APMO 1989

Let $G$ be the centroid of triangle $ABC$, and also the intersection point of $A_1{B}_2$, $A_2{B}_3$, and $A_3{B}_1$. By Menelao's theorem on triangle $B_1{A}_2{A}_3$ and line $A_1{D}_1{C}_2$, $$\frac{A_1{B}_1}{A_1{A}_2}\cdot \frac{D_1{A}_3}{D_1{B}_1}\cdot \frac{C_2{A}_2}{C_2{A}_3}=1⟺\frac{D_1{A}_3}{D_1{B}_1}=2\cdot 3=6⟺\frac{D_1{B}_1}{A_3{B}_1}=\frac{1}{7}.$$ Since $A_{3}G=\frac{2}{3}{A}_3{B}_1$, if $A_3{B}_1=21t$ then $G{A}_3=14t$, $D_1{B}_1=\frac{21t}{7}=3t$, $A_3{D}_1=18t$, and $G{D}_1=A_3{D}_1-A_{3}G=18t-14t=4t$, and $$\frac{G{D}_1}{G{A}_3}=\frac{4}{14}=\frac{2}{7}.$$ Similar results hold for the other medians, therefore $D_1{D}_2{D}_3$ and $A_1{A}_2{A}_3$ are homothetic with center $G$ and ratio $-\frac{2}{7}$. By Menelao's theorem on triangle $A_1{A}_2{B}_2$ and line $C_1{E}_1{A}_3$, $$\frac{C_1{A}_1}{C_1{A}_2}\cdot \frac{E_1{B}_2}{E_1{A}_1}\cdot \frac{A_3{A}_2}{A_3{B}_2}=1⟺\frac{E_1{B}_2}{E_1{A}_1}=3\cdot \frac{1}{2}=\frac{3}{2}⟺\frac{A_1{E}_1}{A_1{B}_2}=\frac{2}{5}.$$ If $A_1{B}_2=15u$, then $A_{1}G=\frac{2}{3}\cdot 15u=10u$ and $G{E}_1=A_{1}G-A_1{E}_1=10u-\frac{2}{5}\cdot 15u=4u$, and $$\frac{G{E}_1}{G{A}_1}=\frac{4}{10}=\frac{2}{5}.$$ Similar results hold for the other medians, therefore $E_1{E}_2{E}_3$ and $A_1{A}_2{A}_3$ are homothetic with center $G$ and ratio $\frac{2}{5}$. Then $D_1{D}_2{D}_3$ and $E_1{E}_2{E}_3$ are homothetic with center $G$ and ratio $-\frac{2}{7}:\frac{2}{5}=-\frac{5}{7}$, and the ratio of their area is ${\left(\frac{5}{7}\right)}^2=\frac{25}{49}$.