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HMMT November, 2017

Combinatorics Difficulty 5.8 AIME, harder Prove it United States

Problem:

Consider five-dimensional Cartesian space
R5={(x1,x2,x3,x4,x5)∣xi∈R} \mathbb{R}^{5}=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right) \mid x_{i} \in \mathbb{R}\right\}
and consider the hyperplanes with the following equations:
- xi=xjx_{i}=x_{j} for every 1≤i<j≤51 \leq i<j \leq 5;
- x1+x2+x3+x4+x5=−1x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=-1;
- x1+x2+x3+x4+x5=0x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=0;
- x1+x2+x3+x4+x5=1x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1.
Into how many regions do these hyperplanes divide R5\mathbb{R}^{5}?

Solution

Solution:

Note that given a set of plane equations Pi(x1,x2,x3,x4,x5)=0P_{i}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=0, for i=1,2,…,ni=1,2, \ldots, n, each region that the planes separate the space into correspond to a nn-tuple of −1-1 and 11, representing the sign of P1,P2,…PnP_{1}, P_{2}, \ldots P_{n} for all points in that region.

Therefore, the first set of planes separate the space into 5!=1205!=120 regions, with each region representing an ordering of the five coordinates by numerical size. Moreover, the next three planes are parallel to each other and perpendicular to all planes in the first set, so these three planes separate each region into 44. Therefore, a total of 4⋅120=4804 \cdot 120=480 regions is created.

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