, 2011

To avoid a large case distinction, we will use the notion of oriented angles. Namely, for two lines $\ell$ and $m$, we denote by $\angle(\ell, m)$ the angle by which one may rotate $\ell$ anticlockwise to obtain a line parallel to $m$. Thus, all oriented angles are considered modulo $180^{\circ}$.

Denote by $T$ the point of tangency of $t$ and $\omega$. Let $A' = t_{b} \cap t_{c}$, $B' = t_{a} \cap t_{c}$, $C' = t_{a} \cap t_{b}$. Introduce the point $A''$ on $\omega$ such that $TA = AA''$ ($A'' \neq T$ unless $TA$ is a diameter). Define the points $B''$ and $C''$ in a similar way.

Since the points $C$ and $B$ are the midpoints of $\operatorname{arcs} TC''$ and $TB''$, respectively, we have
$$\begin{aligned} \angle\left(t, B''C''\right) &= \angle\left(t, TC''\right) + \angle\left(TC'', B''C''\right) = 2\angle(t, TC) + 2\angle\left(TC'', BC''\right) \\ &= 2(\angle(t, TC) + \angle(TC, BC)) = 2\angle(t, BC) = \angle\left(t, t_{a}\right) \end{aligned}$$
It follows that $t_{a}$ and $B''C''$ are parallel. Similarly, $t_{b} \parallel A''C''$ and $t_{c} \parallel A''B''$. Thus, either the triangles $A'B'C'$ and $A''B''C''$ are homothetic, or they are translates of each other. Now we will prove that they are in fact homothetic, and that the center $K$ of the homothety belongs to $\omega$. It would then follow that their circumcircles are also homothetic with respect to $K$ and are therefore tangent at this point, as desired.

We need the two following claims.

Claim 1. The point of intersection $X$ of the lines $B''C$ and $BC''$ lies on $t_{a}$.

Proof. Actually, the points $X$ and $T$ are symmetric about the line $BC$, since the lines $CT$ and $CB''$ are symmetric about this line, as are the lines $BT$ and $BC''$.

Claim 2. The point of intersection $I$ of the lines $BB'$ and $CC'$ lies on the circle $\omega$.

Proof. We consider the case that $t$ is not parallel to the sides of $ABC$; the other cases may be regarded as limit cases. Let $D = t \cap BC$, $E = t \cap AC$, and $F = t \cap AB$.

Due to symmetry, the line $DB$ is one of the angle bisectors of the lines $B'D$ and $FD$; analogously, the line $FB$ is one of the angle bisectors of the lines $B'F$ and $DF$. So $B$ is either the incenter or one of the excenters of the triangle $B'DF$. In any case we have $\angle(BD, DF) + \angle(DF, FB) + \angle\left(B'B, B'D\right) = 90^{\circ}$, so
$$\angle\left(B'B, B'C'\right) = \angle\left(B'B, B'D\right) = 90^{\circ} - \angle(BC, DF) - \angle(DF, BA) = 90^{\circ} - \angle(BC, AB).$$
Analogously, we get $\angle\left(C'C, B'C'\right) = 90^{\circ} - \angle(BC, AC)$. Hence,
$$\angle(BI, CI) = \angle\left(B'B, B'C'\right) + \angle\left(B'C', C'C\right) = \angle(BC, AC) - \angle(BC, AB) = \angle(AB, AC),$$
which means exactly that the points $A, B, I, C$ are concyclic.

Now we can complete the proof. Let $K$ be the second intersection point of $B'B''$ and $\omega$. Applying Pascal's theorem to hexagon $KB''CIBC''$ we get that the points $B' = KB'' \cap IB$ and $X = B''C \cap BC''$ are collinear with the intersection point $S$ of $CI$ and $C''K$. So $S = CI \cap B'X = C'$, and the points $C', C'', K$ are collinear. Thus $K$ is the intersection point of $B'B''$ and $C'C''$ which implies that $K$ is the center of the homothety mapping $A'B'C'$ to $A''B''C''$, and it belongs to $\omega$.

Define the points $T, A', B'$, and $C'$ in the same way as in the previous solution. Let $X, Y$, and $Z$ be the symmetric images of $T$ about the lines $BC, CA$, and $AB$, respectively. Note that the projections of $T$ on these lines form a Simson line of $T$ with respect to $ABC$, therefore the points $X, Y, Z$ are also collinear. Moreover, we have $X \in B'C'$, $Y \in C'A'$, $Z \in A'B'$.

Denote $\alpha = \angle(t, TC) = \angle(BT, BC)$. Using the symmetry in the lines $AC$ and $BC$, we get
$$\angle(BC, BX) = \angle(BT, BC) = \alpha \quad \text{and} \quad \angle\left(XC, XC'\right) = \angle(t, TC) = \angle\left(YC, YC'\right) = \alpha.$$
Since $\angle\left(XC, XC'\right) = \angle\left(YC, YC'\right)$, the points $X, Y, C, C'$ lie on some circle $\omega_{c}$. Define the circles $\omega_{a}$ and $\omega_{b}$ analogously. Let $\omega'$ be the circumcircle of triangle $A'B'C'$.

Now, applying Miquel's theorem to the four lines $A'B', A'C', B'C'$, and $XY$, we obtain that the circles $\omega', \omega_{a}, \omega_{b}, \omega_{c}$ intersect at some point $K$. We will show that $K$ lies on $\omega$, and that the tangent lines to $\omega$ and $\omega'$ at this point coincide; this implies the problem statement.

Due to symmetry, we have $XB = TB = ZB$, so the point $B$ is the midpoint of one of the $\operatorname{arcs} XZ$ of circle $\omega_{b}$. Therefore $\angle(KB, KX) = \angle(XZ, XB)$. Analogously, $\angle(KX, KC) = \angle(XC, XY)$. Adding these equalities and using the symmetry in the line $BC$ we get
$$\angle(KB, KC) = \angle(XZ, XB) + \angle(XC, XZ) = \angle(XC, XB) = \angle(TB, TC).$$
Therefore, $K$ lies on $\omega$.

Next, let $k$ be the tangent line to $\omega$ at $K$. We have
$$\begin{aligned} \angle\left(k, KC'\right) &= \angle(k, KC) + \angle\left(KC, KC'\right) = \angle(KB, BC) + \angle\left(XC, XC'\right) \\ &= (\angle(KB, BX) - \angle(BC, BX)) + \alpha = \angle\left(KB', B'X\right) - \alpha + \alpha = \angle\left(KB', B'C'\right) \end{aligned}$$
which means exactly that $k$ is tangent to $\omega'$.