Geometry Problem Concerning Trisection Points on a Convex Quadrilateral Let $ABCD$ be a convex quadrilateral, and let $P$, $Q$, $R$, $S$, $T$, $U$, $V$, and $W$ be the trisection points of the sides of $ABCD$, as shown. ![enter image description here]( If the area of quadrilateral $ABCD$ is 180, then find the area of hexagon $AQRCUV$.

**Solution:** Since $BQ = AB/3$, $BR = BC/3$, and $\angle QBR = \angle ABC$, triangles $QBR$ and $ABC$ are SAS-similar. Furthermore, since $Q$ and $R$ are trisection points, the side lengths are in a $3:1$ ratio so the areas are in a $9:1$ ratio. This gives $[BQR] = \frac{[ABC]}{9}.$

![enter image description here](

Likewise, triangles $ACD$ and $VUD$ are similar, and $[VUD] = \frac{[ACD]}{9}.$ Therefore, $[BQR] + [VUD] = \frac{[ABC] + [ACD]}{9} = \frac{[ABCD]}{9} = \frac{180}{9} = 20.$ The remainder of quadrilateral $ABCD$ is hexagon $AQRCUV$, so it has area $180 - 20 = \boxed{160}$.