Tucker Circles

Let three equal lines $P_1Q_1$, $P_2Q_2$, and $P_3Q_3$ be drawn Antiparallel to the sides of a triangle so that two (say $P_2Q_2$ and $P_3Q_3$) are on the same side of the third line as $A_2P_2Q_3A_3$. Then $P_2Q_3P_3Q_2$ is an isosceles Trapezoid, i.e., $P_3Q_2$, $P_1Q_3$, and $P_2Q_1$ are parallel to the respective sides. The Midpoints $C_1$, $C_2$, and $C_3$ of the antiparallels are on the respective symmedians and divide them proportionally.

If $T$ divides $KO$ in the same ratio, $TC_1$, $TC_2$, $TC_3$ are parallel to the radii $OA_1$, $OA_2$, and $OA_3$ and equal. Since the antiparallels are perpendicular to the symmedians, they are equal chords of a circle with center $T$ which passes through the six given points. This circle is called the Tucker circle.

If

$$c\equiv {\overline{KC_1}\over\overline{KA_1}}={\overline{KC_... ...{KC_3}\over\overline{KA_3}}={\overline{KT}\over\overline{KO}},$$

then the radius of the Tucker circle is

$$R\sqrt{c^2+(1-c)^2\tan\omega},$$

where $\omega$ is the Brocard Angle.

The Cosine Circle, Lemoine Circle, and Taylor Circle are Tucker circles.

See also Antiparallel, Brocard Angle, Cosine Circle, Lemoine Circle, Taylor Circle

References

Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, pp. 271-277 and 300-301, 1929.