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My question is: " Find the total number of non similar triangles which can be formed such that all the angles of the triangles are integers"

My attempt: Let $x$, $y$ and $z$ be the angles of the triangles. So $x+y+z=180$, where $x,y,z \ge1$. Total =$\binom{n-1}{r-1}=\binom{179}{2}=15931$. How to proceed further? We have to subtract the total number of similar triangles from this number.

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  • $\begingroup$ Note that if $x, y, z$ are side lengths of some triangle (say, with $x \leq y \leq z$), then $z < x + y$$, and that the reverse is true too. $\endgroup$ – Travis Willse Mar 5 '16 at 11:59
  • $\begingroup$ A first issue is that in this way you count twice many triangles... : for example if (x=2 and y=3) you generate (2,3,175), that you generate also taking (x=2, y=175)... $\endgroup$ – Jean Marie Mar 5 '16 at 12:00
  • $\begingroup$ yes exactly.. i m thinking same.. $\endgroup$ – Pratyush Mar 5 '16 at 12:02
  • $\begingroup$ I think straight line can also be considered as summation of angles of straight line is also $180$ geometrically so it adds $3$ more solutions so adding $3$ we get $15934$ and this divide by $2$ we get $7967$ i think thats the way its solved $\endgroup$ – Archis Welankar Mar 5 '16 at 12:47
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I'm assuming you measure angles in degrees, that degenerate triangles are forbidden, and that mirror congruent triangles are considered congruent.

The similarity type of a triangle is then determined by its angles $x$, $y$, $z$, whereby $x\leq y\leq z$. It follows that $1\leq x\leq 60$. Given $x$ we can write $y=x+t$ with $t\geq0$, and we have to take care that $$x+t=y\leq z=180-x-(x+t)\ .$$ It follows that $$0\leq t\leq {180-3x\over2}\ ,$$ so that we have $$a(x):=1+\left\lfloor{180-3x\over2}\right\rfloor$$ admissible choices for $t$. For $x=2m$ we obtain $a(2m)=91-3m$, and for $x=2m-1$ we obtain $a(2m-1)=92-3m$. It follows that the total number $N$ of triangles is given by $$N=\sum_{m=1}^{30}(183-6m)=2700\ .$$

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