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I really cannot figure this question out. Can anyone give me a hint please!?

Find an integer $a$, for which $a$, $a+1$ and $a+2$ are the lengths of the sides of an obtuse triangle.

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  • $\begingroup$ hero's formula will be of help $\endgroup$ – SA-255525 Sep 14 '14 at 18:49
  • $\begingroup$ @SA-255525 did you mean Heron's formula :). $\endgroup$ – Varun Iyer Sep 14 '14 at 18:53
  • $\begingroup$ @VarunIyer: Hero of Alexandria aka Heron of Alexandria. Heron's formula is the name more typically used, but Hero's formula also shows up. $\endgroup$ – Semiclassical Sep 14 '14 at 18:55
  • $\begingroup$ @Semiclassical ahh got it. $\endgroup$ – Varun Iyer Sep 14 '14 at 18:57
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From the definition of an obtuse triangle we have that for side lengths $a$, $b$, and $c$:

$$a^2 + b^2 < c^2, c^2 + b^2 < a^2, a^2 + c^2 < b^2$$

We can use any one of them:

So:

$$a^2 + (a+1)^2 < (a+2)^2$$

$$a^2 + a^2 + 2a + 1 < a^2 + 4a + 4$$

$$a^2 - 2a - 3 < 0$$

$$(a-3)(a+1) < 0$$

So now we know that:

$$-1 < a < 3$$

Since $a >0$

$$0 < a < 3$$

Now we must check if values satisfy triangle inequality.

If $a = 1$

$$(1,2,3),\ 1 + 2 = 3$$

So this doesn't satisfy.

If $a = 2$

$$(2,3,4),\ 2 + 3 > 4$$

So this does satisfy.

If $a = 3$

$$(3,4,5),\ 3 + 4 > 5$$

So this does satisfy.

So the values of $a$ that work is $2$ and $3$.

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    $\begingroup$ $a=1$ gives $(1,2,3)$ as the lengths, but that violates the triangle inequality since $1+2=3$ $\endgroup$ – Semiclassical Sep 14 '14 at 18:58
  • $\begingroup$ @Semiclassical woops forgot about that will fix thanks. $\endgroup$ – Varun Iyer Sep 14 '14 at 18:59

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