Let the isosceles trapezoid $ABCD$ have $AB=a, CD=b$, and the axis of symmetry intersects the two bases at $E, F, EF=h$. Construct a semicircle (inward) with $BC$ as the diameter, intersecting $EF$ at $P_{1}, P_{2}$ (if tangent, the two points coincide; if separated, there is no solution), which is the desired result. Let $P_{1} E=x_{1}, P_{1} F=x_{2}$, then $\triangle P_{1} E C \sim \triangle B F P_{1}$, so $x_{1} x_{2}=\frac{a b}{4}$. Also, $x_{1}+x_{2}=h$, then $x_{1}, x_{2}$ are the two roots of the equation $x^{2}-h x+\frac{1}{4} a b=0$.

When $h^{2} \geq a b$, there are solutions $x_{1,2}=\frac{h}{2} \pm \frac{1}{2} \sqrt{h^{2}-a b}$. Similarly consider $P_{20}$.