In the convex quadrilateral $A B C D$, the midpoints of sides $B C$ and $C D$ are $E$ and $F$ respectively. The segments $A E, E F$, and $A F$ divide the quadrilateral into four triangles whose areas are four consecutive integers. What is the maximum possible area of triangle $A B D$?
Let's use Figure 1. Let the areas of the triangles \( ABE, AEF, AFD, FEC \) be the positive integers \( n, n+1, n+2, n+3 \) (not necessarily in this order). Then \( T_{ABCD} = 4n + 6 \). Since \( T_{ABCD} = T_{ABD} + T_{BCD} \), \( T_{ABD} \) is maximized precisely when \( T_{BCD} \) is minimized.

![](https://cdn.mathpix.com/cropped/2024_05_02_7be55654dc2fad49cadcg-1.jpg?height=420&width=553&top_left_y=278&top_left_x=775)

Figure 1

On the other hand, in triangle \( BCD \), \( EF \) is the midline parallel to \( BD \), so triangle \( BCD \) is similar to triangle \( ECF \). The similarity ratio is 2, so the areas of the triangles satisfy \( T_{BCD} = 4 \cdot T_{ECF} \).

Since \( T_{ECF} \) is one of the numbers \( n, n+1, n+2, n+3 \), \( T_{BCD} \) is minimized when \( T_{ECF} = n \), and in this case \( T_{BCD} = 4 \cdot T_{ECF} = 4n \). Then \( T_{ABD} = T_{ABCD} - T_{BCD} = (4n + 6) - 4n = 6 \).

Thus, the maximum possible area of triangle \( ABD \) is 6 square units.

We must also show that there exists a quadrilateral \( ABCD \) that satisfies the conditions. Here are some examples of good constructions from the submitted solutions. (Unfortunately, a significant portion of the submitted solutions forgot to show this, which is why there were relatively many 4-point solutions.)

1. Example: Let \( ABCD \) be a trapezoid with bases \( AB = 6 \) and \( CD = 4 \), and the height between the bases is 2 (Figure 2). Then \( T_{ABD} = 6 \); \( T_{ABCD} = 10 \), \( T_{ABE} = 3 \), \( T_{AFD} = 2 \), and \( T_{ECF} = 1 \), from which \( T_{AEF} = 10 - (3 + 2 + 1) = 4 \).

![](https://cdn.mathpix.com/cropped/2024_05_02_7be55654dc2fad49cadcg-1.jpg?height=285&width=501&top_left_y=1213&top_left_x=801)

2. Example: Now let \( ABCD \) be a trapezoid with bases \( BC = 4 \) and \( AD = 3 \), and the height between the bases is 3 (Figure 3). Then \( T_{ABD} = 6 \); \( T_{ABCD} = 14 \), \( T_{ABE} = 4 \), \( T_{AFD} = 3 \), and \( T_{ECF} = 2 \), from which \( T_{AEF} = 14 - (4 + 3 + 2) = 5 \).

![](https://cdn.mathpix.com/cropped/2024_05_02_7be55654dc2fad49cadcg-1.jpg?height=377&width=352&top_left_y=1641&top_left_x=881)

Figure 3

(Many examples were based on these two figures; the bases were halved, doubled, or divided by \(\sqrt{2}\), while the height was adjusted accordingly: doubled, halved, or multiplied by \(\sqrt{2}\).)

3. Example: The last example (although \( AB \) is still parallel to \( CD \)) is based on the fact that the diagonals of the quadrilateral are perpendicular to each other. Let \( ABCD \) be a quadrilateral whose diagonals intersect perpendicularly at point \( M \),

\[
AM = \frac{6\sqrt{2}}{5}, \quad CM = \frac{4\sqrt{2}}{5}, \quad BM = 3\sqrt{2} \text{ and } DM = 2\sqrt{2}
\]

in length (Figure 4). Then

\[
\begin{aligned}
& T_{ABE} = \frac{T_{ABC}}{2} = \frac{3\sqrt{2} \cdot \left(\frac{6\sqrt{2}}{5} + \frac{4\sqrt{2}}{5}\right)}{2 \cdot 2} = 3 \\
& T_{AFD} = \frac{T_{ACD}}{2} = \frac{2\sqrt{2} \cdot \left(\frac{6\sqrt{2}}{5} + \frac{4\sqrt{2}}{5}\right)}{2 \cdot 2} = 2
\end{aligned}
\]

while

\[
T_{ECF} = \frac{T_{BCD}}{4} = \frac{(2\sqrt{2} + 3\sqrt{2}) \cdot \frac{4\sqrt{2}}{5}}{2 \cdot 4} = 1
\]

Since

\[
T_{ABCD} = \frac{AC \cdot BD}{2} = \frac{5\sqrt{2} \cdot \frac{10\sqrt{2}}{5}}{2} = 10, \quad \text{ hence } \quad T_{AEF} = 10 - (3 + 2 + 1) = 4
\]

Finally,

\[
T_{ABD} = \frac{(2\sqrt{2} + 3\sqrt{2}) \cdot \frac{6\sqrt{2}}{5}}{2} = 6
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_7be55654dc2fad49cadcg-2.jpg?height=279&width=553&top_left_y=630&top_left_x=775)

Figure 4

Based on the work of Györffy Ágoston (Budapesti Fazekas M. Gyak. Ált. Isk. és Gimn., 11th grade), Lukács Lilla Réka (Budapest, Szent István Gimn., 11th grade), Olosz Adél (Pécs, PTE Gyak. Ált. Isk., Gimn. és Szakgimn., 11th grade), and Schrettner Jakab (Szegedi Radnóti Miklós Kís. Gimn., 11th grade).