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Source: Mathematics Regional Olympiad of Mexico Southeast 2019 P5

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AlanLG

#1 h Oct 23, 2021, 7:31 PM

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Let $n$ a natural number and $A=\{1, 2, 3, \cdots, 2^{n+1}-1\}$ . Prove that if we choose $2n+1$ elements differents of the set $A$ , then among them are three distinct number $a,b$ and $c$ such that

$bc<2a^2<4bc$

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#2 h Oct 24, 2021, 1:14 AM

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$A=\left\{1,2,3 \right\}\cup \left\{4,5,6,7 \right\}\cup \left\{8,9,\cdot \cdot \cdot ,15 \right\}\cup \cdot \cdot \cdot \cup \left\{2^n,2^n+1,\cdot \cdot \cdot ,2^{n+1}-1 \right\}$
The number of the sets is $n,$ so by the principle of drawer , we have $3$ elements $b<a<c$ belongs to the same set.
Then $a<2b,c<2a.$
$\Rightarrow bc<a\cdot 2a=2a^2, 4bc>4\cdot \frac{a}{2}\cdot a=2a^2$
$\Rightarrow bc<2a^2<4bc$

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