AoPS - Problem

Source: Chinese TST 2009 2nd P1

Tags: geometry, geometry proposed

Fang-jh

21.03.2009 12:35

In convex pentagon $ ABCDE$, denote by $ AD\cap BE = F,BE\cap CA = G,CA\cap DB = H,DB\cap EC = I,EC\cap AD = J; AI\cap BE = A',BJ%Error. "capCA" is a bad command. = B',CF%Error. "capDB" is a bad command. = C',DG\cap EC = D',EH\cap AD = E'.$ Prove that $ \frac {AB'}{B'C}\cdot\frac {CD'}{D'E}\cdot\frac {EA'}{A'B}\cdot\frac {BC'}{C'D}\cdot\frac {DE'}{E'A} = 1$.

Let $ AI\cap CD=A_1$ and etc.. $ \frac{EA'}{A'B}\cdot \frac{DA_1}{A_1C}=\frac{S_{EIA'}}{S_{BIA'}}\cdot \frac{S_{DIA_1}}{S_{CIA_1}}=\frac{EI}{BI}\cdot \frac{DI}{CI}$. $ \Rightarrow \frac{EA'}{A'B}=\frac{EI}{IC}\cdot \frac{DI}{IB}\cdot \frac{CA_1}{A_1D}=\frac{S_{EHB}}{S_{BHC}}\cdot \frac{S_{EJD}}{S_{EJB}}\cdot \frac{S_{AID}}{S_{AIC}}$ where I donate with $ S_{XYZ}$ the area of the triangle $ \triangle XYZ$. $ \Rightarrow \prod_{\textrm{cyc}}\frac{EA'}{A'B}=\prod_{\textrm{cyc}}\frac{S_{EHB}}{S_{BHC}}\cdot \frac{S_{EJD}}{S_{EJB}}\cdot \frac{S_{AID}}{S_{AIC}}=1$.