6. Given a rectangle $A B C D$ with point $S$ being the intersection of the diagonals. On side $\overline{A D}$, points $E$ and $F$ are chosen such that $|A E|=|E F|=|F D|$. Determine the ratio of the area of pentagon $E B S C F$ to the area of rectangle $A B C D$.
6. Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_63a858253359fff06b75g-4.jpg?height=382&width=626&top_left_y=303&top_left_x=772)

Let $|A B|=|C D|=a$ and $|B C|=|D A|=b$.

Then $|A E|=|E F|=|F D|=\frac{b}{3}$.

1 POINT

$S$ is the intersection of the diagonals, and the diagonals of the rectangle are equal and bisect each other at their intersection.

This means that the isosceles triangles $D A S$ and $B C S$ are congruent, so they have equal heights to the base.

In triangle $B C S$, the height to side $\overline{B C}$ has a length of $\frac{a}{2}$.

1 POINT

Then $p_{B C S}=\frac{1}{2} \cdot b \cdot \frac{a}{2}=\frac{a b}{4}$.

1 POINT

The area of the right triangle $A B E$ is equal to the area of the right triangle $C D F$

1 POINT

and is $p_{A B E}=\frac{1}{2} \cdot a \cdot \frac{b}{3}=\frac{a b}{6}$.

1 POINT

We have: $p_{E B S C F}=p_{A B C D}-\left(p_{A B E}+p_{C D F}+p_{B C S}\right)$

1 POINT

$p_{\text {EBSCF }}=a b-\left(\frac{a b}{6}+\frac{a b}{6}+\frac{a b}{4}\right)=\frac{12-(2+2+3)}{12} \cdot a b=\frac{5}{12} \cdot a b$

1 POINT

Thus, $p_{\text {EBSCF }}=\frac{5}{12} p_{A B C D}$.

1 POINT

The desired ratio of areas is $p_{\text {EBSCF }}: p_{A B C D}=5: 12$.

1 POINT $\qquad$
Note: If a student takes some specific values for the lengths of the sides of the rectangle and thus arrives at the correct ratio, but also has all steps and explanations of the relationships between the lengths of segments, such a solution should be scored with 4 POINTS.