Each side face of a dodecahedron lies in a uniquely determined plane. Those planes cut the space in a finite number of disjoint regions. Find the number of such regions.
Problem
Source: Bundeswettbewerb Mathematik 2016, Round 2 - #4
Tags: geometry, 3D geometry, dodecahedron, combinatorics, combinatorial geometry
08.09.2016 00:31
according to my solution its 195 (just use a generalization of euler)
08.09.2016 00:41
It should be $185$. What exactly are you referring to by Euler?
08.09.2016 00:59
Kezer wrote: It should be $185$. What exactly are you referring to by Euler? He probably means the Euler Characteristic (generalizing Euler's Polyhedron Formula). In this case, we work in $\mathbb{R}^3$ which has Euler Characteristic $-1$ i.e. we have $V-E+F-S=-1$ where $V,E,F,S$ are the numbers of vertices, edges, faces and spaces, respectively. So we only need to determine $V,E,F$ in order to find $S$. Using some geometric observations, we then easily find $V=52, E=300$ and $F=432$ and hence $S=52-300+432+1=185$.
08.09.2016 18:42
Is there any other solution?
08.09.2016 21:57
Yes. Consider the following lemma: Lemma: Suppose there are $n$ planes in space with $n \geq 1$ that divide the space in $r$ regions. Now add a plane $P$. Consider the intersections of $P$ with the other planes. Then we have $k$ disjoint regions on $P$. Now the number of disjoint regions with the $n+1$ planes is $r+k$. Basically, add the planes one by one and we get $185$ after a lot of work.