2003 IMO SL #G4

Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that \[ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}. \]



Invert at $P$ with radius $1$, such that $X^{*}$ denotes the point when $X$ is inverted. By the Inversion-Distance formula, we have $AB=(A^*B^*)(PA)(PB)$, $BC=(B^*C^*)(PB)(PC)$, $AD=(A^*D^*)(PA)(PD)$, and $DC=(D^*C^*)(PD)(PC)$. Also, note $A^*B^*C^*D^*$ is a parallelogram, and hence $$\frac{AB\cdot BC}{AD\cdot DC}=\frac{A^*B^*\cdot PA\cdot PB\cdot B^*C^*\cdot PB\cdot PC}{A^*D^*\cdot PA\cdot PD\cdot D^*C^*\cdot PD\cdot PC}$$

$$=\frac{PB^2}{PD^2},$$as desired. $\square$ 

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