Study implicitly defined objects

Create configurations you "don't know how to create"


If we are only able to construct an object (or configuration) that satisfies some of the desired properties, but not all, then we can revert to "implicit construction". For example, we are looking for a line that bisects the perimeter and area of a given triangle; we know how to cut a triangle in two pieces with a line, and perhaps also bisect the area or perimeter, but not both area and perimeter. In simple cases, we look for the position of a free point (possibly on a line or circle) in a configuration so that the configuration satisfies one or more additional requirements (and we ignore how this can be achieved by construction).

OK Geometry can sometimes help in solving such problems: It uses numerical methods to change the position of one or more objects in a configuration so that the requirements are satisfied. What is really important here is the subsequent observation of the geometric properties of the solution found, which often provides nice hypotheses for the solution of the original problem.

Creating configurations by implicit construction in OK Geometry is fairly simple:
  1. Construct the configuration (in OK Geometry or import it from a dynamic geometry system) without considering the implicit requirements (i.e., the requirements we cannot satisfy).
  2. In the Sketch Editor, we specify the additional condition to be fulfilled. The possible conditions are listed in the command Advanced|Check property. Conditions can be composed using the Number|Compose conditions command.
  3. In the Sketch Editor, apply the Advanced|Implicit construction command. The command is explained in the OK Geometry Sketch Editor Reference Manual.
  4. Observe the obtained solution. You can use the Observe command or (in OK Geometry Plus) a more sophisticated Advanced Query command or the Analyse object in reference triangle command.
With implicit constructions we can create configurations that we "do not know how to create". But this come with a price. First, there is no guarantee that we will find a solution, even if there is one. And if a solution is found, the dynamic configuration found is less precise than the usual constructions (we are allerted of this fact by one or two red triangles with exclamation marks at the top of the screen). Consequently, the observations are less reliable and some observation methods cannot be applied in such configurations.

We illustrate the use of implicit constructions with a few examples.

Example 1


Problem: For a given triangle ABC and a point P that does not lay on any of the sidelines of the triangle ABC, let ka be the circle through the points B, C, P. Define kb and kc cyclically. The circles ka, kb, kc are, in general, not congruent to each other. Is it possible to position the point P so that the three circles ka, kb, kc are congruent to each other?

1. First, we make a construction with a free point P, without the condition of congruence. We also measure the radii ra, rb, rc of the three circles.

2. We use the command Advanced|Check property|Equivalence|SameValues3 to check if the three radii are equal. We get the condition parameter Same_ra_rb_rc = False.
3. Then we apply the command Advanced|Implicit construction. We ask OK Geometry to position the point P on the plane so that the condition Same_ra_rb_rc is satisfied.


4. The Observe command of the obtained configuration shows that the required condition is satisfied if P is the orthocentre of ABC. It is quite easy to prove the correctness of this hypothesised solution.

Example 2

Problem: How to pack four congruent circles into a triangle, as shown in the figure at the top of this article.

It is quite easy to place 3 congruent circles at the corners of the triangle ABC (For example: Select a point Sa on the bisector of angle A, then draw the circle ka with Circle centre + 1 object command, and finally draw circles kb, kc congruent to ka using the Circle radius + 2 objects command.). With the command Circle 3 objects you can place a circle k between the three circles touching them. So, we can create a construction that satisfies the requirements, except for the congruence of the circle k with the circles ka, kb, kc.

1. First we draw a configuration that satisfies all the requirement except the congruence of the central circle k with the circles ka, kb, kc at the corners. By measurement we obtain the radius ra of the circle ka and the radius rr of the circle k.
2. Using the command Advanced|Check conditions|Equivalence|Same values we check the condition Same_ra_rr = False.

3. Then we apply the command Advanced|Implicit construction. We ask OK Geometry to position the point Sa on the line sa so that the condition Same_ra_rr is satisfied.


4. At this point you can observe and study the obtained configuration. The Triangle analysis or the command Analyse object in reference triangle observes that the point S is the centre X35 of the triangle ABC. (X35 is the harmonic conjugate in X1,X3 of the inversion of the incentre in the circumcircle of ABC). To construct the configuration we need some more data, e.g., the point A' = point of contact of circles ka and k. With Triangle analysis we observe that AA' contains the point S' = X55 (the insimilicentre of the inscribed circle and the circumcircle of ABC.
Following this observations, we can obtain the configuration for a given triangle ABC as follows:
  1. Construct the point S = X35 of ABC (= harmonic conjugate in X1,X3 of the inversion of the incentre in the circumcircle of ABC).
  2. Construct the point S' = X55 of ABC (= insimilicentre of incircle and circumcircle of ABC).
  3. Let p be the image of the line AS' under the dilation with centre S and dilation factor 2.
  4. Then A' is the intersection of p  and sa (the angle bisector at A).
  5. The rest is trivial.
The Advanced query of this configuration shows several other ways to get A' or Sa from A,B,C,S and S'. You can also use the Observe formulae to observe the radii of the four circles in the configuration: rr = r/2 - r/(4·cos A + 4·cos B + 4·cos C -2), where r is the radius of the circumcircle of ABC..


Example 3

Problem: Given an acute triangle ABC and a point P not laying on its sidelines, let A' be the intersection, other than A, of the line AP with the circumcircle of ABC. Define B' and C' cyclically. The triangle A'B'C' is called the circumcevian triangle of P in ABC.

Is it possible that for some point P on the plane the the triangle ABC and the circumcevian of P in ABC share the same nine-point circle?

We proceed as in the previous examples: We draw a configuration that satisfies all the requirement except that the nine-point circles k and k' of the two triangles coincide. (To draw the triangle A'B'C' you can use the command Special|TrianglePoint derived objects|Triangle|Circumcevian triangle. To draw the nine point circle, use the command Special|Triangle derived object|Circle|Nine point circle).

Using the command Advanced|Check conditions|Equivalence|Coincidence of lines/circles check the condition Same_k_k' = False.

Then we apply the command Advanced|Implicit construction. We ask OK Geometry to position the point P in the plane so that the condition Same_k_k' is satisfied.


Observation with Triangle analysis or the command Analyse object in reference triangle shows that the point P is the Walsmith point (X5000) of ABC, which is one of the intersections of the Stevanovich circle and the Euler line of ABC. The formal definition of the Walsmith point is  rather complicated and it is (probably) not easy to prove the observed fact (which is not listed among the properties of the Walsmith point in ETC). 



Download figures (png)

Download constructions (p)