ITT - We solve AND understand questions that are simple to state and think about but not necessarily easy to solve...

ITT - We solve AND understand questions that are simple to state and think about but not necessarily easy to solve. There's no use in just copying and pasting a proof from somewhere, you need to convince everyone that your proof really is correct.
Hopefully this exercise will help everyone, including the answerer, understand these problems and solutions better (after all, at the risk of sounding cliche, if you can explain something difficult in simple terms, then you really have understood it).
Starting off:
Hilbert's 3rd Problem - Given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces that can be reassembled to yield the second?

Is the sun a solid?

>Given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces that can be reassembled to yield the second?

Yes, with the right tools you could in theory slice up a pacific giant octopus into its constituent atoms and produce a somewhat confused adolescent.

If I can quickly verify the correctness of the answer to a problem, can I also solve that problem quickly?

You might want to look up the definition of polyhedra.
Also, the answer is most definitely no, but it is true for polygons (i.e. the 2 dimensional case).

if P=NP (not likely), and "quickly" means "in polynomial time"

Come on buddy, be a bit more rigorous than that. We're talking about maths, we can afford to not be silly and get the wrong answer as you have.

>responding to shitpost seriously
buddy, have a kek and continue with the thread, responses to shitposts are useless

I genuinely wasn't sure if he was making a genuine answer or not. What makes you think that he really didn't think that?

How would you make a perfect circle from a regular square without changing the area?

Is that a question in response to Hilbert's Third Problem?
Because if it is, a circle is not a polygon.

oops, sorry
I was going to ask how a regular square could make an equilateral triangle but then I found this
pretty cool concept

Ah gotcha.
It has been proven to be true only for 2-d, for 3-d and up it's been proven to be false.

Can a circle be divided chords into equal area pieces such that no two pieces are congruent?

I have proved by exhaustion that this cannot be done with 4 chords or less.

>Can a circle be divided chords into equal area pieces such that no two pieces are congruent?
Had trouble understanding what the question is... could you rephrase it somehow?

Draw any chords on a circle. This will suicide the circle into pieces (the number of pieces well be 1+the number of chords+the number of intersections of the chords). Can chords be chosen such that all of these pieces have the same area, yet none of them are congruent?

I think I get it, he's saying that you can't divide a circle with 4 or less chord lines so that all the pieces have the same area but aren't equal

*split the circle

What is the minimum area of an ellipse that encloses the four given circles?

Yes

Thanks for that user.

Is there a name for this problem? Also, mind sharing your proof if you have time (or give hints, whatever you want)?

I tried and failed to disprove it for like 20 minutes, props

The problem appears to be undiscussed. And just to be clear, I don't know the answer to the problem, I only know such a division does not exist in 4 chords or less. I understand exactly how to construct any equal division of the circle by chords and what property determines whether or not such a division can exist. One issue is iterating through all permutations of chords but the biggest problem is that the method I have can only describe the placement of chords by solving a trigonometric equation which can only be solved numerically. Essentially the method involves finding the interior angles between the radii which touch the endpoints of the chords. So it appears to me that this question, while easy to understand, cannot be answered analytically. I will post the details when I get to my computer.

Equation describing the interior angle of a chord from the area of its division:

y-siny = 2piF

Equation describing the separation between the interior angles of two chords from the area of their middle division:

cos(z/2) cos(x-y/2) csc(x-y/2-z/2) (1-cos(y/2) sec(x-y/2))^2-sin(x) cos(y/2) sec(x-y/2)+x = 2pi F*

The picture should adequately explain what I mean by all of this.

How does a bicycle work?

Diophantine equations would be pretty good for this

Now let me give an example to clarify. This division produces 5 pieces, so each piece must be 1/5 the area of the total circle. Thus C1 and C2 both cuts the circle into 2/5 and 3/5 while C3 cuts the circle into 1/5 and 4/5.

So the interior angle of C1 and C2 is y such that

y-siny = 2pi(2/5)
y1 ~ 2.8248
y2 ~ 2.8248

C1 and C2's middle division is a single piece, so F* = 1/5

So the angle between them is x such that

cos(2.8248/2) cos(x-2.8248/2) csc(x-2.8248/2-2.8248/2) (1-cos(2.8248/2) sec(x-2.8248/2))^2-sin(x) cos(2.8248/2) sec(x-2.8248/2)+x = 2pi/5
x ~ 1.53055473106046

Finally, the interior angle of C3 is y3

y-siny = 2pi(1/5)
y3 ~ 2.11314

To check that this division exist all we have to do is to make sure that all the interior angles sum up to less than 2pi:

2.8248 + 2.8248 - 1.53055473106046 + 2.11314 = 6.23218526893954 < 6.28... = 2pi

It's a tight fit, but the division does exist. And it has congruent pieces since C1 and C2 are congruent. So this does not disprove the theory.