How many triangles are there in which the lengths of the sides are whole numbers not greater than 10?
Let the sides of the triangle be $a \geq b \geq c$. We need $b+c > a$, so $b$ must always be greater than $\frac{a}{2}$. The smallest such integer is $\left[\frac{a}{2}\right]+1$. Therefore, the possible values of $b$ are $\left[\frac{a}{2}\right]+1, \left[\frac{a}{2}\right]+2, \ldots, a$; $c$ can range from $a-b+1$ to $b$, so for a given $b$, it can take $2b-a$ different values. Thus, the number of all such integer triangles with the longest side $a$ is:

$$
N_{a}=\left(2\left[\frac{a}{2}\right]-a+2\right)+\left(2\left[\frac{a}{2}\right]-a+4\right)+\left(2\left[\frac{a}{2}\right]-a+6\right)+\ldots+a
$$

This is an arithmetic series, with the number of terms being the number of different possible values of $b$: $a-\left[\frac{a}{2}\right]$, so

$$
N_{a}=\frac{\left(a-\left[\frac{a}{2}\right]\right)\left\{\left(2\left[\frac{a}{2}\right]-a+2\right)+a\right\}}{2}=\left(a-\left[\frac{a}{2}\right]\right)\left(\left[\frac{a}{2}\right]+1\right)
$$

If $a$ is even: $\quad a=2n \quad N_{2n}=n(n+1)$,

if $a$ is odd: $a=2n+1 \quad N_{2n+1}=(n+1)^{2}$.

The number of integer triangles whose sides are no longer than 10 units:

$$
\begin{gathered}
S_{10}=N_{1}+N_{2}+N_{3}+\ldots+N_{10}= \\
=1+2+4+6+9+12+16+20+25+30=125 . \\
\left(\text { Generally } S_{2n}=N_{1}+N_{3}+N_{5}+\ldots+N_{2n-1}+N_{2}+N_{4}+N_{6}+\ldots+N_{2n}=\right. \\
=1^{2}+2^{2}+\ldots+n^{2}+1^{2}+1+2^{2}+2+3^{2}+3 \ldots+n^{2}+n= \\
=2 \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)}{6}(4n+2+3)= \\
=\frac{n(n+1)(4n+5)}{6} \\
\text { and } S_{2n+1}=S_{2n}+N_{2n+1}=\frac{n(n+1)(4n+5)}{6}+(n+1)^{2}= \\
\left.=\frac{n+1}{6}\left(4n^{2}+5n+6n+6\right)=\frac{(n+1)(n+2)(4n+3)}{6}\right) .
\end{gathered}
$$

# Both problems were solved by: Gehér L., Kővári T. 

Gacsányi Sándor (VIII. grade)

Only 112 was solved by: Bognár J., Csernók L., Gősy S., Jankó B., Párkány M., Tarnay Gy. Tarnóczy T., Vörös M.