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2018 Purple Comet Problems
15
Purple Comet 2018 HS problem 15
Purple Comet 2018 HS problem 15
Source:
March 19, 2020
algebra
Problem Statement
Let
a
a
a
and
b
b
b
be real numbers such that
1
a
2
+
3
b
2
=
2018
a
\frac{1}{a^2} +\frac{3}{b^2} = 2018a
a
2
1
+
b
2
3
=
2018
a
and
3
a
2
+
1
b
2
=
290
b
\frac{3}{a^2} +\frac{1}{b^2} = 290b
a
2
3
+
b
2
1
=
290
b
. Then
a
b
b
−
a
=
m
n
\frac{ab}{b-a }= \frac{m}{n}
b
−
a
ab
=
n
m
, where
m
m
m
and
n
n
n
are relatively prime positive integers. Find
m
+
n
m + n
m
+
n
.
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