QUIZ 2. 4 Plane Dividers
First, the maximum number of areas you can separate with \(3\) planes are \(8\), by placing all planes orthogonally.
Caption
Modeling provided by figuro.io
Note that the entire topology is not different even if you tweak some angles unless you place two planes in parallel, as shown below.
However you cannot put \(4\) planes in \(3\)-dimensional space orthogonally. (This can be proved by linear algebra.) So now your goal is how to add one more plane to divide these \(8\) already separated areas as much as you can.
This might be non-intuitive at the first. Now let me give two very strong observations.
- If the new cube passes \(n\) of the \(8\) areas, the number of new areas will be \(2n + (8-n) = n+8\).
- The maximum possible \(n\) is \(7\).
Why the first observation is correct? Because any plane would divide any convex area into \(2\) areas, and all of \(8\) already separated areas are convex.
Why the second observation is correct? If the new plane is able to divide all of \(8\) areas, then this means you can pick up \(8\) points from the plane, where each point has positive/negative x/y/z coordinate.
Any convex space that contains these \(8\) points essentially create \(3\) dimensional space. Because if you put any \(3\) different points to create a plane, there should be a completely opposite sign triples that can't be included on that created plane. I believe there is a mathematically strict proof using dot product(which defines a plane in multi-dimensional space) or matrix dimension, but I am not covering that here.
However, dividing \(7\) areas is possible, as shown below.
Therefore, the answer is \(15\). Generalizing this, I believe the answer for \(n\)-dimensional space with \(n+1\) of \((n-1)\)-dimensional hyperplanes are \(2^{n+1} - 1\).
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