Can you Solve This Putnam 2012 Problem about Acute Triangles?

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SUMMARY

The forum discussion addresses Problem A1 from the 2012 Putnam Competition, which involves demonstrating that among 12 real numbers \(d_1, \ldots, d_{12}\) within the interval (1, 12), there exist indices \(i, j, k\) such that \(d_i, d_j, d_k\) can form the side lengths of an acute triangle. The consensus is that this can be proven relatively easily using properties of triangle inequalities and the characteristics of acute triangles.

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jakncoke1
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Lets take a crack at em.

Prob A1: Let $d_1,...,d_{12}$ be 12 real numbers in the interval (1,12), show that there exists indicides $i,j,k$, such that $d_i,d_j,d_k$ are side lengths of an acute triangle.
 
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This is actually quite easy.
WLOG, Let $d_1 \leq ... \leq d_{12}$ . If we note that for sides a,b,c , the triangle being acute is equivalent to $a^2+b^2 > c^2$ $a \leq b \leq c$

Assume that there exists no $ i < k < j$ such that $(d_j)^2< (d_i)^2 + (d_k)^2.$ There fore, $d_1^2 + d_2^2 \leq d_3^2$ , since $d_k > 1 for all k. d_3^2 > 2$.
Then $d_4^2 > d_3^2 + d_2^2 = d^4 > 2 + 1$
$d_5^2 > d_4^2 + d_3^2 = 3 + 2$
so $d_p ^2 > F(p)$ where F(p) is the pth fibonacci number.
so $d_{12}^2 > F(12)$
or $d_{12}^2 > 144, d_{12} > 12$, which is a contradiction.
 
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