Let $ABC$ be a non-right triangle and let $M$ be the midpoint of $BC$. Let $D$ be a point on $AM$ (D≠A, D≠M). Let ω1 be a circle through $D$ that intersects $BC$ at $B$ and let ω2 be a circle through $D$ that intersects $BC$ at $C$. Let $AB$ intersect ω1 at $B$ and $E$, and let $AC$ intersect ω2 at $C$ and $F$. Prove, that the tangent on ω1 at $E$ and the tangent on ω2 at $F$ intersect on $AM$.
Problem
Source: 2019 Slovenia 2nd TST Problem 3
Tags: Team Selection Test, geometry
14.02.2019 04:56
Solution. The radical axis of $\omega_1$ and $\omega_2$ passes through the midpoint of the common external tangent $BC$, so $AM$ is such a radical axis. Since $A$ must have equal powers respect of both circles, we conclude that $AE\cdot AB=AF\cdot AC$, i.e. $BEFC$ is cyclic. Define $P$ as the intersection point of the tangents to $\omega_1$ and $\omega_2$ through $E$ and $F$, respectively. Let $K$ be the second point where $EF$ meets $\omega_2$. Observe that $$\angle KEA=\angle FEA=\angle FCB=\angle FKC=\angle EKC$$i.e. $AE\parallel KC$, so $(EAF)$ and $\omega_2$ are tangent externally at $F$, so $PF$ touches $(AEF)$. Analogously, $PE$ is tangent to $(AEF)$. In summary, $PE=PF$, so $P$ has equal powers respect of $\omega_1$ and $\omega_2$, thus it lies on $AM$. $\blacksquare$.
19.02.2019 02:52
My proof was similar to above. Following the same logic, we get that $BEFC$ is cyclic. Since triangle $ABC$ is similar to triangle $AFE$, we get AM is actually the symmedian of EAF through A. After that, if we draw a line parallel to BC through A, we notice that this line is actually tangent to (AFE). Since a tangent line, a tangency point and a third point defines a circle, we find that (FDC) can be obtained through a homothety centered at F of (AFE). Similarily for E and (EDB) and (AFE). Now, we finish as above by noticing that a tangent line at F stays tangent throughout the homothety and similarily for E so we just get that the two tangents are also tangents to (AFE) as F and E, thus making them intersect on the Symmedian of EAF through A, so we're done.