25 January 2021
Improved comparison of expressions
With today's
release,
GeoGebra Discovery supports comparison of algebraic expressions
in an improved way. To illustrate this new feature let us imagine a planar construction
which is invariant of scaling, that is, by dragging its free points similar
figures will be obtained for every position of the free points. In fact,
if only compass-and-straightedge steps are allowed, this is usually the case.
For example, given a square. We want to compare the length of its diagonals
(in fact they are equal long) to the length of the side of the square. It is clear
that for all positions of the two free input points (which designate one side of the square)
the ratio of a diagonal and a side is \(\sqrt2\).
It is very simple to try to get this result
symbolically: Just select the Relation
tool and select segments \(f\) and \(j\), then, by clicking on
More..., you get
the required proportion. The same result can be obtained by typing the command
Relation(f,j). In some cases, however, we are interested in more
complicated relations. Here, for example, a student may ask the relationship between,
say, \(f\cdot g\) and \(j\).
GeoGebra requires some tricks here in its internals. First of all, it is clear that \(f\cdot g\) is
a quadratic quantity, because it is a product of two linear quantities. On the other hand,
\(j\) is still a linear quantity. So any correspondence between the quadratic and linear
quantities must be related to some change of the linear expression, namely, its square
should be observed. When asking
Relation(f·g,j), GeoGebra will compare \(f\cdot g\) and \(j^2\),
by symbolically searching for a matching constant \(K\in\mathbb{R}^+\) such that
\[f\cdot g=K\cdot j^2.\]
Let us mention some possible uses. A simple way to do some more experiments in the
square above is to ask Relation(f^2+g^2,j), or, being a bit
more unusual, Relation(f^3+g^3,j^2).
Or even Relation(f^3+j^3,g^2). (Try it! You can use copy-paste
to enter this formula in GeoGebra.)
Some other experiments can be done in the following way:
In a regular pentagon the product of the lengths
of the diagonals from a given vertex can be expressed with the formula \(f\cdot k\cdot l\cdot j\),
see below:
At the same time we can create the circumcircle of the pentagon and consider its radius \(R\).
Now: what is the relationship between the above product and \(R\)? Can we generalize this result?
Entries on topic GeoGebra
- I restart my blog… (28 September 2020)
- A new GeoGebra version with better angle bisectors… (29 September 2020)
- Better language support… (6 October 2020)
- Detection of perpendicular lines… (19 October 2020)
- Points attached to an algebraic curve… (31 October 2020)
- Explore envelopes easily! (11 November 2020)
- Offsets of a trifolium (17 November 2020)
- Ellipsograph of Archimedes as a simple LEGO construction (18 November 2020)
- Pete-Dőtsch theorem (29 December 2020)
- Comparison improvements (25 January 2021)
- ApplyMap (26 January 2021)
- Discovering geometric inequalities (27 October 2021)
- Proving inequalities (28 October 2021)
- Supporting logic in function calculus (2 November 2021)
- Supporting logic with technology: Part 2 (12 February 2022)
- Difficulty of geometry statements (30 July 2024)
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Zoltán Kovács
Linz School of Education
Johannes Kepler University
Altenberger Strasse 69
A-4040 Linz
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