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Geometry Expressions Newsletter
March 2013   
Geometry Expressions 3.1 Beta
Geometry Expressions 3.1 Beta program has started.  Sign up for the beta here.
3.1 introduces major enhancements to our app generating and other capabilities, including:
  • Background images on JavaScript apps
  • Variable control via Advance Button, Media Buttons, Random Generation, or Timer.
  • Visibility of geometry objects controlled by mathematical expressions.
  • SVG graphics output.
  • ceil() and floor() functions. 
Use of these controls allows you to create:

How to... Use controlled visibility with Geometry Expressions 3.1 Beta
One use of controlled visibility is to create a single app which illustrates a sequence of configurations.  As an example, let's say we want to illustrate the construction for the circumcircle of a triangle. Here are the steps:
  • Draw a triangle and constrain the coordinates of its vertices, retaining the default values. 
  • Construct perpendicular bisectors of two sides.
  • Select these lines and from the right click context menu select Visibility Condition.
  • Enter v>=1.   
  • Draw the circle centered at the intersection of the bisectors, through one of the points of the triangle.  
  • Set its visibility condition to v>=2.
  • In the Variable Panel, set the bounds of v to be 0 and 2, and set its value to 0. 
  • Use File/Export/HTML JavaScript App.  
  • Set v to have UI Type/Advance Button and put as a label the text you want to see on the button. 

 The resulting app may be viewed here.

Problem of the Month
spline curve   
ABCD is a trapezoid with parallel sides AB and CD.  E and F are the midpoints of AB and CD respectively.
A closed curve is composed of two Cubic Bezier Splines: one with control polygon EBCF, the other with control polygon FDAE.

Show that the radius of curvature is continuous at E and F.

An app which lets you explore this problem is here.

February Problem Solution
ABCDEFG is a regular heptagon.


I is the intersection of BD and CE, J is the intersection of CF and BE, K is the intersection of CG and BF, L is the intersection of AC and BG.

Show that B,C,I,J,K,L are vertices of a regular heptagon.

An interactive illustration is here.

Can you generalize this result?
ngon solution

Defining a generic n-gon in Geometry Expressions, we observe that the angle BLC is 360/n.  Likewise BKC, BJC, BIC.  This tells us that BLKJIC are concyclic. (The diagram indicates how this may readily be proved by hand).

Now the angle subtended by chord LK at the circumference of this circle is angle LCK, which is equal to ACG, which is 180/n.  Hence LK is the side of a regular ngon inscribed in the circle BLKJIC.  Similarly for sides BL, KJ, JI, IC.

One can show in a similar fashion that for given k the intersections of the diagonal from B to the ith point on the ngon with that from C to the i+kth all lie on a regular n-gon of which BD is a diagonal across k points.  (The example above has k=1).

eBook Bundle

11 Geometry Expressions eBooks are now available as a bundle for the recession-friendly price of $85.95
Bundle includes:
  1. The Tortoise and Achilles
  2. Calculus Explorations  
  3. The Farmer and the Mathematician
  4. Developing Geometry Proofs  
  5. 101 Symbolic Geometry Examples
  6. 101 Conic Sections Examples  
  7. Using Symbolic Geometry to Teach Secondary School Mathematics  
  8. Connecting Algebra and Geometry through Technology
  9. Function Transformations
  10. Exploring with Geometry Expressions
  11. The Farmer and the Mathematician II 

Pythagoras proof
Our Euclid's Elements contains over 120 interactive diagrams.
Available on iTunes

Sample chapters of other iPad eBooks are available here

Interactive math Apps which can be viewed on your iPad are available from Euclid's Muse

See our YouTube channel for videos exploring features in Geometry Expressions.

Watch more videos about creating Apps with Geometry Expressions.

Learn how to Export an Animation with Geometry Expressions.
Quick Links
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