Problem 1065

Solution: Let $M$ and $N$ be the midpoints of sides $AB$ and $CD$ respectively, and let point $P$ and side $CD$ lie on opposite sides of line $AB$. Quadrilateral $MXNY$ is a parallelogram, so $S_{\triangle MXY} = S_{\triangle NXY}$. Let $S_{ABCD} = S$. Then $S_{\triangle ACD} + S_{\triangle ABD} + S_{\triangle ABC} + S_{\triangle BCD} = 2S$. Segment $MX$ is the midline of triangle $ABC$, so lines $MX$ and $CP$ are parallel, meaning points $B$ and $P$ are equidistant from line $MX$.

Therefore, $S_{\triangle PMX} = S_{\triangle BMX} = S_{BCXM} - S_{\triangle BCX} = \frac{3}{4} S_{\triangle ABC} - \frac{1}{2} S_{\triangle ABC} = \frac{1}{4} S_{\triangle ABC}$.

Similarly, points $A$ and $P$ are also equidistant from line $MY$. Therefore, $S_{\triangle PMY} = S_{\triangle AMY} = \frac{1}{4} S_{\triangle ABD}$. Moreover, point $Z$, the intersection of diagonals $BM$ and $PX$ of trapezoid $BPMX$, lies inside trapezoid $BPMX$, and point $T$, the intersection of diagonals of trapezoid $AYMP$, lies inside trapezoid $AYMP$, so point $M$ lies on segment $ZT$, and thus inside triangle $PXY$. Note that

$S_{MXNY} = S - S_{\triangle CNX} - S_{\triangle DNY} - S_{AMYD} - S_{BCXM} = S - \frac{1}{4} S_{\triangle ACD} - \frac{1}{4} S_{\triangle BCD} - \frac{3}{4} S_{\triangle ABD} - \frac{3}{4} S_{\triangle ABC}$.

Therefore,

$$\begin{array}{r} S_{\triangle PXY} = S_{\triangle MXY} + S_{\triangle PMX} + S_{\triangle PMY} = \frac{1}{2} S_{MXNY} + S_{\triangle PMX} + S_{\triangle PMY} = \\ = \frac{1}{2} \left( S - \frac{1}{4} S_{\triangle ACD} - \frac{1}{4} S_{\triangle BCD} - \frac{3}{4} S_{\triangle ABD} - \frac{3}{4} S_{\triangle ABC} \right) + \frac{1}{4} S_{\triangle ABC} + \frac{1}{4} S_{\triangle ABD} = \\ = \frac{1}{2} S - \frac{1}{8} \left( S_{\triangle ACD} + S_{\triangle ABD} + S_{\triangle ABC} + S_{\triangle BCD} \right) = \frac{S}{2} - \frac{2S}{8} = \frac{S}{4} \end{array}$$

As required.