Shestakov C.A.
In a convex quadrilateral $ABCD$, points $E$ and $F$ are the midpoints of sides $BC$ and $CD$ respectively.
Segments $AE$, $AF$, and $EF$ divide the quadrilateral into four triangles, the areas of which are (in some order) consecutive natural numbers. What is the greatest possible value of the area of triangle $ABD$?
Author: Shestakov S.A.

Let the areas of the triangles be $n, n+1, n+2$, and $n+3$. Then $S_{A B C D}=4 n+6$. $E F$ is the midline of triangle $B C D$, so $S_{B C D}=4 S_{E C F} \geq 4 n$. Therefore, $S_{A B D}=S_{A B C D}-S_{B C D} \leq 6$.

We will show that this area can equal 6. Consider an isosceles trapezoid $A B C D$ with bases $A D=6, B C=4$ and height 2. Then

$S_{C F E}=1, S_{A B E}=2, S_{A D F}=3, S_{A E F}=S_{A B C D}-1-2-3=4$, and in this case $S_{A B D}=6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ff42d6d57f9dccbb1f5g-34.jpg?height=205&width=511&top_left_y=272&top_left_x=802)

## Answer

6.