How many ordered triples of integers (x,y,z) satisfy the following system of inequalities?
−x−y−z−x+y+zx−y+zx+y−z≤−2≤2≤2≤2
Answer Choices:
A. 4
B. 8
C. 11
D. 15
E. 17
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First, we rewrite the inequalities to obtain the following:
−x−y−z≤−2,(1)−x+y+z≤2,(2)x−y+z≤2,(3)x+y−z≤2.(4)
We can start by bounding our variables. By adding (2) and (3), we get that
2z=(−x+y+z)+(x−y+z)≤2+2=4⟹z≤2
By symmetry (adding (2), (4) and (3), (4)), we get that x≤2 and y≤2.
By adding (2) and (1), we get that
−2x=(−x+y+z)+(−x−y−z)≤2−2=0⟹x≤0
By symmetry (adding (1), (3) and (1), (4)), we get that y≥0 and z≥0.
Thus
0≤x,y,z≤2⟹x,y,z∈{0,1,2}.
We must still have
x+y+z≥2.
We can note that since our inequalities are symmetric in x,y,z (swapping them won't do much), we can take cases on min(x,y,z).
Case 1. min(x,y,z)=0. In this case, without loss of generality (since we're dealing with symmetry), suppose x=0. Then, our inequalities become
−y−z≤−2y+z≤2−y+z≤2y−z≤2
Note that since y,z∈{0,1,2}, the last two are automatically satisfied. The first two combine to say that y+z=2 (the first becomes y+z≥2), so we have 2 distinct (up to permutation) possibilities: (0,0,2),(0,1,1). Each of these can be re-ordered in 3 ways, for a total of 6 solutions here.
Case 2. min(x,y,z)=1. Without loss of generality suppose x=1. In this case, our inequalities become
y+z≥1y+z≤3−y+z≤1y−z≤1
Note that the first inequality is automatically satisfied as y,z≥1, and the second inequality is satisfied if not both y and z are 2. We note that both possibilities (1,1,1) and (1,2,1) are solutions, for a total of 4 solutions in this case.
Case 3. min(x,y,z)=2. In this case x=y=z=2, and we can verify this works, for an additional solution.
This gives a total of
6+4+1=(C) 11
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