# How many non-congruent triangles with perimeter 11 have integer side lengths? [closed]

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

I failed to solve it. Can anyone help?

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

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Martin R, Em., Community, Kamil Jarosz
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• How did you fail. – fleablood Feb 6 '16 at 7:46
• You just need find $a + b + c = 11$ with $a < b+ c; b< a+c; c<b+a$ – fleablood Feb 6 '16 at 7:51
• Yes I have realized that. But then I am unsure how to proceed. – rugi Feb 6 '16 at 7:53
• Just assume $a<b<c$, then proceed from $c<a+b$. – Alistair Feb 6 '16 at 8:18
• 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". – rugi Feb 6 '16 at 12:13

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.