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2010 Portugal MO
2
Triangle sides quotient - Portuguese MO, Problem 5, 2010
Triangle sides quotient - Portuguese MO, Problem 5, 2010
Source:
August 28, 2010
geometry proposed
geometry
Problem Statement
Show that any triangle has two sides whose lengths
a
a
a
and
b
b
b
satisfy
5
−
1
2
<
a
b
<
5
+
1
2
\frac{\sqrt{5}-1}{2}<\frac{a}{b}<\frac{\sqrt{5}+1}{2}
2
5
−
1
<
b
a
<
2
5
+
1
.
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