/sci/ - Science & Math

let A be the circle with diameter a
let B be the circle with diameter b
let C be the circle with diameter a+b

[(A+B)/2] - [B/2] + [A/2] = the shaded area

>>8239030
You should define A, B, and C as the areas of the circles of their respective letters, not just the circles

>>8239030
Incorrect.

>>8239060
is L the lattice spacing and is it constant for all neighbors.
And then, does "compact support" for V mean that it's zero at V(j) for some j>J?
Also, there seem to be an index _0 missing for the one-particle term, otherwise you got H=<H>N+...

>>8239030
This is the best approach, but just for the sake of fucking around, suppose we wanted to do it like this:

Let $S$ be the shaded region. [math] \displaystyle A = \iint_S 1 \ dA [/math], and by Green's theorem, we can transform this integral into [math]\displaystyle \frac{1}{2} \int_{\partial S}(xdy - ydx)[/math]. Now the line integral can be done piecewise by parametrizing the three half-circle paths, but is there an easy way of doing this? Choosing the origin will make it easy to do one of those half-circle paths, but what about the remaining two?

(2pi(a^2) + pi2ab)/2

>>8239018
π/8(3a+b)(a+b)

>>8239076
It's correct

although it should read

area of [(A+B)/2] - [B/2] + [A/2] = the shaded area

>>8239128
Then tell me what the area would be when a = 1 and b = 2

>>8239143
3/4pi

>>8239143
area of A ≈ 0.79
area of B ≈ 3.14
area of C ≈ 7.07

(7.07/2) - (3.14/2) + (0.79/2) = 2.36

>>8239149
Very well, but you lose points for not having the solution in terms of a and b.

>>8239018
>Give me your hardest "find the area of the shaded region" problems.
provides an elementary school level problem
A(a) = (1/2)*pi*(((a+b)^2)+(a^2)-(b^2)) = pi*((a^2)+ab)
A(b) = (1/2)*pi*(((a+b)^2)-(a^2)+(b^2)) = pi*(ab+(b^2))

Look at it as 4 half circles and its easy.

>>8239158
Kek

>>8239175
>>Give me your hardest "find the area of the shaded region" problems.
>provides an elementary school level problem
THEN POST HARDER ONES.

>>8239158
(3/4pi)*(a/a)*(b/b)

Alright /sci/ let's get serious.

a*(a+b)*pi/4

>>8239018
(πab)/4

1/2 * pi * (a/2)^2 + ((a+b)/2)^2 * pi * 1/2 - (b/2) ^2 *pi * 1/2

1/2 * pi * [ (a/2)^2 + ((a+b)/2)^2 - (b/2) ^2 ]

1/2 * pi * [ a^2 * 1/4 + (a^2 + 2ab + b^2) * 1/4 - b^2 * 1/4 ]

1/8 * pi * [ a^2 + a^2 + 2ab + b^2 - b^2 ]

1/8 * pi * [2a^2 + 2ab ]

pi/4 * [ a^2 + ab ]

pi/4 * a * (a+b)

>>8241038

If /sci/ can't solve this introductory problem you are all nothing but popsci memers who should never discuss QM.

>>8241105
Mathfag here and there are about 20 terms that I do not know the definition for.

This is why math threads are way better when it is an actual complex problem that can be described with elementary terms and not your physics technobabble that could be solved by a toddler with a dictionary.

>>8239018
>>8239060
Take your pedophile cartoons back to >>>/a/.

>>8241127

Then grab a dictionary and let us see your "solution" you petulant brainlet.

Not all mathematical terms are elementary, either. Are you in high school?

>>8241038
>>8241105
>QM

Fuck off. This is a math thread.

>>8241105
Solved it in my head in like 4 minutes. Would've been quicker but I had a phone call from my terminally ill Gran and couldn't stop thinking about the Hodge conjecture.

something like this?

>>8239077
[math]L[/math] is the lattice [math]\left(a\mathbb{Z}/\Lambda\mathbb{Z}\right)^n[/math] where [math]a[/math] is the lattice spacing and [math]\Lambda[/math] is the lattice size. Compact support means the interaction term is zero on all but a finite subset of the lattice as [math]\Lambda \rightarrow \infty[/math].
There are no indices missing. Good job assuming the existence of a number operator you dunce.

>>8239018
(pi/8) * (2a^4 - b^4 + a^2 + b^2 + 2ab)

>>8239018
(Pi/8)(2a^2+2ab)

>>8241375
Right.
>>8241580
This simplifies to the right answer.
>>8241558
You fucked up anon.

It takes me some time...

>>8241626
Fuck, I forgot to square it.

>>8241626
>>8241631
typical engineer
>how can i integrate this...

>>8241520
>two dimensional phase space
I thought physicists liked hard problems.

>>8242549
[math]x[/math] and [math]p[/math] are the two canonical coordinates on the symplectic manifold [math]M[/math] whose lagrangian manifold is the phase space [math]D[/math]. Symplectic manifolds are [math]2n[/math] dimensional where[math]n \in \mathbb{N}[/math].
If you need me to BTFO you again just ring me up.

>>8241520
>>8242581
Take your pedophile cartoons back to >>>/a/.

>>8239060
the proof is trivial

>>8242581
Verbose.

>>8241127
What a load of bs

>>8239018
pi(((A+B)/2)^2)/2 + pi((B)^2)/2 - pi((A)^2)/2

>>8239018
pi(((A+B)/2)^2)/2 - pi((B)^2)/2 + pi((A)^2)/2

Using only chords, divide a circle into equal area pieces with no two pieces congruent. Pic related almost does it except for that small intersection on the left

>>8242839
>Pic related almost does
How the hell do you figure

>>8242880
What do you mean? The structure is fairly simple to understand once you realize each piece has definite area since they divide the circle equally. So each chord is defined by how it divides the circle, and the separation of intersecting chords is defined by the area of its intersections.

>>8239018
>>8239175
I'm in college and I don't even know how to start this problem and how you got to that answer.

Could anyone provide a good explanation or resource? I'd appreciate it.

>>8242939
you have the big circle, diameter is a+b (radius is diameter halved)
on the left you have a smaller circle diameter of a
you have on the right half circle with a diameter of b

area of circle is r^2*pi


do you get it now?

>>8242917
Yeah but what specifically did you do to ensure the pieces were near equal? Just guess and check with different random chords?

I guess what I'm wondering is whether this is a problem with analytic solutions or an exercise in numerical optimization.

>>8242939
Simply plug it into your favorite symbolic geometry software package.

>>8242785
He just means x and p are vectors.

>>8242992
Good question. If we start with the property that the division must be equal then we can see that the number of possibilities for some number of chords is finite. There are only so many ways the chords can intersect. However the problem does not appear to be analytically solvable as the equations that govern the position of a chord based on the areas it divides the circle into can only be solved numerically.

Alright smartasses try this one on for size.

Find the volume of the intersection of eight spheres of radius [math]r[/math] centered at the corners of a cube of side length [math]r[/math].

>>8242581
Yeah but Poincare Bendixon only works in two dimensions. More than that and things get more complicated.

>>8243490
that's what makes the problem so difficult

>>8243493
Darn. I guess I'd need to read up on the differential geometry of symplectic manifolds then. Or just hamiltonian vector fields.

>>8242939
You can't solve it without more information. That's the point of the "you should be able to solve this" meme.

>>8239175
>Look at it as 4 half circles and its easy.
You don't know that that's what they are though.

>>8241631
not defined for all a,b; consider x =1 if [a,b]=[-1,1].

>>8242939

Don't fret. A lot of people know the answer or find it easy, so the answers without explanation make sense to them. Here's how it works (you can only solve this is you assume the line shown is dividing the circle):

The top half of the circle has a diameter of a+b. The area of a circle is pi * r^2. This is a half circle (so, half the area of a full circle. The radius is half the diameter, or 1/2 * (a+b) in this case. Therefore, the area is

pi/2 * ((1/2)(a+b))^2 = pi/2 * (1/2)^2 * (a+b)^2 = pi/2 * 1/4 * (a+b)^2 = pi/8 * (a+b)^2.

Now, we only want the shaded area. All we need to do is subtract the non-shaded region, which is simply half a circle with diameter b. So, let's apply a similar process to the top portion.

Area of non-shaded region of top part (another half circle or semicircle) is pi/2 * (b/2)^2 = pi/2 * b^2 * 1/4 = pi/8 * b^2

Okay. Now we need to look at the bottom half. We want the shaded region of the bottom half, which is simple half of the circle defined with diameter a.

Area of shaded part of bottom half (also a semicircle) = pi/2 * (a/2)^2 = pi/8 * a^2

- We know the whole area of the top half (both shaded and non-shaded) is:

pi/8 * (a+b)^2

- We know the area of the top half that is non-shaded (and thus needs to be subtracted from the area of the top half to leave only the shaded area):

pi/8 * (a+b)^2 - pi/8 * b^2

- We know the area of the bottom half that IS shaded and thus needs to be added:

[pi/8 * (a+b)^2] - [pi/8 * b^2] + [pi/8 * a^2]

The above is the final formula.


---

How do I use TeX here?

>>8242663
Don't you ever grow tired of this? You must be in every second thread on /sci/

>>8239018
>Shaded area= [(piAB^2)/2]-(Bpi^2)+[(Api^2)/2]

Does this look correct?

>>8242772
Underrated meme

this is the hardest one.
bet you cant do it

>>8244675
why do you enable /a/utism on /sci/ence ?

>>8244907
kek

>>8244915
So much irony

>>8244920
back to >>>/a/

>>8244926
u fukin pedophile stop stealing my meme fucking cancer go back >>>/a/

>>8244926
Not an argument
>>>/asp/

>>8244929
s-sorry senpai

>>8239018
impossible with calculus I

>>8245027
fuck, I just realized I'm a cargo culter
lmao, area=integration in my brain

>>8242839
do you know if it's possible?
how about we proof that first

>>8243371
Okay looks like no one is attempting this might as well post the answer because I thought it looked cool.

>>8239018

>>8245432
>Subtle homework help post

>>8245432
the anti derivative of 1/x is ln(x) therfore the limit does not exist.

The area is "infinite".

>>8239018
This is easy.
Let A be the area of the entire circle, so A = pi * (a+b)^2.
Call the areas of the grey and white regions G and W. Then
G = 0.5 * pi * b^2 + (0.5 * A - 0.5 * pi * a^2)
and
W = 0.5 * pi * a^2 + (0.5 * A - 0.5 * pi * b^2).
More simply,
G = pi * (b^2 + a * b)
and
W = pi * (a^2 + a * b).
As expected, G + W = A.

for given a, b:
grey semicircle area = pi/2 * a^2
white semicircle area = pi/2 * b^2
large semicircle area = pi/2 * (a + b)^2 = pi/2 * (a^2 + 2ab + b^2)

grey area = grey semicircle + (large semicircle - white semicircle)
= pi/2 * (a^2 + (a^2 + 2ab + b^2 - b^2))
= pi/2 * (2a^2 + 2ab)
= 2pi * a * (a + b)

>>8245498
Fuck me, I'm retarded. Replace a and b with a/2 and b/2 obviously.

Result is pi/2 * a * (a + b)

>>8245498
>>8245501
And I'm even more retarded because (assuming a, b are the radii of the smaller semicircles):
pi/2 * (2a^2 + 2ab) = pi * (a^2 + 2ab) = pi * a * (a + b)

Since a, b should be the diameters rather than radii, replace a <= a/2 and b <= b/2
pi * a/2 * (a/2 + b/2) = pi * a/2 * (a + b)/2 = pi/4 * a * (a + b)

Should be easy

>>8245557
seems so, only tricky part is the integration necessary for the upper right portion of the shaded region, unless there's an easier method I am missing.

>>8244915
I'm more bored of your repetitive posts than manga scribbles.

>>8245557
[eqn]90-25 \pi +\frac{25}{2}\ \sin ^{-1}\left(\frac{3}{5}\right)[/eqn]

>>8245883
I'm sure there is. This is a sixth grade problem for Chinese kids, so no calc probably

Find the area of the region in terms of R or r given that: the radius of the big circle is R, the radius of the small circle is r, and the center of the small circle intersects the big circle.

>>8245976
Really?

I don't know what I'm missing.

>>8246102
[eqn]
\begin{cases}
\pi R^2 & R\leq \frac{r}{2} \\
\frac{1}{2} \left(-r \sqrt{4 R^2-r^2}-2 R^2 \tan ^{-1}\left(\frac{r \sqrt{4 R^2-r^2}}{r^2-2 R^2}\right)+2 r^2 \cot ^{-1}\left(\frac{r}{\sqrt{4 R^2-r^2}}\right)+2 \pi R^2\right) & \frac{r}{2}<R<\frac{r}{\sqrt{2}} \\
\frac{1}{2} r \left(\pi r+r-2 \sqrt{2} R\right) & R=\frac{r}{\sqrt{2}} \\
-\frac{1}{2} r \sqrt{4 R^2-r^2}-R^2 \tan ^{-1}\left(\frac{r \sqrt{4 R^2-r^2}}{r^2-2 R^2}\right)+r^2 \cot ^{-1}\left(\frac{r}{\sqrt{4 R^2-r^2}}\right) & \text{otherwise}
\end{cases}
[/eqn]

>>8239018
Why are people saying this is impossible without calculus?

It's literally just difference and sums of areas of semicircles.

[math] \frac{1}{2} \pi ( \frac{a+b}{2} )^2 - \frac{1}{2} \pi ( \frac{b}{2} )^2 + \frac{1}{2} \pi ( \frac{a}{2} )^2 [/math]

>>8246116
One can use the formula for the area of a circle cut by a chord.

http://mathworld.wolfram.com/CircularSegment.html

The Chinese kids probably learned that.

>>8245557
Oh boy, hope you guys are ready for this.
As someone pointed out, only the upper right shaded area takes work. I won't simply calculations with the hope that the work is clearer.
Call the area of the right circle above the cord A. Noting that the line forms an angle of a = arctan(1/2), we see that
A = ((2*(pi/2 - a))/(2*pi))*(pi*5^2) - 2*(0.5*5*cos(a)*5*sin(a))
= 17.67871794...
Let S = 25*(1 - pi/4), the area of the lower right shaded region. Let T be the area of the shaded upper right region.
Then
T = A + 2*S - 25
= 3.40880977.

>>8246102
Assuming no circle completely overlaps the other:
Find the two points of intersection, draw a line between them (the chord), and just add the two areas "above" the cord.

>>8246284
The area of the entire rectangle minus the area of the two circles is ~42.

You'd expect the area of the shaded region to be a little less than half of that, no?

>>8246230
Because /sci/ is infested with unenlightened engineer half-wits.

>>8246304
Using my answer, we get that the shaded area is about 19: the "easy" shaded regions are about 5 and the smaller one is about 3.
Maybe you misread my post.

>>8246313
Yeah I did my bad.

>>8245557

>>8245557
Not exaggerating when I say you should be able to do this in your head.
I didn't write anything down I just used my browser calc to get an approximate answer.
it's 18.7776607

>>8246952
protip: you can divide the shaded area into 3 and a half equal portions. You can calculate a single portion by subtracting 1/4 of the area of a circle from a square.

>>8246953
Not that easy

>>8246956
Is that easy

>>8246970
no, you gigantic retard, that top right shaded area isn't evenly divided into two pieces

>>8246929
>Primary 6 question

>>8246979
Meant for >>8245557

>>8242839
Take a circle, divide it in half exactly by a horizontal chord. Divide the upper semi-circle in equal-area halves with a horizontal chord. Now just divide the lower semi-circle into two pieces of equal area with a chord that isn't horizontal.

>>8239018
are all you niggers trolling again?

ratio of a and b is not defined, you cant fucking solve this.

>>8239018
pi/4 * (a+ab+b)^2

>>8247631
pi/4 * (a+ab-b)^2**

>>8247634
pi/8 * (a+ab-b)^2*******

>>8247627
It doesn't need to be you fucking mong.

Shaded area
[eqn]= \pi \left[ \left(\frac{a+b}{2} \right)^2 + \left(\frac{a}{2} \right)^2 - \left(\frac{b}{2} \right)^2 \right] \\
= \frac{ \pi}{4}(a^2 + 2ab + b^2 + a^2 - b^2) \\
= \frac{ \pi}{4}(2a^2 + 2ab) \\
= \frac{a(a + b) \pi}{2}[/eqn]

>>8247627
>>8247642
Oops. Let's try again.

Shaded area
[eqn]= \frac{ \pi}{2} \left[ \left(\frac{a+b}{2} \right)^2 + \left(\frac{a}{2} \right)^2 - \left(\frac{b}{2} \right)^2 \right] \\
= \frac{ \pi}{8}(a^2 + 2ab + b^2 + a^2 - b^2) \\
= \frac{ \pi}{8}(2a^2 + 2ab) \\
= \frac{a(a + b) \pi}{4}[/eqn]

>>8246952
your approximate answer is off senpai

Here's a problem I gave my Calc1 students.

Evaluate the following exactly: [math]\displaystyle \int_0^{\pi} \frac{x \sin(x)}{1-\cos^2(x)} dx [/math]

Protip: Even wolframalpha can't do this; ask it and it'll give you a decimal approximation. But you can find an exact answer using nothing more than Calc1.

Here's a problem I gave my Calc1 students.

Evaluate the following exactly: [math]\displaystyle \int_0^{\pi} \frac{x \sin(x)}{1+\cos^2(x)} dx [/math]

Protip: Even wolframalpha can't do this; ask it and it'll give you a decimal approximation. But you can find an exact answer using nothing more than Calc1.

>>8250283
[eqn]\frac{\pi ^2}{4}[/eqn]

>>8250299
Clever lad.

>>8242939
the area of the lower half-circle with radius a/2
+ the entire half of the large circle with radius a+b
- the upper half-circle with radius b/2

>>8250375
(a+b)/2 I meant obv

>>8239018
Assume r1 is a/2 and r2 is (a+b)/2.
Then the shaded area A is equal to:
A = pi*r1*r2

>>8247619
That will result in the topmost piece being the same exact shape as the bottom-most piece.

>>8245073
Proving its possible would be just as difficult as finding a solution.

>>8250283
>>8250299
I'm stumped. How do you do this one?

>>8251050
One cannot find the indefinite integral, but one can express the given definite integral in terms of itself and solve for it.

Try the u-substitution [math]u=\pi-x[/math], consider what it does to the bounds (and identities on sin, cos under this), and use properties of integrals to transform the given integral in an expression involving itself.

>>8241038
I don't know these fags or what they're talking about

>>8241520
I don't know these fags either

try this

>>8245557
Has anyone gotten this yet?

>>8252748
Not thinking of this question and coming back to it later sure cleared my mind. I know how to calculate it now.

Ratio of 3 sides of triangle is [math]1:2:√5[/math]
Smallest angle of triangle = [math]\tan^{-1}{0.5}[/math]
Height of triangle inscribed in >>8246929 is [math]5 \left( \frac{2}{√5} \right)^2 = 4[/math]

Top right area: [math]0.5(5)(10) - 0.5(5)(4) - (0.5)(5)(5)2 \tan^{-1}{0.5} = 25 - 10 - 25 \tan^{-1}{0.5} = 15 - 25 \tan^{-1}{0.5}[/math]

Remaining area: [math]3/4 (100 - 25 \pi) = 75 - 18.75 \pi[/math]

Total area = [math] 90 - 18.75 \pi - 25 \tan^{-1}{0.5}[/math]

>>8252433
>the challenge question is easier than the main question
Shiggy diggy doo

>>8252433
>>8252811
Anyway:
From point D, by tangent-secant theorem, and letting length of AB = x,
8 + √[8(8 + 0.25x)] = 8 + √(64 + 2x) = 0.5x
x - 16 = √(256 + 8x)
x^2 - 32x + 256 = 256 + 8x
x^2 = 40x
x = 40 since x > 0

Challenge question:
Let radius of circle E = y
Shaded area = 0.5π(2y)^2 - 2(0.5πy^2) = 2πy^2 - πy^2 = πy^2 = area of circle E.

>>8252433
is DEF an equilateral triangle?

No one even wants to give a shot at >>8242839 ?

>>8252885

>>8239018
A is 1/2 of B

>>8252934
Is that supposed to be a serious answer?

>>8241067
We've got a winner.

>>8239018
If you can't do this you might as well take up women's studies and give up on ever achieving a STEM degree because you clearly lack the talent.

>>8254788
Same area as the cross-section of your mouth :^)

try this on for size bitches

>>8255531
TPO = 8
Let TP = r

r + PO = 8
r + √(2r^2) = 8
r + (√2)r = 8
(√2 + 1)r = 8
r = 8/(√2 + 1) = 8(√2 - 1)/(2 - 1) = 8(√2 - 1)

Shaded area = {0.25π(8)^2 - [8(√2 - 1)]^2 - 0.75π[8(√2 - 1)]^2}/2 = [16π - 64(3 - 2√2) - 48π(3 - 2√2)]/2 = 8π - 96 + 64√2 - 72π + 48π√2 = (48π + 64)√2 - 64π - 96

>>8245557
https://www.youtube.com/watch?v=xnE_sO7PbBs

>>8244915
But anon science and autism go together really well

>>8245557
I think I've got the algorithm for finding the the white area in the upper right corner:
1. We draw a diameter on the second (right) circle, so that it crosses the upper and lower side of the rectangle
2. The diameter will intersect the line from the 2 opposite angles AC (lower left and upper right) in point Q
3. We find the length of AC (it's trivial)
4. We find the length of QC, it's part of the rectangular triangle QCF
5. QCF is similar to ABC, so QC/AC = FC/AB
6. We find the area of the trapezoid in red
7. Find the area of the segment created by line RY
8. Find the area of triangle ROQ, it's similar to triangle QFC
9. Find the area of the black space in the lower right, little cube area (F1)B(R1)O minus the sector (F1)O(R1)
Now subtract all the areas we just found from the area of the bigger cube (A1)BC(A2) - [ (F1)BCQ + (A1)(F1)OR + area of sector RY + the black area ]
10. SOLUTION: Find the area of triangle ABC then subtract the area we found in step 9

I haven't explained everything rigorously, because it would be too much writing for trivial stuff.

>>8255531
Solution for shaded area:
1. Find radius for the smaller circle
2. Find the area of the smaller circle
3. Find the area of the quadrant of the bigger circle
4. Construct rectangle ODPC and find it's area
5. Subtract the area of the quadrant of the smaller circle from the one of the rectangle in the previous step
6. SOLUTION: From the bigger quadrant, subtract the area from the previous step and the area of the smaller circle, whatever number you get, divide by 2

This one was way too easy.

>>8239079
Nigga just learned greens theorem

>>8239018
Had to dig through the /sci/ archive for a copy of this picture. It's my favorite variant of this meme.


Also, what the fuck, I was just looking at a 2012 thread and it was full of pictures of Mochizuki. How long have we been on this IUTeich thing, I could have sworn it was at most a couple years ago.

>>8256101
>>8255557
HAH clever, but try this then!

>>8256852
>AM-GM inequality question
Here's your (You)

>>8242772
>Math meme words
Let's get this started.

>trivial
>Clearly,
>(Why?)

Find the area of the Mandelbrot Set. You should be able to do this.

>>8254788
As my arousal approaches its maximum level, the area tends to infinity.

>>8256071
>"I have't explained everything rigorously because it would be too much writing for trivial stuff"
>proceeds to write more than the "rigorous" answers in this thread
how about you just admit you don't know how to do it

This thread is baby tier right. Genuinly curious since i'm medfag.

>>8257507
It is.

>>8256101
I know, right? The challenge questions are way easier than the main questions for some reason.

>>8257307
> implying

>>8245557
You do realize there is a full circle on both sides of the diagonal, right? Then we'll just calculate the piece in the lower right corner and in the middle, subtract them from the area of the triangle - the area of the circle, divide by two and then subtract this area from triangle - circle to get the black region's total area. No need for calc, no need for fancy methods.

>>8257851
you're missing the left small piece

>>8257857
No, I take both the small pieces at once and use symmetry so that I can divide by two to get the area of a small piece.

>>8257851
>>8257862
And this seems to give 7/8*(triangle - circle)

Prove me wrong

The following is an excellent exercise in Calculus I and II (derivatives, integrals, and optimization).

Given a ketchup cup of base radius [eqn]R[/eqn] and side length [eqn]L[/eqn], find the optimum angle [eqn]\theta[/eqn] to fan out your ketchup cup that maximizes the volume [eqn]V[/eqn] of [eqn]sweet delicious ketchup[/eqn].

>>8258197
To clarify, find the equation for V, then optimize theta. Don't optimize the cross-sectional area.

>>8258197
b-b-b-but senpai. ketchup being a semisolid can be piled higher than the rim of the ketchup cup.

>>8258211
Then just use vodka.

This is the answer

>>8258222

That'll make my fries soggy

>>8241067
>this

>>8242839
>>8252885

I think doing something like this would work

hoperylfetttly i dint' misread somethings

>>8260536
This is the proper way to cut a pizza into three equal pieces.

>>8260536
ok never mind I just realized these 2 pieces would be congruent

>>8242839
Well I still haven't found an answer yet but it cannot be done for 2 to 7 pieces and I'm fairly certain it cannot be done for any even number of pieces and less certain that it can be done for any number of pieces at all

>>8260536
>>8260588
first divide the pizza in half. the first half should be divided using a cut parallel to the the first cut. the second half should be cut at an angle.

challenge is probably more about making sure the cuts are even unfortunately

>>8242839
Ok I don't know if your still in the thread but I came up with a super weak "proof" on why it can't be done

Consider the area of the circle bounded by the chord and the minor arc, a "cap". The cap must always encompass an area that is an integer multiple of 1 over the total number pieces you want. eg if you want 9 pieces, it must be 1/9 2/9 3/9 etc. A piece that is m/9 must then be split into m pieces of area 1/9, for example. If you want n pieces you obviously cannot have 2 1/n pieces, as they will be congruent.

I think a good way to think about it is to consider the number of intersections required to make the n pieces, which will have to a maximum of n/2 - 1 intersections. 0 intersections never works as you'll be left with at least 2 1/n area caps. In fact, the only 2 solutions with only one 1/n cap are the maximum intersections and 2nd to maximum intersections solutions. The maximum intersection solution won't work as it will always have rotational symmetry, so you always need to use the next solution with 1 area 1/n cap.

Eventually it comes down to that you can't 2 caps that encompass the same area (or else they will Only be able to be split congruently), and the fact that purely different area'd caps won't be able to split each other into equal areas. I don't have a formal proof yet but this is essentially the gist of it. Sorry if this didn't make much sense but I haven't really done any circle theorems or formal proofs for a while now

>>8260623
This won't work as explained in >>8260632. You can never have 2 isolated "end caps" as they will have equal area and will be congruent

>>8260632
Also the idea behind the fact that it cannot be split into an even number of pieces is that you will always end with at least two 1/n area pieces when trying to do so. Forming pieces with only intersecting chords in the remaining area will always create an even number of pieces

dam

>>8260791
Side of square = diameter of circle = CD = √(BE*BD) = √s
Let G be the midpoint of AD.
Then CFG and CGD are congruent right angled triangles with right angle at F and D respectively.
∠FCD = ∠FCG + ∠GCD = 2 arcsin 1/√5
sin ∠FCD = sin (2 arcsin 1/√5) = 2 sin (arcsin 1/√5) cos (arcsin 1/√5) = 2(1/√5)(2/√5) = 4/5
sin ∠BCF = cos ∠FCD = √[1 - (4/5)^2] = √(9/25) = 3/5
A(DEF) = 0.5A(DBF) = 0.5[A(DCBF) - A(DCB)] = 0.5[0.5S sin BCF + 0.5S sin FCD - 0.5S] = 0.25S(3/5 + 4/5 - 1) = 0.25S(2/5) = 0.1S

>>8260895
>diameter of circle = CD
no

>>8260904
Sorry, diameter of semicircle.

Though come to think of I don't know why I even included it when I didn't even use it in the equation. But whatever.

>>8241067
OK, now can someone explain, in English why this is the solution i.e. As in the rationale behind the 7 expressions in this post?

>and now you're fucked

This is why it's so hard to find good math teachers.

hey can someone point out where my algebra went wrong?
compared to
>>8241067
>>8241067
can't find it : (


radius of entire circle = (a+b)/2

area of entire circle = π((a+b)/2)^2

half of that is a semi circle
(1/8)(a + b)^2

subtracted out the area of the semi circle b, and added in a

(1/8)π(a+b)^2 + π(a/2)^2 - π(b/2)^2

(1/8)π [(a+b)^2 + 2a^2 - 2b^2]

(1/8)π [ 3a^2 + 2ab - b^2 ]

>>8258197
[eqn] \theta_{1,2} = \frac{\pi}{2} + \sin ^{-1} \left ( -\frac{R}{2L} \pm \sqrt{\left ( \frac{R}{2L} \right )^2+2}~ \right )
[/eqn]
probably made a mistake along the way and am too lazy to check which one of them is a minimum and which is a maximum

>>8255557
PO = sqrt(2r^2) -- im missing something?

>>8261704
just looked at it again and found a mistake.
also it's probably the solution with the + since for [math] R \rightarrow 0 [/math] it gives the correct solution of a 45° triangle.
so it should be
[eqn]
\theta = \frac{\pi}{2} + \sin ^{-1} \left ( -\frac{R}{2L} + \sqrt{\left ( \frac{R}{2L} \right )^2+\frac{1}{2}}~ \right )
[/eqn]

>>8261759
Pythagora's theorem nigga.

>>8261772
i mixed up the r with the wrong radius haha, my bad

>>8258205
I let θ' = θ + π/2.

I got V as
(π/3)(3lr^2 cos θ' + 3l^2r sin θ' cos θ' + l^3 sin^2 θ' cos θ')

and dV/dθ' as
(π/3)[3l^2r - (3lr^2 + l^3) sin θ' - 6l^2r sin^2 θ' - 2l^3 sin^3 θ']
or
(-lπ/3)[-3lr + (3r^2 + l^2) sin θ' + 6lr sin^2 θ' + 2l^2 sin^3 θ']

I want to know if there's any mistake in my working.

>>8254788
-1/12