Let $EFGH,ABCD$ and $E_1F_1G_1H_1$ be three convex quadrilaterals satisfying: i) The points $E,F,G$ and $H$ lie on the sides $AB,BC,CD$ and $DA$ respectively, and $\frac{AE}{EB}\cdot\frac{BF}{FC}\cdot \frac{CG}{GD}\cdot \frac{DH}{HA}=1$; ii) The points $A,B,C$ and $D$ lie on sides $H_1E_1,E_1F_1,F_1,G_1$ and $G_1H_1$ respectively, and $E_1F_1||EF,F_1G_1||FG,G_1H_1||GH,H_1E_1||HE$. Suppose that $\frac{E_1A}{AH_1}=\lambda$. Find an expression for $\frac{F_1C}{CG_1}$ in terms of $\lambda$. Xiong Bin
Problem
Source: Chinese MO 2004
Tags: geometry proposed, geometry
yetti
27.09.2011 08:38
Posted before at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=31052, no reply. Let $O \equiv AC \cap BD$ and $P = EF \cap AC$. By Menelaus theorem for $\triangle ABC, \triangle ACD$ $\Longrightarrow$ $\frac{\overline{PA}}{\overline{PC}} = \frac{\overline{EA}}{\overline{EB}} \cdot \frac{\overline{FB}}{\overline{FC}} = \frac{\overline{GD}}{\overline{GC}} \cdot \frac{\overline{HD}}{\overline{HA}}$ $\Longrightarrow$ $P, H, G$ are collinear $\Longrightarrow$ $AC, EF, GH$ concur at $P$. (Likewise, $BD, FG, HE$ concur at $Q$.) Let $X \equiv E_1BF_1 \cap AOC$, $Y \equiv G_1DH_1 \cap AOC$ and $U \equiv E_1BF_1 \cap G_1DH_1$. Parallel to $HEQ$ through $B$ and parallel to $GHP$ through $X$ meet at $K$. $\triangle KXB \sim \triangle HPE$ are centrally similar with center $A$ $\Longrightarrow$ $K \in DHA$. Let $K_1 \equiv H_1AE_1 \cap KX$. Since $K_1E_1 \parallel KB$ and $DH_1Y \parallel KK_1X$ $\Longrightarrow$ $\frac{\overline{XE_1}}{\overline{XB}} = \frac{\overline{XK_1}}{\overline{XK}} = \frac{\overline{YH_1}}{\overline{YD}}.$ By Menelaus theorem for $\triangle UE_1H_1, \triangle UBD$ cut by common transversal $OAXY$, $-\lambda = \frac{\overline{AE_1}}{\overline{AH_1}} = \frac{\overline{YU}}{\overline{YH_1}} \cdot \frac{\overline{XE_1}}{\overline{XU}} = \frac{\overline{YU}}{\overline{YD}} \cdot \frac{\overline{XB}}{\overline{XU}} = \frac{\overline{OB}}{\overline{OD}}$. Similarly, $\frac{\overline{CF_1}}{\overline{CG_1}} = \frac{\overline{OB}}{\overline{OD}} = -\lambda$.