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2023 Poland - Second Round
4
4
Part of
2023 Poland - Second Round
Problems
(1)
System of equations, but no need for solutions?
Source: Polish Math Olympiad 2nd stage 2023 P4
2/11/2023
Given pairwise different real numbers
a
,
b
,
c
,
d
,
e
a,b,c,d,e
a
,
b
,
c
,
d
,
e
such that
{
a
b
+
b
=
a
c
+
a
,
b
c
+
c
=
b
d
+
b
,
c
d
+
d
=
c
e
+
c
,
d
e
+
e
=
d
a
+
d
.
\left\{ \begin{array}{ll} ab + b = ac + a, \\ bc + c = bd + b, \\ cd + d = ce + c, \\ de + e = da + d. \end{array} \right.
⎩
⎨
⎧
ab
+
b
=
a
c
+
a
,
b
c
+
c
=
b
d
+
b
,
c
d
+
d
=
ce
+
c
,
d
e
+
e
=
d
a
+
d
.
Prove that
a
b
c
d
e
=
1
abcde=1
ab
c
d
e
=
1
.
algebra
system of equations