Problem 1744

1. Homothety and Initial Setup:
By homothety at $P$, it suffices to consider the case when the radius of $\Gamma$ is equal to the common semiperimeter of the triangles $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, and $\triangle PDA$. Let $K \overset{\text{def}}{=} A_1C_1 \cap B_1D_1$. Denote by $X_{AB}$, the point on $AB$, where the incircle of $ABCD$ is tangent to $AB$. Similarly, define $X_{BC}$, $X_{CD}$, and $X_{DA}$.

2. Excircles and Tangency Points:
Denote by $\omega_{AB}$, $\omega_{BC}$, $\omega_{CD}$, $\omega_{DA}$, the $P$-excircles of $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$, respectively. Then $A_1, B_1, C_1$, and $D_1$ are their tangency points with the rays $PA, PB, PC$, and $PD$, so the excircles are cyclically tangent at those points.

3. Tangency Points on Sides:
Let $T_{AB} \overset{\text{def}}{=} \omega_{AB} \cap AB$. Similarly, define $T_{BC}$, $T_{CD}$, $T_{DA}$. Next, draw circles $\Omega_A \equiv \odot (A, AA_1=AT_{AB}=AT_{DA})$ and similarly $\Omega_B, \Omega_C, \Omega_D$. Note that $\{\Omega_A, \Omega_B\}, \{\Omega_A, \Omega_D\}, \{\Omega_C, \Omega_B\}, \{\Omega_C, \Omega_D\}$ are pairs of tangent circles. Also, $\Omega_A, \Omega_B, \Omega_C, \Omega_D$ are all internally tangent to $\Gamma$ at $A_1, B_1, C_1, D_1$ respectively.

4. **Congruent Triangles and Circle $\gamma$**:
Note that $AX_{AB} = AX_{DA}$ and $AT_{AB} = AT_{DA}$. Combining these relations, we conclude $X_{AB}T_{AB} = X_{DA}T_{DA}$. This implies that $\Delta IX_{AB}T_{AB}$ and $\Delta IX_{DA}T_{DA}$ are congruent. So $IT_{AB} = IT_{DA}$. Hence, we come to the conclusion that $\{T_{AB}, T_{BC}, T_{CD}, T_{DA}\}$ lie on a circle centered at $I$. Call this circle $\gamma$.

5. Cyclic Quadrilateral and Monge's Theorem:
Next, we define the cyclic quadrilateral $T_{AB}T_{BC}T_{CD}T_{DA}$. Write $X \overset{\text{def}}{=} \overline{T_{AB}T_{DA}} \cap \overline{T_{BC}T_{CD}}$ and $Y \overset{\text{def}}{=} \overline{T_{DA}T_{CD}} \cap \overline{T_{BC}T_{AB}}$.

Using Monge's theorem on the triples $\{\Omega_D, \Omega_C, \Omega_B\}$, $\{\Omega_D, \Omega_A, \Omega_B\}$, and $\{\Omega_D, \Gamma, \Omega_B\}$, we have that $B_1, D_1, X$ are collinear and $X$ is the exsimilicenter of the circles $\{\Omega_D, \Omega_B\}$. Similarly, $Y$ lies on $A_1C_1$ and is the exsimilicenter of $\{\Omega_A, \Omega_C\}$.

6. Inversion and Radical Axis:
Consider the inversion $\Phi (X, \sqrt{XT_{AB} \cdot XT_{DA}})$. Let $\gamma_X$ be the circle of inversion. Note that this inversion swaps $\{\Omega_B, \Omega_D\}$ and $\{B_1, D_1\}$ and fixes $\gamma$. Hence, $\Phi$ also fixes $\Gamma, \Omega_A, \Omega_C$ and the tangency points of the last two circles with $\Gamma$. Hence, $C_1, A_1$ are fixed under this inversion and we conclude that $X \equiv \overline{C_1C_1} \cap \overline{A_1A_1}$ and $A_1, C_1 \in \gamma_X$. Similarly, defining $\gamma_Y$, we get that $Y \equiv \overline{B_1B_1} \cap \overline{D_1D_1}$ and $B_1, D_1 \in \gamma_Y$.

7. Final Concurrency:
Now we finish off by showing that $P, I, K$ lie on the radical axis of $\{\gamma_X, \gamma_Y\}$. Note that $P, I$ are centers of two circles, namely $\Gamma$ and $\gamma$ which are orthogonal to both $\gamma_X$ and $\gamma_Y$. So $\overline{PI}$ is the radical axis of $\{\gamma_X, \gamma_Y\}$.

Lastly, note that
$$\operatorname{Pow}_{K,\gamma_X} = KA_1 \cdot KC_1 = KB_1 \cdot KD_1 = \operatorname{Pow}_{K,\gamma_Y}$$
So $K \in \overline{PI}$. We are done. $\blacksquare$