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File: 1445824295254.gif - (2035.94 KB, 370x319) Thumbnail displayed, click image for full size.
2084804 No.3439310

Happy Pi Day!

No.3439317

Thank you, you too! Adorable gif BTW.

No.3439667
File: The_Squared_Circle.gif - (10.88 KB, 701x700) Thumbnail displayed, click image for full size.
11140

>>3439317
<3

Have some belated Pi Day puzzles, Lulz. If a circle has a radius r, then what is the sidelength of the square with the same area as the circle? If a sphere has a radius R, then what is the sidelength of the cube with the same volume as the sphere?

Challenge mode: describe how to square a circle using only a compass, straightedge, and a marked cylinder (and a flat surface to mark on and a marking instrument ofc). I've done this before with a soup can for shits and giggles.

No.3439747

>>3439667

> If a circle has a radius r, then what is the sidelength of the square with the same area as the circle?

sqr(pi) x r

>If a sphere has a radius R, then what is the sidelength of the cube with the same volume as the sphere?

cubic root of 4/3 pi r^3

>I've done this before with a soup can for shits and giggles.

No you haven't, because squaring a circle with a straight edge and compass is one of the geometry problems that has been proven impossible. You can only approximate the solution. An exact solution would take as many steps as there are digits in Pi.

No.3439796

Can someone explain me how the wooden stick actually cross the fence post on the lower right of the picture?

No.3439805
File: squaring-the-circle.png - (38.97 KB, 570x533) Thumbnail displayed, click image for full size.
39901

>>3439747

>No you haven't, because squaring a circle with a straight edge and compass is one of the geometry problems that has been proven impossible.

You're thinking of a purely geometric construction, which isn't what he asked for.
construct A̅B̅ = 1 (diameter of cylinder)
construct B̅C̅ = π (rolled circumference of cylinder) [note: this is the step you can't do in geometric construction]
construct midpoint of A̅C̅: D
construct circle centered on D with radius A̅D̅
construct a perpendicular line on A̅D̅ at point B
mark the intersection of this perpendicular and the circle: E
△ACE, △ABE, and △BCE are similar triangles
B̅E̅/A̅B̅ = B̅C̅/B̅E̅
B̅E̅ = √(B̅C̅)=√π
Bisect this length to construct a square with sides (√π)/2 and area π/4, the same as a circle with radius 1/2.

No.3439807
File: calculus_98661_large.jpg - (166.36 KB, 800x450) Thumbnail displayed, click image for full size.
170351

>>3439805
Now you're thinking with calculus

No.3439810

>>3439805

>You're thinking of a purely geometric construction, which isn't what he asked for.
> using only a compass, straightedge

https://en.wikipedia.org/wiki/Squaring_the_circle

>Squaring the circle is a problem proposed by ancient geometers. It is the challenge of constructing a square with the same area as a given circle by using only a finite number of steps with compass and straightedge.
>In 1882, the task was proven to be impossible, as a consequence of the Lindemann–Weierstrass theorem which proves that pi (π) is a transcendental, rather than an algebraic irrational number; that is, it is not the root of any polynomial with rational coefficients.
No.3439812

>>3439810

>using only a compass, straightedge, and a marked cylinder (and a flat surface to mark on and a marking instrument ofc)
>and a marked cylinder

Don't quote the guy if you're going to cut off the important bit. Come back when you're not mad about being wrong.

No.3439814

>>3439812
Well I misunderstood what he meant about a marked cylinder. Sorry about that.

No.3439886
File: 1487026679265.png - (11.49 KB, 313x188) Thumbnail displayed, click image for full size.
11766

>>3439805
I couldn't have explained it better myself. When I did it, I used the radius for A̅B̅ and half the circumference for B̅C̅, then B̅E̅ is the desired side length. My compass wouldn't have opened enough to do the construction your way. Did you make that pic?

No.3439894

https://youtu.be/T7GYcXDpIBE?list=PLT6TE4ef8Z_CkrfyIuWAMbl4EoJa0KDVW

No.3439898

>>3439886

>Did you make that pic?

With this site and over 100 hours in paint.
https://www.math10.com/en/geometry/geogebra/geogebra.html

No.3439946
File: geogebra-export.png - (41.10 KB, 842x597) Thumbnail displayed, click image for full size.
42083

>>3439796
It's an optical illusion. I found the video: https://youtu.be/O-UVkVG6bC4

>>3439898
That's pretty cool. It even calculates area, so you can check yourself at the end.

No.3440092

>>3439965
That, and the stick has that crook in it towards the end he's tilting down. This makes it so he doesn't have to tilt the stick back as much to clear the posts. But it looks like he needs to tilt it back more because of perspective.

No.3440093
File: 17-gon.png - (152.48 KB, 842x597) Thumbnail displayed, click image for full size.
156143

I constructed a regular heptadecagon (17 sides). This is the next regular polygon after the pentagon with a prime number of sides that can be constructed with only a compass and straightedge.

No.3440376
File: 220px-Carl_Friedrich_Gauss_1840_by_Jensen.jpg - (12.32 KB, 220x280) Thumbnail displayed, click image for full size.
12616

Bitches don't know about my heptadecagon.

No.3440421
File: 1486946660256.jpg - (54.99 KB, 563x548) Thumbnail displayed, click image for full size.
56311

>>3440376
https://youtu.be/87uo2TPrsl8
I was following this guy's construction, but he messes up at 6:09. The diagram shown at 6:45 is the corrected version. As long as you're mindful of this, the video's a pretty chill guide.

No.3440498

>>3440421
That makes me wonder. I've only ever heard of square root of negative one being used in a practical sense, do any other negative square roots get used? Like sqrt(-5) or sqrt(-7), etc.

No.3440503

>>3440498
2 x 2 = 4
-2 x -2 = 4
2 x 2 x 2 = 8
-2 x -2 x -2 = -8
sqrt(4) = 2
Cube root of -8 is -2
You can have negative values in odd roots.
Its physically impossible to have a negative of a square root be correct, unless its 0.

No.3440514
File: K0IOrdQ60ixinOdXfI6x9pmEL0vQoIMdnQrRWI1xS9-MLFFsId7_qW8v3y0QW5KSm-n_wV3A0a8XewXFGLAPU3pd.png - (11.96 KB, 512x288) Thumbnail displayed, click image for full size.
12245

>>3440503
What? Yes you can. i = sqrt(-1). https://www.youtube.com/watch?v=rYG1D5lUE4I
What I'm wondering what happens if you have sqrt(-5), or any other negative number.

And... I just figured it out. sqrt(A*B) = sqrt(A)*sqrt(B). Negative 5 is the same as negative 1 times positive 5. So sqrt(-5) = sqrt(-1) * sqrt(5). Or... i*sqrt(5). Wolfram Alpha hath confirmed it! https://www.wolframalpha.com/input/?i=sqrt(-5)

No.3440522

>>3440514

>sqrt(A*B) = sqrt(A)*sqrt(B)

Yes. And this is just a special case of the more general exponentiation property: (ab)^n = (a^n)(b^n). Setting n = 1/2 gives you square roots.

No.3440569

>>3440503

>Cube root of -8 is -2

also 1-i√3 and 1+i√3

No.3440581
File: 631planimtr.jpg - (173.05 KB, 550x470) Thumbnail displayed, click image for full size.
177199

If you can't solve the squaring circle problem with just a straight edge and a compass, could you construct a planimeter out of a straight edge and compass to measure the area of the circle?

After that is should be trivial to take the square root of the resulting line and construct the square.

No.3440593

"How can (1+sqrt(-3))^(1/5) + (1 - sqrt(-3))^*1/5) be real?"

Let z := 1 + sqrt(-3) = 1 + sqrt(3) * i with i being the complex unit. Notice that

(1+sqrt(-3))^(1/5) + (1 - sqrt(-3))^*1/5) = z^(1/5) + (conjugate(z))^(1/5)

with

conjugate(a + b * i) = a - b * i

being the complex conjugate (a,b real numbers).

It can be shown (will do below) that

(conjugate(z))^(1/5) = conjugate(z^1/5)

Noticing that for any real number a, b

(a + b * i) + conjugate(a + b * i)
= (a + b * i) + (a - b * i)
= 2 * a

is a real number, we conclude

(1+sqrt(-3))^(1/5) + (1 - sqrt(-3))^*1/5)
= z^(1/5) + (conjugate(z))^(1/5)
= z^(1/5) + conjugate(z^(1/5))
= w + conjugate(w)

is a real number if we set w := z^(1/5), q.e.d.

Now, what's left to show is that for any complex number z, we have

(conjugate(z))^(1/5) = conjugate(z^(1/5)).

You can write z as a * exp(i * phi)

with a = |z| (the norm of z) and phi = arg(z), both being real numbers. Observe then that

z^(1/5) = a^(1/5) * exp(i * phi/5)

and

conjugate(z^(1/5)) = conjugate(a^(1/5) * exp(i * phi/5))
= a^(1/5) * conjugate(exp(i * phi/5))
= a^(1/5) * exp(-i * phi/5)
= (a * exp(-i * phi)^(1/5)
= (conjugate(z))^1/5

which was to shown.

Yeah, math major. Hope that helps.

No.3440607
File: scaling.png - (25.72 KB, 842x597) Thumbnail displayed, click image for full size.
26337

>>3440569
(-8)^(1/3) = [8(-1)]^(1/3) = [8e^(pi*i)]^(1/3) = [8e^(pi*i - 2pi*i)]^(1/3) = [8e^(pi*i + 2pi*i)]^(1/3)

So you get the roots: 2e^(pi*i/3) = 1 + i√3, 2e^(-pi*i/3) = 1 - i√3, and 2e^(pi*i) = -2.

>>3440581
Once you've found the angle BAE in >>3439805, it's game over. You can use this angle to square any circle through scaling. Pic related.

>>3440593
Using the results above, the expression becomes: (1 + i√3)^(1/5) + (1 - i√3)^(1/5) = [2e^(pi*i/3)]^(1/5) + [2e^(-pi*i/3)]^(1/5). Here, both terms have 5 distinct roots each. Each root of a term has a corresponding root of the other term that is its complex conjugate. This means that the imaginary parts of some of the roots cancel when added, but not for all 25 pairings.

No.3440608

Attention everybody in this thread :

If you all are so smart, what are you doing here on a furry smut board ?

Go be smart somewhere else, thanks.

No.3440609

>>3440606

>Once you've found the angle BAE

But constructing that angle is impossible without rolling the circle open using the marked cylinder, which we don't have because the only tools allowed are a compass and a ruler.

It's a problem of measuring the circumference and plotting down a line of exactly pi times the diameter.

No.3440612
>> 3440607
>> [2e^(pi*i/3)]^(1/5) + [2e^(-pi*i/3)]^(1/5)

This can be written as
2^(1/5) * [exp(pi*i/15) + exp(-pi*i/15)]
= 2^(1/5) * (cos(pi/15) + i * sin(pi/15) + cos(pi/15) - i * sin(pi/15)]
= 2^(1/5) * 2 * cos(pi/15)

which is real.

In fact, since

cos(pi/15) = 1/8 * (sqrt(30 + 6 * sqrt(5)) + sqrt(5) - 1)

The whole thing in the original question is

2^(-9/5) * (sqrt(30 + 6 * sqrt(5)) + sqrt(5) - 1)

No.3440625

>>3440609
Then you can't build a planimeter or whatever. Using the compass or straightedge as anything other than a compass or straightedge is also considered cheating.

>>3440612
That's only for the principle roots. There are 4 other roots you aren't accounting for. For each term. Scroll to the bottom: http://www.wolframalpha.com/input/?i=(1%2Bsqrt(3)i)%5E(1%2F5)

No.3440627

>>3440625
Link is broken; just paste this into wolfram: (1+sqrt(-3))^(1/5)

No.3440628

>>3440625

>Then you can't build a planimeter or whatever. Using the compass or straightedge as anything other than a compass or straightedge is also considered cheating.

True, however it's not against the letter of the question because only compass and straightedge are still being used.

Alternatively, instead of constructing a planimeter by modifying the instruments, could you use a ruler and a compass as they are in some unconventional way to emulate a planimeter?

No.3440630

>>3440625
*principal roots
refering to this: https://en.wikipedia.org/wiki/Principal_value

So for [1 + sqrt(3)i]^(1/5), you get the principal root 2^(1/5)e^(pi*i/15) as well as 2^(1/5)e^[(pi*i/3 + 2pi*i)/5] = 2^(1/5)e^(7pi*i/15), 2^(1/5)e^[(pi*i/3 - 2pi*i)/5] = 2^(1/5)e^(-pi*i/3), 2^(1/5)e^[(pi*i/3 + 4pi*i)/5] = 2^(1/5)e^(13pi*i/15), and 2^(1/5)e^[(pi*i/3 - 4pi*i)/5] = 2^(1/5)e^(-11pi*i/15).

>>3440628

>could you use a ruler and a compass as they are in some unconventional way to emulate a planimeter?

Pretty sure you can't.


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