Two circles $\omega_1$ and $\omega_2$, which have centers $O_1$ and $O_2$, respectively, intersect at $A$ and $B$. A line $\ell$ that passes through $B$ cuts to $\omega_1$ again at $C$ and cuts to $\omega_2$ again in $D$, so that points $C, B, D$ appear in that order. The tangents of $\omega_1$ and $\omega_2$ in $C$ and $D$, respectively, intersect in $E$. Line $AE$ intersects again to the circumscribed circumference of the triangle $AO_1O_2$ in $F$. Try that the length of the $EF$ segment is constant, that is, it does not depend on the choice of $\ell$.
Problem
Source: 2016 Peru Cono Sur TST P6
Tags: circles, fixed, Segment, geometry
daniel0105
27.02.2020 17:17
Since we have a spiral similarity between the triangles $AO_1O_2$ and $ACD$, we see that point $A$ is the miquel point of quadrilateral $O_1O_2DC$. Call the intersection point of lines $CO_1$ and $DO_2$ $G$. Obviously, point $G$ is on the circumcircle of triangle $AO_1O_2$. Since $A$ is the miquel point, we also see that $AGDC$ is cyclic. In addition, since the lines $CE$ and $DE$ make right angles with $CG$ and $GD$, respectively, $E$ is on the circumcicle of $AGDC$ and is diametrically opposite to $G$. Now, we will prove that the center of this circle lies on the circumcircle of $AO_1O_2$. This is easy since $\angle GEA = \angle GCA = \angle O_1AC$ which gives $\angle AO_1G = 2\angle GEA$. Call the intersection of $GE$ and the circumcircle of$AO_1O_2$ $H$. Since $\angle GAE = 90^\circ$, $\angle AGH = 90 - \angle GEA$. And since $\angle AO_1G = 2\angle GEA = \angle GHA$, we see that $GHA$ is an isosceles triangle. Thus, $H$ is the center of the larger circumcircle. Now since $GH$ is perpendicular to $FH$, $FG = FE$, and since $FG$ is a diameter of the circumcircle of $AO_1O_2$ which is constant, we are done.