Space cutting

[Darya]

Hi, everyone! I got this problem I've been stuck with.

N places are located in space such that no 2 are parallel, every 3 cross at exactly 1 point, no 4 cross at the same point.
1. In how many parts do those N planes cut the space?
2. What the formula would be like for (k-1) hyperplanes in R^k space?

I solved a similar problem for lines in 2d space (no two lines are parallel and no 3 intersect at 1 point) Every new line divides the space in n+1 new pieces, so the formula I got is: number of pieces= 1+n(n+1)/2.
How can I apply the same logic for question 1? Having a hard time to imagine all those planes, I created a table:
number of planes 0 1 2 3 4
pieces of space 1 2 4 6 10

Am I doing it right? How could I progress? Any tips?

 

[Cubist]

I agree with your 2d results and formula. But I think your result in 3d for 3 planes isn't correct (I think it's 8 not 6). Imagine that the first plane contains the x and z axes. The second contains the y and z axes. The third plane contains the x and y axes. So they cross at the origin. A cube placed with its centre at the origin would be chopped into 8 by the planes (because each cube vertex would end up in a different space).

BTW: I don't know if you, or a kind moderator, can edit your question but I was very confused about this word for far too long (due to old age probably) until I finally read lower down and found the word "plane" ...

N places are located in space such that no 2 are parallel, every 3 cross at exactly 1 point, no 4 cross at the same point.

EDIT: If the cube vertices were simply at (1,1,1), (1,1,-1), (1,-1,-1), (1,-1,1), (-1,1,1), (-1,1,-1), (-1,-1,-1), (-1,-1,1)

 

[Darya]

I agree with your 2d results and formula. But I think your result in 3d for 3 planes isn't correct (I think it's 8 not 6). Imagine that the first plane contains the x and z axes. The second contains the y and z axes. The third plane contains the x and y axes. So they cross at the origin. A cube placed with its centre at the origin would be chopped into 8 by the planes (because each cube vertex would end up in a different space).

BTW: I don't know if you, or a kind moderator, can edit your question but I was very confused about this word for far too long (due to old age probably) until I finally read lower down and found the word "plane" ...


EDIT: If the cube vertices were simply at (1,1,1), (1,1,-1), (1,-1,-1), (1,-1,1), (-1,1,1), (-1,1,-1), (-1,-1,-1), (-1,-1,1)

Ohhhh, so sorry for the typo!! It's PLANES, guys. Unfortunately, it seems I can't edit it anymore.
Yesss, now I realize it's 8. I guess my imagination is too bad.
What about 4 planes? Is it gonna be 16? Now I see a pattern but I'm not sure I can already go for deriving the formula.
Thanks so much for the answer)

 

[Cubist]

I had a go at drawing 4 planes in a 3d modelling program (blender)...



The red, green, and blue planes are orthogonal. I threw the gray plane in at an angle and offset, and then added white spheres to the different spaces that I could see.

Count the spheres below. The left side is a view from above the gray plane, and the right side is a view from below. Don't forget to count the hidden sphere (not shown) that exists in the middle of the planes (within a tetrahedron)



If you can see anything I missed/ did wrong just let me know and I can edit this. Next step, is there a clever way to extend this to extra planes without having to model/ count? (I can't think of one, not yet anyway)

 

[Dr.Peterson]

Ohhhh, so sorry for the typo!! It's PLANES, guys. Unfortunately, it seems I can't edit it anymore.
Yesss, now I realize it's 8. I guess my imagination is too bad.
What about 4 planes? Is it gonna be 16? Now I see a pattern but I'm not sure I can already go for deriving the formula.
Thanks so much for the answer)

Here's one thought: Suppose you have n planes already, and you add the n+1st. How many regions is that new plane divided into by the other planes? And what does that tell you about the number of new regions?

But the next number after 8 is not 16! Don't jump to conclusions about a formula. In fact, a little thought should make it clear why it isn't 16; or you can apply what you learn from my suggestion above.

 

[Cubist]

Here's one thought: Suppose you have n planes already, and you add the n+1st. How many regions is that new plane divided into by the other planes? And what does that tell you about the number of new regions?

Nice! I got it now.

It helped me to think about this by imagining the new plane being added far away from any already enclosed regions (like the enclosed tetrahedron shape that is formed with n=4). This means that on the far-side of the new plane, all the pre-existing planes are diverging from each other. But then with a bit of extra thought, I realised it doesn't matter where the new plane is added.

 

[Darya]

Here's one thought: Suppose you have n planes already, and you add the n+1st. How many regions is that new plane divided into by the other planes? And what does that tell you about the number of new regions?

But the next number after 8 is not 16! Don't jump to conclusions about a formula. In fact, a little thought should make it clear why it isn't 16; or you can apply what you learn from my suggestion above.

I realized that with adding new planes, a number of pieces of space is growing with 2 parametres: the same linear growth as we've saw with lines and another one because it's +1 dimension. The other parameter shows up after we add the third plane. Then, we see from the model that number of pieces is 15.
Let a number of pieces = [MATH]x_n[/MATH] and number of planes =[MATH]n[/MATH]. [MATH]x_1=x_0+1], x_2=x_1+2, x_3=x_2+3+1, x_4=x_3+4+3[/MATH]. I'm I on the right way?
I have a really bad imagination, so the @Cubist's last explanation did't help me.

 

[Cubist]

I have a really bad imagination, so the @Cubist's last explanation did't help me.

You don't have a bad imagination! It can be hard to give clues (without giving too much away) so I was being vague.

Think about the 2d scenario that you've already answered. Think about how this might relate to intersection lines drawn on the surface of a new plane that has just been added (each line is an intersection with a previous plane, in other words consider the new plane as a 2d space with lines on it). Perhaps re-read post #5 which is a great clue!

 

[Cubist]

@Darya are you still struggling? Refer to the images in post#4. The gray plane has just been added as the 4th plane, and I've emphasised with black lines where the previous 3 planes intersect it...



...can you see this looks like the 2d scenario? You get 7 extra volume spaces with the addition of this 4th plane. Can you generalize?

 

[Darya]

@Darya are you still struggling? Refer to the images in post#4. The gray plane has just been added as the 4th plane, and I've emphasised with black lines where the previous 3 planes intersect it...

View attachment 22046

...can you see this looks like the 2d scenario? You get 7 extra volume spaces with the addition of this 4th plane. Can you generalize?

Thank you so sooo much!! Yesterday I realized the intuition and saw the pattern but was really tired. Today I finished the algebraic expression. Its similar to the problem with lines. I was just confused why it came out with the sum up to (n-1), not n.
In unsimplified form:
[MATH]x_n=1+\Sigma_0^{n-1}(n(n+1)/2+1)[/MATH]The x_0 plus the sum of the formula for lines.
Once more, thanks a loottt! Hope you'll have a nice day!

 

[Cubist]

Well done! Don't forget to answer the second part of the question (and ask if you need help, but you may be able to continue on your own now).