<s>[INST] Triangle Geometry question find minumum value of $n-m$ In the triangle shown, for $\angle A$ to be the largest angle of the triangle, it must be that $m<x<n$. What is the least possible value of $n-m$, expressed as a common fraction?

I found that $x<4.5$ and $x>1.6$ so I thought the answer was $4.5-1.6=\frac{29}{10}$ but this answer is wrong. Could someone help me see what I did wrong?
 [/INST] By triangle inequality we know that
1) $3x + (x+4) > x+9$
2) $(x+4)+(x+9) > 3x$
3) $(x+9) + 3x > x+4$.
1) gives us $3x > 5$ or $x > \frac 53$.
2) gives us $x < 13$
And 3) gives us $3x > -5$ which is a no brainer as $x > 0$.
The fact that A is the largest angle gives us that $x+9$ is the largest side.  Obviously $x + 9 > x+4$ so that doesn't help us but $x + 9 > 3x$ tells us that $x < \frac 92$.
Putting that together we get $\frac 53 < x < \frac 92$
Since we must have $x > m$ and we can have $x $ as close to $\frac 53$ as we like, the very most that $m$ can be is $\frac 53$.  Likewise the very least $n$ can be as we must have $m > $ and $x$ may be as close to $\frac 92$ as we want.
So the very least $n-m$ can be will occur when $n$ is its least possible value $\frac 92$ and $m$ is its greatest possible value of $\frac 53$ so
$m - n = \frac 92 - \frac 53 = \frac {17}{6}$
You problem was rounding $\frac 53 = 1.6$..... Why in the HECK did you do that????????
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