3. Let $A B C D$ be a rectangle with area 1 , and let $E$ lie on side $C D$. What is the area of the triangle formed by the centroids of triangles $A B E, B C E$, and $A D E$ ?
Solution:
$1 / 9$

Let the centroids of $A B E, B C E$, and $A D E$ be denoted by $X, Y$, and $Z$, respectively. Let $d(P, Q R)$ denote the distance from $P$ to line $Q R$. Since the centroid lies two-thirds of the distance from each vertex to the midpoint of the opposite edge, $d(X, A B)=$ $d(Y, C D)=d(Z, C D)=\frac{1}{3} B C$, so $Y Z$ is parallel to $C D$ and $d(X, Y Z)=B C-$ $\frac{2}{3} B C=\frac{1}{3} B C$. Likewise, $d(Z, A D)=\frac{1}{3} D E$ and $d(Y, B C)=\frac{1}{3} C E$, so that since $Y Z$ is perpendicular to $A D$ and $B C$, we have that $Y Z=C D-\frac{1}{3}(D E+C E)=\frac{2}{3} C D$. Therefore, the area of $X Y Z$ is $\frac{1}{2}\left(\frac{1}{3} B C\right)\left(\frac{2}{3} C D\right)=\frac{1}{9} B C \cdot C D=\frac{1}{9}$.