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I have been trying without success to prove by contradiction the following problem: Given 5 segments $x_1\leq x_2\leq x_3\leq x_4\leq x_5$ each three of which are sides of a triangle. Prove that there exists an acute angled triangle with sides chosen from the five given segments.

If we assume that all are obtuse angled triangles then we have:

$6x_5^2\geq 3x_3^2+3x_2^2+3x_1^2+3x_4^2$

$ 3x_4^2\geq 2x_3^2+2x_1^2+2x_2^2$

$x_3^2\geq x_2^2+x_1^2 $

However, I can't see the contradiction.

Jack D'Aurizio
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parkhyeyoo
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2 Answers2

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You are on the right track. By assuming that all the possible triples are associated with an obtuse triangle, it follows that:

$$ x_3^2 \geq x_1^2+x_2^2,\quad x_4^2\geq x_3^2+x_2^2 \geq x_1^2+2x_2^2,\quad x_5^2\geq x_4^2+x_3^2 \geq 2x_1^2+3x_2^2$$ but the last inequality contradicts: $$ x_5^2 \leq (x_1+x_2)^2 = x_1^2+2x_1 x_2+x_2^2 \leq x_1^2+3x_2^2.$$

Jack D'Aurizio
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    Part of what makes this work is that you sacrifice the symmetry of expressions such as $2x_3^2+2x_1^2+2x_2^2$ in order to reduce one side of each inequality to something in terms of just $x_1$ and $x_2$, which gives you more useful information. – David K May 18 '15 at 14:04
  • Thanks. After making use of all 10 inequalities I obtained $3x_5^2\geq 5x_1^2+5x_2^2$. Can I get a contradiction from this inequality? – parkhyeyoo May 18 '15 at 14:19
  • @parkhyeyoo: nope, since $x_5=\sqrt{10}$ and $x_1=x_2=\sqrt{3}$ are the sides of a triangle. – Jack D'Aurizio May 18 '15 at 14:22
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Suppose that there are no acute triangles. Then

  • $x_1^2+x_2^2\le x_3^2$
  • $x_2^2+x_3^2\le x_4^2$
  • $x_3^2+x_4^2\le x_5^2$

But

$$(x_1+x_2)^2=x_1^2+x_2^2+2x_1x_2\le x_3^2+(x_1^2+x_2^2)\le x_3^2+x_4^2\le x_5^2$$

which contradicts the triangle inequality.

ajotatxe
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  • After making use of all 10 inequalities I obtained $3x_5^2\geq 5x_1^2+5x_2^2$. Can I get a contradiction from this inequality? – parkhyeyoo May 18 '15 at 14:18