AoPS - Problem

Source: MOP 2005 Homework - Blue Group #21

Tags: trigonometry, geometry unsolved, geometry

Konigsberg

18.05.2014 19:29

Points $E$, $F$, $G$, and $H$ lie on sides $AB$, $BC$, $CD$, and $DA$ of a convex quadrilateral $ABCD$ such that $\frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \frac{CG}{GD} \cdot \frac{DH}{HA}=1$. Points $A$, $B$, $C$, and $D$ lie on sides $H_1E_1$, $E_1F_1$, $F_1G_1$, and $G_1H_1$ of a convex quadrilateral $E_1F_1G_1H_1$ such that $E_1F_1 \parallel EF$, $F_1G_1 \parallel FG$, $G_1H_1 \parallel GH$, and $H_1E_1 \parallel HE$. Given that $\frac{E_1A}{AH_1}=a$, express $\frac{F_1C}{CG_1}$ in terms of $a$.

Since $\frac{AE}{EB}\frac{BF}{FC}=\frac{AH}{HD}\frac{DG}{GC}$, we have $FE,CA,GH$ concurrent at $Y$. Let $AC$ intersect $BD$ at $X$. Draw the line through $E_1$ parallel to $DH_1$ which intersects $DA$ at $Z$. Then $\frac{\sin\angle EYA}{\sin\angle AYH}\frac{\sin\angle AYH}{\sin\angle AHY}\frac{\sin\angle AHE}{\sin\angle AEH}=1$. But note that $\angle E_1BA=\angle YEA,\angle E_1AB=\angle AEH,\angle E_1AZ=\angle AHE,\angle E_1ZA=\angle AHY$, thus $\angle E_1BZ=\angle AYE,\angle E_1ZB=\angle AYH$, so $BZ\parallel AC$. Thus $\frac{E_1A}{AH_1}=\frac{ZA}{AD}=\frac{BX}{XD}$. Similarly $\frac{F_1C}{CG_1}=\frac{BX}{XD}=a$.