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Given side lengths of a triangle $a, b, c$, judge if this is a acute or obtuse triangle.

One idea came into my mind is using cosine formula, but I wonder if we can do this without using trigonometry. Thank you.

JSCB
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2 Answers2

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The border case between acute and obtuse is right, and a right triangle satisfies the Pythagorean theorem. Thus, denoting the two shorter sides by $a$ and $b$ and the longest one by $c$, the criterion is

$$a^2+b^2\lessgtr c^2\;.$$

joriki
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  • This answer came immediately to mind, but my justification was an appeal to the cosine law that the OP referred to, and wanted to avoid. Is there another way to see this inequality? – Richard Sullivan Sep 24 '12 at 12:23
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    @Richard: I'm not sure I understand the question. Isn't what I wrote in the answer enough justification? To continuously deform an obtuse triangle into an acute triangle, you have to go through a right triangle, so in the space of triples of side lengths the side lengths of right triangles form the boundary between the side lengths of obtuse triangles and the side lengths of acute triangles. It only remains to decide which ones are on which side of the boundary, and it seems clear enough that the obtuse ones have $\lt$ and the acute ones have $\gt$. – joriki Sep 24 '12 at 12:29
  • This is the content of Propositions 12 and 13 in Book II of Euclid's Elements. No cosines there! (Of course, his statements are equivalent to the cosine law...) – Per Erik Manne Sep 24 '12 at 12:31
  • @joriki Yes, I suppose that's enough. Maybe I'm a bit slow this morning. – Richard Sullivan Sep 24 '12 at 12:43
  • @jasoncube: You're welcome! – joriki Sep 25 '12 at 08:03
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Let $\,a,b,c\,$ be the triangle's sides' lengths. If there exists a relation of the form

$$a^2>b^2+c^2$$

then you know the angle in front of the side with length $\,a\,$ has to be greater than $\,90^\circ\,$...can you see why?

DonAntonio
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