How many different shaped triangles are there whose angles measured in degrees are integers?
I. solution. The question is the same as asking how many ways there are to write 180 as the sum of three positive integers, if the order of the numbers does not matter.

If the order did matter, the number of all possibilities would be $\binom{179}{2}$, since if we write 180 ones next to each other, we need to choose two out of the 179 spaces between the ones to divide the 180 ones into three parts.

For $a+b+c=180$, and $a, b, c$ are distinct positive integers, then these three numbers have $3!=6$ different orders. Therefore, denoting the number of different shaped, non-equilateral triangles by $x$, there are $6x$ such triangles if the order of the angles matters. For $a+a+b=180$, and $a, b$ are distinct positive integers, then these three numbers have 3 different orders. Therefore, denoting the number of isosceles but not equilateral, different shaped triangles by $y$, there are $3y$ such triangles if the order of the angles matters. There is 1 equilateral triangle.

Based on these:

$$
\binom{179}{2}=\frac{179 \cdot 178}{2}=15931=6 x+3 y+1
$$

The number of isosceles but not equilateral triangles is 88, since the size of the angles at the base can be any integer from 1 to 89 except 60. Thus, $y=88$. Substituting this into (1) gives $15931=6 x+3 \cdot 88+1$, from which $x=2611$.

Thus, the number of different shaped triangles is: $x+y+1=2611+88+1=2700$.

Remark. If we wanted to count how many ways there are to break down $6k$ into the sum of three positive integers, then by a similar line of thought we get that

$$
\frac{(6 k-1)(6 k-2)}{2}=6 x+3 y+1, \quad \text { from which } \quad 18 k^{2}-9 k=6 x+3 y
$$

where $x$ is the number of decompositions into three different numbers, and $y$ is the number of decompositions into two different numbers. On the other hand, by the line of thought of the first solution: $3 k-1=y+1$, from which $y=3 k-2$. Substituting this into the previous equation and rearranging gives $x=3 k^{2}-3 k+1$. Finally, the number of sought decompositions is:

$$
x+y+1=3 k^{2}-3 k+1+3 k-2+1=3 k^{2}
$$

II. solution. Let the angles of a triangle be generally $\alpha \leq \beta \leq \gamma$. Count the triangles according to the size of the smallest angle, $\alpha$. (We do not write the "degree" unit.)

We know that $\alpha \leq \beta \leq\left[90-\frac{\alpha}{2}\right]$, because if $\beta$ were larger than this, then $\gamma$ would also be, but then the sum of the three angles would be greater than 180.

Thus, for a given $\alpha=k$, the size of $\beta$ in the triangle has $\left(\left[90-\frac{k}{2}\right]-k\right)+1$ possibilities. Since $\alpha$ is at most 60, the total number of triangles is:

$$
\begin{aligned}
& \sum_{\alpha=1}^{60}\left(\left[90-\frac{\alpha}{2}\right]-\alpha\right)+1=\left(\sum_{\alpha=1}^{60}\left[90-\frac{\alpha}{2}\right]\right)-\sum_{\alpha=1}^{60} \alpha+60= \\
= & 2 \cdot(89+88+\ldots+60)-(1+2+\ldots+60)+60=2 \cdot \frac{30 \cdot 149}{2}-\frac{61 \cdot 60}{2}+60= \\
= & 4470-1830+60=2700 .
\end{aligned}
$$

III. solution. Let the angles of a triangle be generally $\alpha \leq \beta \leq \gamma$. Count the triangles according to the size of the middle angle, $\beta$. (We do not write the "degree" unit.)

$\beta$ is at least 1 and at most $\left[\frac{180-1}{2}\right]=89$.

Case I: $1 \leq \beta \leq 60$, then $1 \leq \alpha \leq \beta$, from which $\gamma$ is already determined. This gives for $\alpha$, and thus for the number of triangles

$$
\sum_{\beta=1}^{60} \beta=1+2+\ldots+59+60=\frac{60(60+1)}{2}=1830
$$

possibilities.

Case II: $61 \leq \beta \leq 89$, then $\beta \leq \gamma \leq 180-(\beta+1)$, since $\alpha$ is at least 1. Since $\alpha$ is already determined from this, the number of $\gamma$ and thus the number of triangles is

$$
\begin{aligned}
& \sum_{\beta=61}^{89}((180-(\beta+1))-\beta)+1=\sum_{\beta=61}^{89} 180-2 \beta= \\
= & (89-61+1) \cdot 180-2 \cdot \frac{(89-61+1) \cdot(89+61)}{2}=5220-4350=870 .
\end{aligned}
$$

This totals $1830+870=2700$ possibilities, so there are this many triangles.

Remark. The triangles can also be counted according to the largest angle, although in a slightly more complicated way.

IV. solution. Let $a_{i}, b_{i}$, and $c_{i}$ denote, respectively, the number of ways to break down the positive integer $i$ into the sum of one, two, or three positive integers, where the order of the terms does not matter.

Clearly, $a_{i}=1$ for all $i$.

It is also true that $b_{2 k}=b_{2 k+1}=k$, since the smaller number is an integer between 1 and $k$ in both cases.

If a number greater than 3 is broken down into the sum of three positive integers: $k=x+y+z$, then $k-3=(x-1)+(y-1)+(z-1)$, where the three terms are non-negative. Thus, we have obtained a decomposition of the number 3 less, which is either a one-term (if exactly two of $x, y$, and $z$ are 1); or a two-term (if exactly one of the three numbers is 1); or a three-term (if all three numbers are greater than 1).

Conversely: from a one-, two-, or three-term decomposition of the number $k-3$, we get a three-term decomposition of $k$.

Based on the above, it is true that:

$$
c_{k}=a_{k-3}+b_{k-3}+c_{k-3}
$$

Let now $k=6 n$. Then $c_{6 n-3 \text { further decomposed, and then in the result the } c_{i} \text { terms always further decomposed }}$ we get that

$$
c_{6 n}=a_{6 n-3}+b_{6 n-3}+a_{6 n-6}+b_{6 n-6}+\ldots+a_{6}+b_{6}+a_{3}+b_{3}+c_{3}
$$

Since 3 can only be decomposed into the sum of three positive integers in one way $(3=1+1+1), c_{3}=1$. Furthermore, $b_{3}=1$, $b_{6 k}=3 k$ and $b_{6 k+3}=3 k+1$. Based on these, arranging the terms:

$$
\begin{aligned}
c_{6 n}= & \left(a_{3}+a_{6}+\ldots+a_{6 n-6}+a_{6 n-3}\right)+\left(b_{3}+b_{6}+\ldots+b_{6 n-6}+b_{6 n-3}\right)+c_{3}= \\
= & (2 n-1)+((1)+(3 \cdot 1)+(3 \cdot 1+1)+(3 \cdot 2)+(3 \cdot 2+1)+\ldots+ \\
& +(3 \cdot(n-1))+(3 \cdot(n-1)+1))+1= \\
= & (2 n-1)+(n \cdot 1+2 \cdot 3 \cdot(1+2+\ldots+(n-1)))+1= \\
= & 2 n-1+n+6 \cdot \frac{n(n-1)}{2}+1=3 n+3 n(n-1)=3 n