Rectangle of the Year

We can use proof by contradiction. Assume that $A,B,C,$ and $D$ are all integers. Let the horizontal side of the rectangle be $a$ , the vertical side of the rectangle be $b$ , the horizontal leg of triangle C be $e$ , and the vertical leg of triangle C be $f$ . We can see that $ef=2C$ , $(a-e)b=2B$ , and $(b-f)a=2A$ . Multiplying the second and third equations together gives us $ab(ab-af-eb+ef)=4AB$ Adding the second and third equations together tells us that $-af-eb=2A+2B-2ab$ . Substituting this and the first equation into our other formula gives us $ab(ab+2A+2B-2ab+2C)=4AB \implies (ab)^2+(-2A-2B-2C)ab +4AB=0 \implies ab=A+B+C \pm \sqrt{(A+B+C)^2-4AB}$ We know that $ab=A+B+C+D$ , so this simplifies to $D^2+4AB=(A+B+C)^2 \implies D^2+4AB=(2018-D)^2 \implies D=1009-\frac{AB}{1009}$ . $D$ must be an integer and $1009$ is prime, so $A$ or $B$ must be a multiple of $1009$ . $2018$ or any larger multiples couldn't satisfy $A+B+C+D=2018$ , so we can set $A=1009$ . This implies $C=0$ which is a contradiction because $C$ must be positive. Thus we know that $A,B,C,$ and $D$ cannot all be integers.

If you factor $2018$ , you get $2\cdot1009$ .

For all areas to be integers, the triangles A and B have to both have at least one even length leg.

If the height of B is 2, then the base of A is 1009 and the height of A is 1, so A does not have integer area.

The same argument applies if the base of A is 2, in which case the height of B is 1009 and base of B is 1, leaving B with non-integer area.

So A and B can't both have an integer area at the same time making it impossible for all the areas to be integers.

This is the approach I took. I assumed that the values of A, B, C, and D were integers, to begin with, and then used proof by contradiction to prove that this can not be the case. Here is my solution:

$2018 = 2 × 1009$ , meaning that $(x, y) = (1009, 2)$ or vice versa. From the picture I drew, $p + q = 2$ and $r + s = 1009$ .

$Area(B) = (y × r) / 2$ , but since $y$ is a prime number, the only way for $B$ to be an integer is for $r$ to be divisible by 2 or in other words, even.

If $r$ is even, then $s$ is must clearly be odd.

$Area(C) = (q × s) / 2$ , and for $C$ to be an integer, $q$ must be even.

However, that can't be true, especially since $p + q = 2$ where $q$ must be a member of the positive integers less than 2 and greater than 0. Having $q$ being a member of the positive reals won't help either since then the area of $C$ will end up not being an integer.

We have a contradiction meaning $A, B, C, D$ can not all be integers!