, 2024

Solution:

By the alternate segment theorem we have that:
$$180^\circ = \angle DCY + \angle CYD + \angle YDC = \angle DCY + \angle DXB + \angle YXD = \angle DCY + \angle YXB$$
Therefore opposite angle of $BXYC$ are supplementary and so $CYXB$ cyclic.
One can apply power of a point at $K$ :
$$KT^2 = KB \cdot KC = KX \cdot KY = KD^2 \Longrightarrow KT = KD.$$

Figure 1: The proposer's solution using a new point $Q$
(Alternatively you can sidestep power of a point by observing that $K$ is the radical centre of the incircle $DEF$, the circumcircle $\Gamma$ and the circle $CYXB$ and so $KT^2 = KD^2$.)
Now let $AI$ meet $\Omega$ at $M$, the midpoint of $\widehat{BC}$ not containing $A$. Let the tangent at $M$ meet $KT$ at $Q$. Observe that $QM \parallel KD$ so $\angle TKD = \angle TQM$ and also $KT = KD$, $QT = QM$ hence $\triangle TKD \sim \triangle TQM$. As $T, K, Q$ are collinear, this means that $T, D, M$ are collinear so $TD$ and $AI$ meet at $M$ which lies on $\Omega$.

Solution:

The role of $X$ and $Y$ in this problem is secondary. Draw any circle through $B$ and $C$ which meets the incircle at $X$ and $Y$ and you determine the same point $K$. This is because $K$ is the radical centre of the three circles in play. Therefore $K$ is the intersection of $BC$ with the radical axis of the circumcircle $\Omega$ and the incircle $DEF$, and so is independent of the choice of circle through $B$ and $C$ which gives rise to $X$ and $Y$. In the problem as posed, this is disguised, with angle properties of $X$ and $Y$ giving rise to the circle $BXYC$.
You can now work with the simplified diagram shown in Figure 2.
Somehow we must use a characterization of $M$ in order to finish. In the proposer's solution, we used the tangent to the circumcircle $\Omega$ at $M$ being parallel to $BC$. In this alternative, we use the fact that the internal angle bisector of angle $\angle CTB$ meets $\Omega$ again at $M$.
Triangle $TKD$ is isosceles with apex $K$ so
$$\angle KDT = \angle DTK = \angle DTB + \angle BTK.$$

Figure 2: Illustration of an angle chase
By the alternate segment theorem, $\angle BTK = \angle BCT = \angle DCT$. Now angle $\angle KDT$ is an exterior angle of triangle $DCT$ so $\angle CTD = \angle DTB$.
Therefore the line $TD$ is the internal angle bisector of angle $\angle CTB$ and so must pass through $M$, the midpoint of the arc $\widehat{BC}$ of $\Omega$ which does not contain $T$.

Solution:

The centre of direct enlargement from the incircle to the circumcircle gives another way to finish the proof. This enlargement carries $D$ to $M$ since the tangent lines to their associated circles are parallel. The centre of enlargement therefore lies on the line $MD$. Let $MD$ meet the incircle again at $U$ and the circumcircle again at $V$. Draw the tangent to the incircle at $U$ to meet $BC$ at $W$ so triangle $WDU$ is isosceles with apex $W$ and has equal base angles $\angle WDU$ and $\angle DUW$. The enlargement carries the line $WU$ to the tangent line to the circumcircle at $V$ which meets $BC$ at $S$. Enlargements carry lines to parallel lines, so $\angle DVK' = \angle DUW = \angle WDU = \angle K'DV$. Therefore triangle $K'DV$ is isosceles with apex $K'$.

Figure 3: The three points $K$ on the radical axis.

Figure 4: The enlargement of $W$ gives $K'$.
This identifies $S$ as the intersection of the radical axis of the two circles with $BC$, so $K' = K$ and $V = T$ and the proof is complete.

Solution:

Let $\phi$ be the inversion with center $K$ and radius $KD$. Note that this inversion maps incircle of $ABC$ to itself and $K, X, Y$ are collinear, hence $\phi(X) = Y$. Also $\phi(D) = D$, so $\phi$ maps circle $XBD$ to circle $YDB'$, where $B' := \phi(B)$ is the point on $BC$ different from $B$ such that $\angle BXD = \angle DYB'$, hence $\phi(B) = C$. From $\phi(B) = C$ we get that $KD = KT$ as $KD^2 = KB \cdot KC = KT^2$ since $KD$ is the radius of inversion.
The rest of the solutions is the same as in the other solutions.