Show that if you choose any $12$ real numbers between $1$ and $12$, three of them must be the sides of an acute triangle.
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Note three positive real numbers $x \leq y \leq z$ are the sides of an acute triangle if and only if $z \leq \sqrt{x^2 + y^2}$ – ferson2020 Mar 27 '13 at 17:51
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@ferson2020 That should be an inequality, to rule out a right triangle. – Alex Becker Mar 27 '13 at 17:54
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Note that three numbers $a\leq b\leq c$ are the sides of an acute triangle iff $a^2+b^2>c^2$. Suppose no triple of $d_i$ among $1< d_1,\leq \cdots\leq d_{12}< 12$ are the sides of an acute triangle. Then $1<d_1^2\leq \cdots \leq d_{12}^2<144$, and we have $1<d_1^2,1<d_2^2$ and $d_{i}^2+d_{i+1}^2\leq d_{i+2}^2$. But these last three inequalities imply that $d_i^2>F_i$, the $i^{th}$ Fibonacci number, and $F_{12}=144$. Thus we have a contradiction.
Alex Becker
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Can you explain how $d_i^2>F_i$ is implied by those previous three inequalities? – Hillbilly Tim Mar 27 '13 at 19:46
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1@HillbillyTim Well $F_i$ is defined by $F_1=1,F_2=1$ and $F_i+F_{i+1}=F_{i+2}$. The proof of the inequality is then simple induction: if $d_i^2>F_i$ and $d_{i+1}^2>F_{i+1}$ then $$d_{i+2}^2\geq d_i^2+d_{i+1}^2>F_i+F_{i+1}=F_{i+2}.$$ – Alex Becker Mar 27 '13 at 19:53