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How many non-congruent triangles with perimeter 11 have integer side lengths?

I failed to solve it. Can anyone help?

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closed as off-topic by heropup, Martin R, Em., user228113, Kamil Jarosz Feb 6 '16 at 9:13

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  • 2
    $\begingroup$ How did you fail. $\endgroup$ – fleablood Feb 6 '16 at 7:46
  • $\begingroup$ You just need find $a + b + c = 11$ with $a < b+ c; b< a+c; c<b+a$ $\endgroup$ – fleablood Feb 6 '16 at 7:51
  • $\begingroup$ Yes I have realized that. But then I am unsure how to proceed. $\endgroup$ – rugi Feb 6 '16 at 7:53
  • $\begingroup$ Just assume $a<b<c$, then proceed from $c<a+b$. $\endgroup$ – Alistair Feb 6 '16 at 8:18
  • $\begingroup$ This is a contest question and has a valid solution which fleablood has demonstrated below. Also math.stackexchange.com/questions/170319/… shows a formula for determining this type of problem. Hence it is surely not off topic on the basis "This question is missing context or other details". $\endgroup$ – rugi Feb 6 '16 at 12:13
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The triangle inequality says two sides must be larger than the third.

Let $a \le b \le c$

So $a + b > 5.5$ and $c < 5.5$ and $c = 11 - a - b$

If $a = 1$, then $b > 4.5$ so $b \ge 5$ and $c \le 5$ so: $a = 1;b=5;c=5$

If $a = 2$ then $3.5 < 4 \le b $ and $c \le 5$ so $a = 2; b= 4; c =5$.

If $a = 3$ then $2.5 < 3 \le b$ and $c \le 5$ so if $b = 3$ then $c = 5$ and if $b=4$ then $c = 4$.

If $a = 4$ then $1.5 < 4 \le b$ and $c \le 3$ which is impossible.

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