How many triangles are possible with no side greater than $4$ units?

Question

Consider a triangle having integer sides such that so side is greater than $4$ units.How many such triangles are possible?

I could not solve it by trying to use combinatorics.

So, how to do it?

Thanks fro any help.

The original question came from here: http://www.examrace.com/d/pdf/efaf18b/NSEJS-Solved-Paper-2012.pdf

Answer 1

The solutions in the PDF linked are incorrect. They have calculated $4\times3\times2$ which is only counting scalene triangles including rotations and reflections and then it excludes 7 without even listing them all. I would strong recommend you get official solutions from the Indian Association of Physics Teachers rather than from 3rd party sites.

My solution:

Scalene triangles: $\space^4C_3=4$, however $(1,2,3),(1,2,4),(1,3,4)$ are not allowed.

Isoceles trianges: $\space^4P_2=12$, however $(1,1,2),(1,1,3),(1,1,4),(2,2,4)$ are not allowed. So $8$ more.

Equilateral: $\space^4C_1$, so $4$ more.

Total: $1+8+4=13$ triangles.

List: $(1, 1, 1), (1, 2, 2), (1, 3, 3), (1, 4, 4), (2, 2, 2), (2, 2, 3), (2, 3, 3), (2, 3, 4), (2, 4, 4), (3, 3, 3), (3, 3, 4), (3, 4, 4), (4, 4, 4)$

Answer 2

With numbers this small, just list them by hand. Let the sides be $a,b,c$ with $a \le b \le c$ How many are there with $c=1?$ With $c=2?$ Etc. If the limit were higher, I would solve it by hand for small $c$, then look for a general expression. It should not be more than quadratic in $c$. Then summing over $c$ will be no more than cubic.

Answer 3

More generally, the number of integer triangles with a greatest side length of $c$ is given by

$$n=\left\lceil \frac{c+1}{2}\right\rceil \cdot \left\lfloor\frac{c+1}{2} \right\rfloor $$

Proof

Let $(i, j ,k)$ be a sequence of non negative integers that gives the lengths of the sides of a triangle. We restrict our attention to incongruent triangles and therefore triples where $i\le j\le k$ and $k\le i + j$. The associated triangle has perimeter, $p=i+j+k$.

Let $A$ denote the set of all triangular triples, Let $B$ denote the set of all non-degenerate triples. For both of these sets $\mathscr A$, let $\mathscr A(p)$ denote the subset of triples in $\mathscr A$ with sum equal to $p$.

For each non-negative integer $k$, let $L_\alpha(k)$ denote the number of triangular triples of each type with longest side $k$.

In non-degenerate triples $2k<(i+j)+k=p\le k+k+k=3k$. Thus for a fixed value of $k$, $2k+1\le p \le 3k$; and for a fixed value of $p$, $p/3\le k \le (p-1)/2$. Because $k$ is an integer

$$\lceil p/3\rceil\le k \le \lfloor (p-1)/2 \rfloor$$

Let $f(p, k)$ denote the number of triangular triples (degenerate or not) with perimeter $p$ and largest entry $k$. Clearly the largest value for $j$ is $k$ in which case $i$ takes its smallest possible value, $p-2k$. Since $2i\le i+j=p-k$, $i\le(p-k)/2$; so the largest possible value of $i$ is $\lfloor (p-k)/2\rfloor$. Hence

$$f(p, k)=\lfloor (p-k)/2 \rfloor -(p-2k) + 1 = \lfloor (p-k)/2-(p-2k)+1\rfloor$$ $$=\lfloor (3k-p+2)/2 \rfloor$$

Then

$$L_b(k)=\sum_{p=2k+1}^{3k}\lfloor(3k-p+2)/2 \rfloor=\sum_{m=2}^{k+1}\lfloor m/2 \rfloor$$

If $k=2q$, $L_b(k)=2\{1+2+\cdots+q\}=q(q+1)$. If $k=2q+1$, $L_b(k)=q(q+1)+(q+1)$. Thus for all $k$ $$L_b(k)=\lfloor(k+1)/2\rfloor\cdot \lceil(k+1)/2 \rceil$$