9. In a math competition, there were three problems: A, B, and C. Among the 25 participating students, each student solved at least one problem; among those who did not solve problem A, the number of students who solved problem B is twice the number of students who solved problem C; the number of students who only solved problem A is one more than the number of students who solved problem A among the remaining students; among the students who only solved one problem, half did not solve problem A. How many students solved only problem B?
9. Let the sets of students who solved problems A, B, and C be $A$, $B$, and $C$ respectively. Then, the three sets $A$, $B$, and $C$ can be divided into 7 parts, with the number of elements in each part being $a$, $b$, $c$, $d$, $e$, $f$, and $g$, as shown in the figure. From the problem, we have
$$
\left\{\begin{array}{l}
a+b+c+d+e+f+g=25 \\
c+d=2(d+g) \\
a=b+e+f+1 \\
a=c+g .
\end{array}\right.
$$

We can derive $3 c+a=26, 3 a+d=26$, then $a=26-3 c \geqslant c, d=26-3 a=9 c-52 \geqslant 0$, which gives $\frac{52}{9} \leqslant c \leqslant \frac{26}{4}$. Therefore, $c=6$.