In a triangle, the measure of each side is an integer, and the perimeter is given: $k$. Determine the possible smallest and largest values for the sides $a, b, c$, if I. $a<b<c$; II. $a \leqq b \leqq c$.
$$
\begin{aligned}
& a+b+c=k \\
& 0a, m-1>2, m>3, \text{ i.e., } k \geq 9 \text{ in this case. Thus, }

$$
a_{\min }=2, \text { if } k=9,11,13, \ldots
$$

For even $k$, however, $a$ and $(c+b)$, and thus $a$ and $(c-b)$, have the same parity, so from (2a) we have

$$
a \geq (c-b)+2 \geq 3
$$

and for $k=2 m \geq 12$, similarly to the previous case, the triple

$$
a=3, \quad b=m-2, \quad c=m-1,
$$

satisfies the conditions, so

$$
a_{\min }=3, \text { if } k=12,14,16, \ldots
$$

In the triples (5) and (6), we have also found the largest value of $c$, since from (4) $c2, & r>3, & r>2 \\
k=9,12,15, \ldots & k=13,16, \ldots & k=11,14, \ldots
\end{array}
$$

Finally, $b_{\min }$ is obtained when, on the one hand, $a+b$ is the smallest, i.e., $c$ is the largest, and on the other hand, within the value of $(a+b)_{\min }$, $a$ takes its largest value, $a^{*}$. As we have seen, $c_{\text {max }}$ is 1 or 2 less than the value of $(a+b)_{\min }$, and similarly, $a^{*}$ is 1 or 2 less than $b_{\min }$ depending on the parity of $(a+b)_{\min }$, i.e.,

$$
\begin{array}{cc}
c_{\max } \geq (a+b)_{\min }-2, & a^{*} \geq b_{\min }-2, \quad \text { so } \\
k=(a+b)_{\min }+c_{\max } \geq 2(a+b)_{\min }-2=2\left(a^{*}+b_{\min }\right)-2 \geq 4 b_{\min }-6 \\
b_{\min } \leq \frac{k+6}{4}, & b_{\min }=\left[\frac{k+6}{4}\right]
\end{array}
$$

Depending on the four possible remainders of $k: 4$ division, the appropriate triple is:

| For $k=4 s$: | $s$, | $s+1$, | $2 s-1$, | provided | $s>2$, | $k \geq 12$, |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| For $k=4 s+1$: | $s$, | $s+1$, | $2 s$, | provided | $s>1$, | $k \geq 9$, |
| For $k=4 s+2$: | $s$, | $s+2$, | $2 s$, | provided | $s>2$, | $k \geq 14$, |
| For $k=4 s+3$: | $s$, | $s+2$, | $2 s+1$, | provided | $s>1$, | $k \geq 11$. |

In summary, under condition (2),

$$
\begin{aligned}
3-\left(k-2\left[\frac{k}{2}\right]\right) & \leq a \leq\left[\frac{k-3}{3}\right] \\
{\left[\frac{k+6}{4}\right] } & \leq b \leq\left[\frac{k-3}{2}\right], \\
{\left[\frac{k+5}{3}\right] } & \leq c \leq\left[\frac{k-1}{2}\right]
\end{aligned}
$$

For $k=9,11,12$, and 14, there is a unique solution, where the smallest and largest values of each side are equal.

II. Essentially, we obtain the desired values under condition (3) in a similar manner. Instead of (2a),

$$
c-b \geq 0
$$

is used, so from (4) $a \geq 1$. The value $a_{\min }=1$ is obtained for $k=2 m+1(\geq 3)$ with the values $b_{\max }=c_{\max }=m$ in a triangle, and for $k=2 m(\geq 6)$, $a_{\min }=2, b_{\max }=c_{\max }=m-1$; in a common expression,

$$
a_{\min }=2-\left(k-2\left[\frac{k}{2}\right]\right), \quad c_{\max }=b_{\max }=\left[\frac{k-1}{2}\right]
$$

Furthermore, from (3),

$$
\begin{aligned}
k & \geq 3 a, & k & \leq 3 c \\
a_{\max } & =\left[\frac{k}{3}\right], & c_{\min } & =\left[\frac{k+2}{3}\right]
\end{aligned}
$$

Finally, since in the second inequality of (8) $a^{*} \geq b_{\min }-1$ is used, as the value of $b_{\min }-a^{*}$ is 0 or 1, similarly to the above,

$$
\begin{aligned}
& k \geq 2 a^{*}+2 b_{\min }-2 \geq 4 b_{\min }-4, \\
& b_{\min } \leq \frac{k+4}{4}, \quad b_{\min }=\left[\frac{k+4}{4}\right] .
\end{aligned}
$$

The triples that prove the actual occurrence of the found values are:

$$
\begin{array}{l|ll|lll|llll}
k & 2 m+1 & 2 m & 3 r & 3 r+1 & 3 r+2 & 4 s & 4 s+1 & 4 s+2 & 4 s+3 \\
a & 1 & 2 & r & r & r & s & s & s+1 & s+1 \\
b & m & m-1 & r & r & r+1 & s+1 & s+1 & s+1 & s+1 \\
c & m & m-1 & r & r+1 & r+1 & 2 s-1 & 2 s & 2 s & 2 s+1 \\
& k \geq 3 & k \geq 6 & k \geq 3 & k \geq 7 & k \geq 5 & k \geq 8 & k \geq 5 & k \geq 6 & k \geq 3
\end{array}
$$

In summary, under condition (3),

$$
\left.\begin{array}{rl}
2- & \left(k-2\left[\frac{k}{2}\right]\right) \leq a \leq\left[\frac{k}{3}\right] \\
{\left[\frac{k+4}{4}\right] \leq b \leq\left[\frac{k-1}{2}\right]} \\
{\left[\frac{k+2}{3}\right] \leq c \leq\left[\frac{k-1}{2}\right]}
\end{array}\right\} \text { if } k=3, k \leq 5 .
$$

András Kemény (Budapest, Berzenyi D. Gymnasium, III. grade)

Éva Romhányi (Szeged, Radnóti M. Gymnasium, II. grade)