Let $\Gamma, \Gamma_1, \Gamma_2$ be mutually tangent circles. The three circles are also tangent to a line $l$. Let $\Gamma, \Gamma_1$ be tangent to each other at $B_1$, $\Gamma, \Gamma_2$ be tangent to each other at $B_2$, $\Gamma_1, \Gamma_2$ be tangent to each other at $C$. $\Gamma, \Gamma_1, \Gamma_2$ are tangent to $l$ at $A, A_1, A_2$ respectively, where $A$ is between $A_1,A_2$. Let $D_1 = A_1C \cap A_2B_2, D_2 = A_2C \cap A_1B_1$. Prove that $D_1D_2$ is parallel to $l$.
Problem
Source: China TST 2023 Problem 16
Tags: geometry
khina
29.03.2023 21:25
Let $A_1B_1 \cap A_2B_2 = X$. By homotheties we may see that $X$ is the top point of $\Gamma$. By Ceva it suffices to prove that $CX$ is the median of $CA_1A_2$, which is readily equivalent to $X$ being on the radical axis of $\Gamma_1, \Gamma_2$. But since $AX$ is a diameter of $\Gamma$, we have that an inversion from $X$ sends $A_i$ to $B_i$ for $i \in \{1, 2\}$, so thus $A_1A_2B_1B_2$ is cyclic and radical axis on $(A_1A_2B_1B_2), \Gamma_1, \Gamma_2$ finishes.
CANBANKAN
31.03.2023 00:43
Here is a solution by 几何大神杨皓哲. Since he is a new user, he can't post images and latex, so I will post for him.
Here is an alternative solution that I smh did not notice to use Ceva.
Define $A'$, $X_1$, $X_2$ as the point diametrically of $A$ in $\Gamma$, $A_1$ in $\Gamma_1$, $A_2$ in $\Gamma_2$, respectively. By homothety, $A_1, B_1, A'$ are collinear and $A_2, B_2, A'$ are collinear. Moreover $\triangle A'B_2A \sim \triangle A'AA_2$ from two right angles. Therefore, $A'B_2 \cdot A'A_2 = A'A^2 = A'B_1 \cdot A'A_1$, meaning $A_1, B_1, A_2, B_2$ are concyclic.
Define $D_1' = A_1C \cap \overline{AB_1X_1}$ and $D_2' = A_2C \cap \overline{AB_2X_2}$. We know that $\measuredangle D_1'B_1D_2' = \measuredangle D_1'B_2D_2' = \measuredangle D_1'CD_2' = 90^{\circ}$. So the five points are concyclic. Then, $\measuredangle CB_2D_1' = \measuredangle CB_1D_1' = \frac 12 \widehat{CX_1} = \frac 12 \widehat{CA_2} = \measuredangle AB_2A_2$, meaning $D_1', B_2, A_2$ are collinear. So $D_1' = D_1$.
Now apply Reim wrt $(A_1A_2B_1B_2)$ and $(B_1B_2D_1D_2C)$ on lines $A_1D_2$ and $A_2D_1$ yields the result.
HaO-R-Zhe
31.03.2023 00:50
Hi CANBANKAN and Khina guess who I am. Thanks CANBANKAN for posting lol
hhy13
31.03.2023 03:58
HAOZHE ORZ