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Any ideas?

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Thread replies: 15
Thread images: 2

File: The-Mystery.png (11KB, 555x455px) Image search: [Google]
The-Mystery.png
11KB, 555x455px
>Image is drawn to scale
>3 very interesting numbers
>Almost a perfect pythagorean triple

Why?
>>
>>8857791
>Almost
It's fucking nothing
>>
>>8857798
The question was "Why?"... Not "what"
>>
>>8857791
Google "phi^2 + e^2"
Now Google "pi^2"

It's off by 0.14, which is too big to be explained by rounding error
>>
>>8857810
>why?
Coincidence, there's nothing special behind this
>>
>>8857826
>>8857791
This is not any type of "error"
>>
File: doge.jpg (30KB, 403x432px) Image search: [Google]
doge.jpg
30KB, 403x432px
>>8857791
Layman here. Hi!

I assume your saying that these numbers, one of which i recognise as Pi, equal a pythagorian triple?

My immediate guess is that its something to do with Modular forms, which i conjecture actually represent the limitations of space and the laws governed within it.

http://www.dailymotion.com/video/x223gx8_bbc-horizon-1996-fermat-s-last-theorem_shortfilms

Go to 14:04

Bye!
>>
>>8857854
Oh I see, the inequality is intentional.
Want to hear your thoughts on why 0=1
>>
>>8857791
using the distance equation. (e^2 - φ^2)^1/2 approximately equals 2.1842669503. Not even close...
>>
>>8857867
lmfao, that is not the distance equation...

it is an addition sign between horizontal and vertical changes, not a subtraction.
>>
>>8857877
fuck me. No wonder my mom says I should kill myself. 3.16339850283
>>
>>8857888
stahp it, you're fine senpai
you can be my special little snowflake instead.
>>
>>8857791
>Why?
probably something to do with the nature of higher dimensions that is bleeding through to our 3d geometry.
>>
Fun fact: if you try to create a real triangle by filling in the missing part (i.e solving the P theorem [math]\phi ^2+e^2 = \pi ^2[/math] ) by adding the missing part to one of the sides ([math]e[/math] for the example), you get

[math]\phi ^2+\left(e-\left(e-\sqrt{\pi ^2-\phi ^2}\right)\right)^2=\pi ^2[/math] , or simplified [math]\phi ^2+(\sqrt{\pi ^2-\phi ^2})^2=\pi ^2[/math] , which is a real and existing triangle minus the [math]e[/math]
>>
>>8858063
>pi = pi
Wow, spooky.
Thread posts: 15
Thread images: 2


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