Optimisation based constructions

Create optimisation based constructions and investigate their geometric properties

Geometric objects are not always defined by their geometric properties. Sometimes we look for objects that satisfy an extremality condition; for example, we look for a polyline of minimum length that satisfies certain conditions, or a triangle of maximum area that satisfies certain conditions, etc. In simple cases, we look for a position of a free point (possibly on a line or a circle) such that the resulting configuration satisfies an optimisation criterion. OK Geometry can help in solving such problems: It uses numerical methods to find the (local) optimal solution, but what is really important is the observation of the geometric properties of the solution found, which often provides nice hypotheses for the solution of the optimisation problem.

Generating a hypothesis for solutions of optimisation problems in OK Geometry is fairly straightforward:
  1. Construct the configuration (in OK Geometry or import it from a dynamic geometry system) without considering the optimisation requirement.
  2. In Sketch Editor, measure or calculate the quantity to be optimised.
  3. In Sketch Editor apply the Advanced|Optimise command. The command is explained in the OK Geometry (Basic) 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.
We illustrate the process first with a familiar and simple example, followed by two less trivial examples.

Example 1


Problem: Given are points A and B on the same side of a line k. Where to position a point C on the line k, so that the length of the path from A to C and then to B is as short as possible.

First, we make a construction with a free point C on the line k. We also measure the total length Len_ACB of the polyline connecting points A, C, and B.

Apply the command Advanced|Optimisation. We look for the position of the point C on the line k, which minimises the parameter Len_ABC.

The Observe command indicates that the angle between k and AC is congruent to the angle between k  and BC. This is the hypothesised solution of the problem.

The proposed solution needs an argumentation, a proof. You can get the proof, with almost no words needed, by mirroring the point B in the line k.

 

Example 2

Problem: In a given triangle ABC inscribe an ellipse with maximum area.

If A', B', C' are the points of contact on the sides BC, CA AB of an ellipse inscribed in the triangle ABC, then the lines AA', BB', CC' concur in a point (by Brianchon's theorem). The point P of concurrence thus uniquelly determines the inscribed ellipse. 

For a given point P, you can draw the inscribed ellipse in the triangle ABC using the command Special|TrianglePoint derived objects|Conics|Inconic in OK Geometry Plus Sketch Editor.
To determine the area of the ellipse, we draw the axes of the ellipse (command Points|Conic points) and measure the semi-axes a and b. Then we calculate the Area = πab.



For an arbitrary point P inside the triangle ABC we construct the inellipse and calculate its area.

With the command Advanced|Optimisation we position the point P in the plane so that the area  of the inellipse is maximal.

Once OK Geometry has found the position of P that maximises the area of the inellipse, we observe the configuration, in particular the point P. The Observe command shows that P is the centroid of the triangle ABC. The command Analyse object in reference triangle furthermore shows that the obtained ellipse is the Steiner's inellipse, which touches the sides of the triangle at their midpoint.


Example 3

Problem: Given two intersecting lines and a point P not laying on any of them, find the shortest line segment that passes through P and whose endpoints lie on the two lines.
 


First, we create a construction that satisfies all requirements except optimisation. We denote the intersection of the lines as A and the endpoints of the segment we are looking for as B and C. Note that B is a free point on one of the given lines, while C is a constructed point. We also measure the length of the line segment Len_BC = |BC|. 

We use the  Advanced|Optimisation command to position the point B on one of the lines so that Len_BC reaches its minimum. Then we examine the position of the point P in the triangle ABC.

The advanced query of the point P (which has to be marked as unknown) and likewise the Analyse object in reference triangle applied to the point P in the triangle ABC show that the segment BC is shortest when P is the mirror image of the base P' of the altitude from the vertex A in the midpoint of BC. In other words, when |BP| = |P'C|, where P' is the base of the A-altitude in ABC. This fact can be proved analytically (a synthetic proof of this fact is not known to the author of these lines).

The found property of the solution leads to the next problem: how to construct (by Euclidean tools) the line segment BC for a given pair of lines and a point P. Well, there is no such general construction. The proof is in the figure above. A little calculation shows that the condition |BP| = |P'C| is satisfied if x = |AC| = 1+22/3. The solution to this construction problem is thus equivalent to the Delian problem of doubling the cube, which is known to be unsolvable.


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