Hints for construction tasks




Use properties as hints for solving construction tasks
 

In construction tasks it is necessary to specify construction steps (equivalent to a compass and ruler or a square set operations) that lead from given objects to the object to be constructed. Dynamic geometry systems can help solve geometric tasks in a variety of ways, from examining the of loci of points to visually checking constructions by dragging points.

We show here a method specific to OK Geometry that often provides not only useful hints for construction, but also a pleiad of different, often unexpected, solutions to construction tasks.

Here is the general approach:
  1. 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.
  2. Use OK Geometry to observe the configuration.
  3. Choose the observed properties that relate the object(s) to be constructed to the given ones. At this stage, it is very helpful to use the query buttons (magenta buttons with question marks on the right side of the display) - to display only the properties that contain a specified object.
  4. Work out the construction steps.
Notes.
  1. If no satisfactory property is found, it often helps to add  new points or objects to the configuration that can be constructed to the given ones.
  2. Of course the observed properties used in the construction should be proved to validate the construction.
  3. The Plus mode of OK Geometry contains a much more powerful mechanism for relating the objects to be constructed to the initially given ones. See this link.
We illustrate the method with a simple example.


Given is a line k and points A, B on the same side of k. 
How to construct a circle through A and B that touches the line k ?

In general, there are, obviously, two different solutions to the task.

First, you need a configuration to observe. You can draw first the given objects (points A,B and line k) and use the command Circle 3obj to position the circle(s) you are looking for. Or you can start with a circle and then position the points A, B and the line k accordingly.

The obvious strategy for solving this construction task is to find the position of the centre of the circle you are looking for, or the position of a third point on its circumference - the obvious choice is the point of contact of the circle with the line k. We eventually add some object that can be easily constructed from the initially given objects. In the configuration(s) OK Geometry detects properties that lead to different solutions to our construction task. We show some properties and the corresponding constructions. Proving some of them can be a challenging task.

In the following figures, we use the following notation:
D and D' are the points of contact of the circles to be constructed with the line k,
C is the intersection of the lines AB and k,
F is the intersection of k and the bisector of AB,
G and G' are the centres of the circles we are looking for,
E is the midpoint of GG'.

For each construction we present
- the property that is the clue for the construction,
- the construction steps,
- an animation of the construction.

We also provide the OK Geometry files for the construction. Constructions are implemented as scenes. To view the scenes use the green squares button, use also the N button to (un)hide transparent objects.


Here are some constructions obtained from the properties.

Construction 1


 

Triangles ACD and DCB are similar.
Consequently, |CD|2 = |CA|·|CB| .

The construction is based on a well known property of right triangles.
 

Steps:

  1. Let C be the intersection of line AB and k.
  2. Draw a semicircle kBC with BC as diameter.
  3. Draw the line p through A and perpendicular to AB.
  4. Let P be the intesection point of p and kBC.
  5. Draw the circle k(C,P).
  6. Let D and D' be the intersections of k(C,P) and the line k.
  7. The circle through A,B,D (and A,B,D') is the solution.


Animation
 


Construction 2

 
|CF|2 = |FA|2 + |CD|2

The construction is based on Pythagoras' theorem for right triangles.
 
Steps:

  1. Let p be the bisector of AB.

  2. Let F be the intersection of lines p and k.

  3. Draw a semicircle kCF with CF as diameter.

  4. Let H be a point on kCF so that |FA|=|FH|

  5. Draw the circle k(C,H).

  6. Let D and D' be the intersection points of k and k(C,H).

  7. The circles through A,B,D and A,B,D' are the sought circles.

    Animation

     

Construction 3

|CA|·|CB| is the power of C wrt. the sought circle.
 

The construction is based on power theorem. In our case the power of the point C is the same for all circles passing through A,B and equals the square of the tangent segment from C to the circle.
Steps:

  1. Let C be the intersection of line AB and k.

  2. Draw the circle with diameter AB. Let P be its centre.

  3. Draw the circle with PC as diameter.

  4. Let Q be one of the intersections of circles with diameters AB and PC.

  5. Draw the circle k(C,Q).

  6. Let D and D' be the two intersections of the line k and the circle k(C,Q).

  7. The circles through A,B,D and A,B,D' touch the line k.



Construction 4

 
The line FA is tangent to the circle through D,A,D'.
Also, |CD| = |CD'| .

 

Steps:

  1. Let C be the intersection of line k and line AB. Let F be the intersection of line k and bisector of AB.

  2. Let p be the line through A that is perpendicular to FA.

  3. Let q be the line through C, perpendicular to k.

  4. Let G be the intersection of p and q.

  5. Draw the circle k(G,A).

  6. The circle k(G,A) intersects the line k in points D and D'.

  7. The circles through A,B,D and throught A,B,D' touch the line k.

    Animation
     


Construction 5

 
The centre K of the circle through G,B,G' lays on the tangent to the circle through ABF.
The line EC is perpendicular to k.

Steps:
  1. Let C be the intersection of line k and AB. Let F be the intersection of k and bisector p of AB.
  2. Let E be the intersection of p and the perpendicular to k through C.
  3. Draw the line m through E that is parallel to AB.
  4. Draw the tangent n to circle ABF through B. Note that ∢(n,AB) = ∢BFA.
  5. Let K be the intersection of m and n.
  6. Let G and G' be the intersections of the circle k(K,B) with p.
  7. The circles k(G,B) and k(G',B) are the tangent to k and pass through A nad B.

 



Download figures (png)

Download constructions (p)