32.57 For a triangle with sides of lengths $a, b, c$ which are all integers, and $a \leqslant b \leqslant c$. If $b=$ 10, then the number of such triangles is
(A) 10.
(B) 55.
(C) $10^{2}$.
(D) infinitely many.
(China Jiangsu Province Suzhou City High School Mathematics Competition, 1990)
[Solution] Consider the general case, i.e., when $b=n$, then $a=k(1 \leqslant k \leqslant n)$, and $b \leqslant c<a+b$,

thus $n \leqslant c<n+k$, at this time, $c$ has exactly $k$ values, i.e., $c=n, n+1, \cdots, n+k-1$, hence we have
\begin{tabular}{c|c|c|c}
\hline$a$ & $b$ & $c$ & Number of triangles \\
\hline 1 & $n$ & $n$ & 1 \\
2 & $n$ & $n, n+1$ & 2 \\
3 & $n$ & $\vdots$ & 3 \\
$\vdots$ & $\vdots$ & $n, n+1, \cdots, n+k-1$ & $\vdots$ \\
$k$ & $n$ & $\vdots$ & $k$ \\
$\vdots$ & $\vdots$ & $n, n+1, \cdots, 2 n-1$ & $n$ \\
$n$ & $n$ & & $n$ \\
\hline
\end{tabular}

Thus, the total number of triangles in this case is $1+2+\cdots+n=\frac{n(n+1)}{2}$.
When $b=10$, the number of triangles is $\frac{1}{2} \cdot 10 \cdot(10+1)=55$, so the answer is $(B)$.