$20 \cdot 27$ From a point $P$ outside a circle, a tangent is drawn to the circle, with length $t$. Another secant from $P$ divides the circle into two unequal arcs, of lengths $m$ and $n$. It is known that $t$ is the mean proportional between $m$ and $n$, and the circumference of the circle is $10$, with $m$ and $t$ being integers. Then $t$ can have
(A) no solutions.
(B) one solution.
(C) two solutions.
(D) three solutions.
(E) infinitely many solutions.
(11th American High School Mathematics Examination, 1960)
［Solution］Given $t^{2}=m n$.
Also, $m+n=10$, so $n=10-m$.
Thus, $t^{2}=m(10-m)$ or $t=\sqrt{m(10-m)}$.
When $m=1$, $\quad t=3$;
When $m=2$, $t=4$;
When $m=3,4,6,7$, $t$ is not an integer;
When $m=5$, then $m=n$ which contradicts the problem statement;
When $m=8,9$, by the symmetry of $m(10-m)$, it is the same as when $m=1,2$.
Therefore, the answer is $(C)$.