A Geometry problem from OMK 2017 The following problem appeared in the National Mathematical Olympiad, or known short as OMK in Malaysia.
Let ABCD be a convex quadrilateral. Let points M and N be the midpoints of BC and CD respectively. Triangle AMN divides quadrilateral ABCD into 4 non-overlapping sections, such that the areas of the sections are of consecutive integers. (If the area of the smallest section is E, then the 2nd smallest section is E+1, the 3rd smallest is E+2, and the largest is E+3.) What is the maximum possible area of triangle ABD? (Note:The problem is edited slightly to prevent certain confusion)
I have seemingly found a solution to the problem:
Let E denote the area of the smallest of the 4 sections. Thus, the total area of quadrilateral is 4E+6. Note that:
$$Area of triangle ABD=Area of quadrilateral ABCD-Area of triangle BCD$$
If the area of triangle ABD should be maximum, the area of triangle BCD must be minimum. Finally, notice that triangle BCD is similar to triangle MCN, and since triangle MCN is one of the 4 sections, the smallest area of triangle MCN must be E. Thus, the minimum area of triangle BCD is $E×2^2=4E$ and thus the maximum area of triangle ABD is 6.
Questions:
1) Does my solution contain any absurdity or wrong calculations somewhere?
2) Is there an alternative solution? (Preferably an elementary solution)
3) Any improvement suggestions? (Improvements in explanation, term usages, etc.)
 A: Your solution is perfect, however I tried – just for fun – to see if such a quadrilateral can indeed exist.
In general, we can construct a quadrilateral $ABCD$ where the areas of triangles $CMN$, $ABM$ and $ADN$ are respectively $n$, $n+a$ and $n+b$. To this end, choose a segment $MN$ at will and a point $C$ such that its distance $CH$ from $MN$ satisfies ${1\over2}MN\cdot CH=n$. Points $B$ and $D$ are then given by the request that $M$ and $N$ be the midpoints of $CB$ and $CD$.
Draw then two lines $EK$ and $FI$, parallel to $CD$ and $CB$, such that their distances $CK$ and $CI$ from $C$ satisfy ${1\over2}CM\cdot CI=n+a$ and ${1\over2}CN\cdot CK=n+b$. Let $E$ and $F$ be the intersections of those lines with the line through $C$ parallel to $MN$. If $A$ is the intersection point of $EK$ and $FI$, then triangles $ABM$ and $ADN$ have areas respectively $n+a$ and $n+b$.
By similar triangles you have then $CF=(1+a/n)MN$ and $CE=(1+b/n)MN$. But triangle $AEF$ is also similar to $CMN$, so the distance from $A$ to line $MN$ is $(1+a/n+b/n)CH$ and the area of triangle $AMN$ turns out to be $n+a+b$.
In our case we need $a+b=3$, then we can choose $a=1$ and $b=2$ (or the other way around) to obtain areas $n$, $n+1$, $n+2$ and $n+3$. That works for every $n$.

EDIT.
The area $S_{ABD}$ of triangle $ABD$ can then be computed as $2(a+b)$ and the areas of the four triangles are consecutive numbers if $0$, $a$, $b$, $a+b$ are consecutive numbers.
The solution outlined above, $\{a,b\}=\{1,2\}$, gives $S_{ABD}=6$, but there is another possible configuration where the areas of the four triangles are consecutive numbers: $\{a,b\}=\{-1,2\}$, giving $S_{ABD}=2$.
Values of $a$ and $b$ leading to $S_{ABD}\le0$ must be discarded, for in that case the resulting quadrilateral is not convex. Hence we can confirm that the maximum value of $S_{ABD}$ is $6$, while the minimum is $2$.
