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2002 Stanford Mathematics Tournament
5
5
Part of
2002 Stanford Mathematics Tournament
Problems
(1)
SMT 2002 Algebra #5
Source:
6/27/2012
Solve for
a
,
b
,
c
a, b, c
a
,
b
,
c
given that
a
≤
b
≤
c
a \le b \le c
a
≤
b
≤
c
, and
a
+
b
+
c
=
−
1
a+b+c=-1
a
+
b
+
c
=
−
1
a
b
+
b
c
+
a
c
=
−
4
ab+bc+ac=-4
ab
+
b
c
+
a
c
=
−
4
a
b
c
=
−
2
abc=-2
ab
c
=
−
2