhombus.
|AE| = |BF| = |CG| = |DH|
Prove: EG is orthogonal to FH.





ABCD is a right deltoid, i.e., |AB|=|AD|, |CB|=|CD|, B = D = 90°. AF bisects BAC, BG bisects CBD, CH bisects DCA, DE bisects ADB.  Prove: EG is orthogonal to FH.
ABCD is a right deltoid, i.e., |AB|=|AD|, |CB|=|CD|, B = D = 90°. DE is orthogonal to AB
EG is orthogonal to CD.
Prove: EG is orthogonal to BD.


 

Example 2


The figure shows the Ascella triangle A'B'C' of the triangle ABC.  The sidelines of the Ascella triangle are the radical axes of the incircle of ABC and the respective vertices A, B, C. The lines AA', BB', CC' are not concurrent (even though it may appear so in the figure). 

The incircle is only one of the characteristic circles of a triangle. If we replace the incircle by other characteristic circles in the construction, we get interesting triangles A'B'C' - among them there may be triangles that are perspective to the reference triangle ABC.

For this purpose we create a generic construction. In the construction, the step of constructing the incircle is replaced by a rule from the command Generic|Triangle objects|Circles (we include all available circles). We check the concurrency of lines AA', BB', CC' and declare the result as parameter Concu_AA'BB'CC'. We also define J as the intersection of lines BB' and CC'. if Concu_AA'BB'CC' is true, then J is the perspector of the triangles ABC and A'B'C'. The generic construction contains 81 examples. 

The Gen button launches the form for managing the examples of the generic construction. Checkmark the TCircle1 rule for the circle. To find examples where the triangles ABC and A'B'C' are perspective, checkmark the condition Concu_AA'BB'CC', select the Scan mode "Fast", and click the Start Scan button. OK Geometry finds 14 examples that satisfy the perspectivity condition. You can Review them one by one. 

To examine the centres of perspectivity for all found examples, start the Triangle analysis module while the Gen form is still active. Set the point J as the examined object and the set the Examples to examine to "review". The Study button starts the observation process for all examples in which the perspectivity of ABC and A'B'C' occurs.

Here is a brief list of results:

Triangle ABC circle Centre of perspectivity of ABC and A'B'C'
Circumcircle X6: SYMMEDIAN POINT (LEMOINE POINT, GREBE POINT)
Brocard circle X39: BROCARD MIDPOINT
Bevan circle X57: ISOGONAL CONJUGATE OF X(9)
Droz-Farny first circle Intersection of X(2)X(3) and X(54)X(70)
Droz-Farny second circle Intersection of X(6)X(235) and X(64)X(1885)
Parry circle X3: CIRCUMCENTRE (degenerate case)
2nd Brocard circle X695: ISOGONAL CONJUGATE OF X(384)
Stevanovich circle X3: CIRCUMCENTRE (degenerate case)
A-Neuberg circle 1st vertex of Brocard triangle
B-Neuberg circle 2nd vertex of Brocard triangle
C-Neuberg circle 3rd vertex of Brocard triangle
Dao-Moses-Telv circle X3: CIRCUMCENTRE (degenerate case)
Montesdeoca-Lemoine outer circle Intersection of X(3)X(6) and X(3235)X(3301)
Montesdeoca-Lemoine inner circle Intersection of X(4)X(6) and X(3235)X(3299)



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