A secant meets one circle at points $A_1$, $B_1$։, this secant meets a second circle at points $A_2$, $B_2$. Another secant meets the first circle at points $C_1$, $D_1$ and meets the second circle at points $C_2$, $D_2$. Prove that point $A_1C_1 \cap B_2D_2$, $A_1C_1 \cap A_2C_2$, $A_2C_2 \cap B_1D_1$, $B_2D_2 \cap B_1D_1$ lie on a circle coaxial with two given circles.
Problem
Source: 2021 Sharygin Geometry Olympiad Finals grades X-XI p5
Tags: Concyclic, coaxial circles, geometry
khanhnx
28.09.2024 18:28
Let $X \equiv A_1C_1 \cap B_2D_2, Y \equiv A_1C_1 \cap A_2C_2, Z \equiv A_2C_2 \cap B_1D_1, T \equiv B_1D_1 \cap B_2D_2$. We have $\angle{YC_1C_2} = \angle{TB_1B_2}$ and $\angle{YC_2C_1} = \angle{TB_2B_1}$. Then $\triangle YC_1C_2 \sim \triangle TB_1B_2$ or $\dfrac{YC_1}{YC_2} = \dfrac{TB_1}{TB_2}$. Similarly, we have $\dfrac{YA_1}{YA_2} = \dfrac{TD_1}{TD_2}$. So $$\dfrac{\mathcal{P}_{Y / (A_1B_1C_1D_1)}}{\mathcal{P}_{Y / (A_2B_2C_2D_2)}} = \dfrac{YA_1 \cdot YC_1}{YA_2 \cdot YC_2} = \dfrac{TD_1 \cdot TB_1}{TD_2 \cdot TB_2} = \dfrac{\mathcal{P}_{T / (A_1B_1C_1D_1)}}{\mathcal{P}_{T / (A_2B_2C_2D_2)}}$$Similarly, we have $\dfrac{\mathcal{P}_{X / (A_1B_1C_1D_1)}}{\mathcal{P}_{X / (A_2B_2C_2D_2)}} = \dfrac{\mathcal{P}_{Z / (A_1B_1C_1D_1)}}{\mathcal{P}_{Z / (A_2B_2C_2D_2)}}$. Let the line through $Z$ and parallels to $A_1C_1$ intersects $A_1B_1, C_1D_1$ at $U, V$. It's easy to see that $A_1, C_1, U, V$ lie on a circle. We have $$\dfrac{\mathcal{P}_{Y / (A_1B_1C_1D_1)}}{\mathcal{P}_{Z / (A_1B_1C_1D_1)}} = \dfrac{YA_1 \cdot YC_1}{ZB_1 \cdot ZD_1} = \dfrac{YA_1 \cdot YC_1}{ZU \cdot ZV} = \dfrac{YA_2 \cdot YC_2}{ZA_2 \cdot ZC_2} = \dfrac{\mathcal{P}_{Y / (A_2B_2C_2D_2)}}{\mathcal{P}_{Z / (A_2B_2C_2D_2)}}$$or $\dfrac{\mathcal{P}_{Y / (A_1B_1C_1D_1)}}{\mathcal{P}_{Y / (A_2B_2C_2D_2)}} = \dfrac{\mathcal{P}_{Z / (A_1B_1C_1D_1)}}{\mathcal{P}_{Z / (A_2B_2C_2D_2)}}$. From this, we have $X, Y, Z, T$ have equal ratio of power with $(A_1B_1C_1D_1)$ and $(A_2B_2C_2D_2)$ or $(A_1B_1C_1D_1), (A_2B_2C_2D_2)$ and $(XYZT)$ are coaxial