I've already asked you that question in a previous post.
Why are you so aggressive ?
- As I said this is BS.
- Where did you learn math, Arthur, by the Jain Academy?
- Now here you’re beyond the level of Tautology reaching the area of Stupidology.
- By all due respect Arthur. This is the most stupid and irrational critic I ever got for this derivation. Not even Hush was able to go that far.
- b is not defined twice, Ignorant!
- There is no point to read math books if you don’t understand them.
- (If you know what the word Exact means)
- With that you’re even exceeding yourself in ignorance.
- A VALUE has no unit, moron!
- Are you aware of what you’re saying!
- I guess not. Otherwise you never had posted it.
And finally
- You should try to go over your jealousy and enjoy the beauty of 4b = π
A series of personal attacks ("BS", "Ignorant", "moron", "stupidology", "jealousy")
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Kepler triangle :
The Kepler triangle is defined as a right triangle whose sides are in geometric progression. If the sides are a, ar, ar^2, then Pythagoras gives r = √ϕ
Base = 1, Height = √ϕ and hypothenuse = ϕ is one representation of this triangle.
If i multiply the base by 7 for ex we have :
Base = 7, Height = 7 * √ϕ and hypothenuse = 7 * ϕ
if base = 1/ϕ, height = 1/ϕ * √ϕ, hypothenuse = 1/ϕ * ϕ
so base = 1/ϕ, height = 1 / √ϕ, hypothenuse = 1
so like 1, √ϕ, ϕ
1/ϕ, 1 / √ϕ, 1 is another one.
There is another unique property of the Kepler triangle is :
Hypothenuse * base = height^2
So we can check that with the 3 previous example :
ϕ * 1 = (√ϕ)^2
(7 * ϕ) * 7 = (7 * √ϕ)^2
1 * 1/ϕ = (1 / √ϕ)^2
When the hypothenuse is equal to 1 we have :
1 * base = height^2
with height = b we have the triangle :
Hypothenuse = 1, Height = b, Base = b^2
So there is only one solution to that for b = height = hypothenuse / √ϕ (The ratio of the progression) = 1 / √ϕ
So the triangle, 1, b, b^2 is a Kepler triangle. There is no debate about that.
see
https://en.wikipedia.org/wiki/Kepler_triangle
That's the geometric constraint in your demonstration.
The solution of b for this triangle can be calculate also with the pythagorean theorem.
On the "tool for calculation"
You claim the triangle is just a "tool for calculation", not an additional hypothesis. But a calculation tool must be justified by the problem. You are not deriving the triangle from the square of perimeter; you are imposing it.
If I may illustrate: suppose I define x = π / 4 Then I arbitrarily decide that x and x^3 are the sides of a right triangle with hypothenuse 1. This would give
x^6+x^2=1 , a different value for x, and thus a different π. Would that be a valid proof? No, because the triangle is arbitrary. The same applies here.
The only way your derivation would be valid is if you could prove that the square of perimeter π necessarily gives rise to a right triangle with sides b and b^2 and hypothenuse 1. You have not done that. You simply assumed it.
b is defined twice.
But it is. First, you define b = π / 4
Second, you impose that b and b^2 are the legs of a right triangle with hypothenuse 1. That second condition is not a consequence of the first; it is an independent constraint. So yes, b is being constrained by two separate conditions. That is not an insult; it is a factual observation.
On units and dimensions
Here, you are confusing units (metres, feet) with dimensions (length, area). In pure mathematics, numbers are dimensionless, that is true. But in geometry, a side has dimension L, and an area has dimension L^2. When you write b^2 (an area) and treat it as a length in Pythagoras' theorem, you are mixing quantities of different natures.
The equation (b^2)^2+b^2=1 is not scale-invariant: if you double all lengths, b becomes 2b, the equation becomes 16b^4+4b^2=1, which is not the same. This proves that your construction depends on the arbitrary choice of scale which is not a property of genuine geometry.
On the tautology
You claim it is not a tautology. But a tautology in this context means: the conclusion is already contained in the premises. You chose a geometric constraint that forces b=1/√ϕ, and then you conclude π=4/√ϕ. You are not discovering π; you are constructing a system that gives that result. That is circular reasoning.
I am not attacking you personally. I am pointing out a logical gap in your derivation. If you can demonstrate that the square with perimeter π necessarily implies the existence of a right triangle with sides b,b^2 and hypothenuse 1, then your proof would be valid. Until then, the derivation remains an arbitrary construction, not a proof.
I invite you to respond on the mathematical points, without insults. I am genuinely interested in a rigorous discussion.