Suggest a Monte Carlo method of proving Pythaogrean Theorem?

Question

How would it be possible to do a computerized Monte Carlo simulation to prove the theorem, without implicit assumption of the law in the first place?

Drawing an imaginary target with 3 squares forming a right triangle and hitting it randomly and counting the times it lands on the squares is self-referential, i.e. circular.

SAH, your proposal already starts with "rignt triangles with known sides", which makes the proof a circular one. Is there a way to define orthgonality in abstract data space without appealing to Pythagorean theorem?

Alexander, I'm puzzled about something. If the "machine" does not "know" the distances, how does it determine that AB > AC? Image manipulation, so that effectively it's comparing photographs? Let me think on this for a while here.

Also, this is meant to be a "data space" analogue to the 5th Postulate problem. What does it mean to have "Euclidian geometry" in abstract data space? If a computer were to have an abstract sentience but not familiar with Euclidian geometry, where would it start?

Answer 1

Lets formalize the problem:

We have geometry machine, which can scan and write
plastic cards representing points. It can perform the
following operations:

Button '_New point' writes arbitrary point on blank card
(axiom of choice).

Button 'point on _Circle' after scanning two cards A and B
produces new arbitrary point C on circle with diameter AB
(compass available).

Button '_Farther' than scans cards A,B and C, and flashes
green light if AB>AC (distances can be compared)

Button '_Random point' scans cards A and B, and writes
new card X, randomly distributed within circle with
center A and radius AB.

Note that the machine [**edit: does know, but] never tells the distance between any points,

for all we know it might be not even represented
by real numbers [**edit: , but members of some ordered set].

Also, despite the existence of straight lines is implied (diameter) we do not have a ruler, and also cannot
verify if and how many parallel lines are there.

With such a machine in our disposal we cannot prove,
but can verify Pithagorean theorem for this machine:

Produce two _N cards A and B.
Produce third card C=_C(A,B).
The trangle ABC is inscribed into circle with diameter AB,
angle C is right.

Produce many points Xi=_R(A,B), and use _F(A,C,Xi) to
find out how many of them are inside circle of radius
b, this is pi b^2/ pi c^2 = b^2/c^2. Use similar
procedure to produce a^2/c^2.

Check if a^2/c^2 + b^2/c^2 = 1.

If the machine implements Euclidean geometry,
the result of this test will be positive.

[Edit:
All right, it's not what you want. My intention was
simply to point out that Monte-Carlo generator must
replace something specific.
For example it can produce uniformely distributed
points from radius R circle, replacing the fith postulate.
Then we can check if Area(r1)/Area(r2) = r1^2/r2^2. If true,
then given all other axioms Pithagorean thorem is there,
otherwise it is not. If you do not tell us what is removed and
replaced by MCG, then indeed the problem is circular
in itself as stated.

Finally, the described 'machine' tests something independent
of Pithagorean theorem. The test will come out positive
for hyperbolic and sperical cases as well.]

Answer 2

it seems like if you work backwards you can do this.

lets agree that if we have a 3 column table of numbers, we can write a program that looks for the relationships among them.

lets also agree that we will start with a random series of right triangles whose sides are known.

lets construct a program that:
1.randomly selects a triangle and records the sides ordered from smallest to largest.
2.normalizes/scales the sides by dividing the all the sides by the largest side
3. creates a table of the results [each row represents 1 randomly selected right triange and the 3rd column is a 1]
4.since the 1 is a constant, looks for a relationship between the first 2 columns and 1.

to go from sin**2 + cos**2 =1 to pythagorean is simple.