[sugar] [IAEP] Sugar DIgest 2008-09-22

[Edward Cherlin]

I have a proof of the Pythagorean Theorem in Dr. Geo II, and could
easily create several more. (There are more than 200 independent
proofs of the theorem.)

On Thu, Sep 25, 2008 at 9:49 AM, Karl Ramberg <karlramberg at gmail.com> wrote:
> Walter Bender wrote:
>>
>> === Sugar Digest ===
>>
>> 1. Trisecting angles: The French mathematician Évariste Galois
>> published three papers in 1830 that laid the foundations of an
>> algebraic proof of why is it not possible to trisect *every* angle in
>> a compass and straightedge construction, something the Ancient Greeks
>> knew, but could not prove. However, what is often overlooked is that
>> the Greeks could trisect angles, using a different set set of
>> instruments. What does this history lesson have to do with Sugar Labs?
>> Two separate but related discussions have dominated the OLPC-Sur list
>> this past week: the Microsoft announcement regarding a Windows XP
>> pilot in Peru and the lack of a square root function in Turtle Art,
>> both of which can be seen through the lens of abstract
>> algebra—apologies in advance for overreaching with this analogy.
>>
>> Let me summarize the Turtle Art discussion first. Some teachers in
>> Uruguay are teaching the Pythagorean Theorem and were stymied by the
>> lack of a square root function in Turtle Art. They wanted to
>> demonstrate that the length of the diagonal of a square is equal to
>> the square root of the sum of the square of each side. In psuedocode,
>> they wanted to build the following construct:
>>
>> repeat 4 (forward 100 right 90)
>> right 45
>> forward sqrt ((100*100) + (100*100))

Socrates proves this to an uneducated slave boy in the Crito, using
tile dissection.

Draw a square, find the midpoints of the sides, and connect each of
those midpoints to the other three. This divides the original square
into four smaller, equal-sized squares, and each of those smaller
squares into two congruent right triangles. The smaller squares are
the squares on the sides of the right triangles, each square
containing two triangles. The four diagonals of the smaller squares
are the sides of a square of intermediate size. This is the square on
the hypotenuse, and is made of four of the same size right triangles.
The sum of the squares on the sides is the same, four right triangles.

> In Etoys it is pretty straight forward to make this script, look at attached
> picture.
>
> Karl
>
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