Advanced query of objects
How is an object related to a subset of objects in a construction? How to construct an object from given objects?
Given a configuration, OK Geometry lists the observed properties of the objects in the configuration. By subsequently clicking the query buttons (magenta buttons on the right side of the displayed construction, e.g. ?•, ?? ) you prompt OK Geometry to display only properties that contain a given object.In Plus mode, the query buttons have an additional functionality. In the configuration, we can declare a set of unknown objects - the remaining objects are, obviously, called known objects. Usually, we consider as known those objects that we know how to construct from the given objects. The advanced query function tries to relate a specified unknown object to a objects that are geometrically derived from known objects.
The advanced query is a great help in hypothesising solutions to construction tasks. It can also reveal surprising relationships between objects in a configuration.
For a detailed description of how to use the advanced query, see the OK Geometry Plus Reference Manual.
Here is the general approach to using the advanced query in construction tasks:
- Obtain, in some way, a configuration with the given objects and the object(s) to be constructed. Sometimes this can accomplished by starting with the final object(s) and then positioning the given objects accordingly. Another way is to use advanced construction commands in OK Geometry.
- Mark all unknown objects using the button •? on the right hand toolbar of the display.
- Use OK Geometry to observe the configuration (command Observe).
- Query one or more unknown objects using the query commands (magenta buttons with question marks on the right side of the display). You will get a list of properties that relate the queried object to known objects. The list may contain object that are not part of the original configuration (auxiliary objects, e.g. point _3617 or Y3617). To get information about such objects, position the mouse curson on the object (or click on the label in the property above the construction). Note that the notation of auxiliary points is temporary and changes each time you run the Observe command.
Example 1
Given is a triangle ABC and a point A' on side BC. Construct a point B' on the side CA and a point C' on the side AB, so that the triangle A'B'C' is equilateral.
To solve this fairly simple construction task with advanced query we first need a configuration. We easily obtain this by drawing first the equilateral triangle A'B'C' and circumscribing a triangle ABC around A'B'C'.
The next step is to mark all the unknown objects. In our case the unknown objects are the points B' and C' and the triangle A'B'C'.
To solve this fairly simple construction task with advanced query we first need a configuration. We easily obtain this by drawing first the equilateral triangle A'B'C' and circumscribing a triangle ABC around A'B'C'.
The next step is to mark all the unknown objects. In our case the unknown objects are the points B' and C' and the triangle A'B'C'.
We first Observe the configuration. Then we query the point B' (or C'). The query reveals that B' is on the side CA and on the line through points Y559, Y759, and Y83, where
- Y559 is the image of A' rotated by 60° around A,
- Y759 is the mirror image od A' in line AB, and
- Y83 is the image of A' rotated by 60° around B.
Advanced query often shows unexpected solutions. Here is one such solution for our construction task:
B' (and analogously C') is the intersection of the line CA with the circle through A', C and Y1013 = the 1st isodynamic point (X15) of the triangle ABC.
B' (and analogously C') is the intersection of the line CA with the circle through A', C and Y1013 = the 1st isodynamic point (X15) of the triangle ABC.
Example 2
Construct a triangle ABC with the given positions of the vertex A, the incentre I of ABC and the circumcentre O of ABC.
Like in the previous example, we first create a complete configuration. Then we mark all unknown objects, in our case the vertices B and C and the triangle ABC.
Then we Observe the configuration. Finally we apply the advanced query to points B (and C).
Like in the previous example, we first create a complete configuration. Then we mark all unknown objects, in our case the vertices B and C and the triangle ABC.
Then we Observe the configuration. Finally we apply the advanced query to points B (and C).
Here is the solution as proposed by OK Geometry. According to this, we first obtain the point Y389, which is the intersection of the circumcircle of ABC and the bisector of the segment AI (of the two intersections, we take the one on the opposite side of CA as B).
The vertex B is the intersection, other than Y389, of the line IY389 and the circumcircle of ABC.
The vertex C is constructed in an analogous way.
Note that OK Geometry merely reports the observations. The correctness of the solution should be established by the user.
The vertex B is the intersection, other than Y389, of the line IY389 and the circumcircle of ABC.
The vertex C is constructed in an analogous way.
Note that OK Geometry merely reports the observations. The correctness of the solution should be established by the user.
Example 3
Consider the famous Malfatti problem of constructing three circles ka, kb, kc inside a given triangle ABC. The circle ka should be externally tangent to the circles kb and kc and to the sides AB and CA. Analogous requirements apply for the circles kb and kc.
The configuration to be studied can be obtained by starting from three circles that are mutually externally tangent to each other. We then construct the sides of the triangle ABC. Alternatively, we can start from the triangle ABC and use in OK Geometry Plus the command for Malfatti triangle (whose vertices are the centres of the Malfatti circles of triangle).
We present two of the many constructions proposed by OK Geometry.
In the first construction we consider the centre I of the circle inscribed into ABC and the centres A'', B'', C'' of the circles inscribed in the triangles BCI, CAI, and ABI. A query of the centre A' of the circle ka shows that A' lays (obviously) on the line AI and on the circle through B'', I, and Y372, where Y372 is a vertex of the square having A''B'' as diagonal.
The second construction shows how to obtain a common tangent line to ka and kb. In the figure
L = intersection of the circle k(B,I) and the segment AB,
M = intersection of the circle k(A,I) and the segment AB,
D = midpoint of L and M,
A'' = incentre of the triangle BCI,
B'' = incentre of the triangle CAI,
E = intersection of lines CI and A''B'',
DE = a common tangent to ka and kb.
It is not easy to prove the correctness of the two constructions.
Downolad constructions (p)