Problem 24 of Fermat Contest 2023

“Quadrilateral ABCD has $\angle BCD = \angle DAB = 90^{\circ}$ . The perimeter of ABCD is 224 and its area is 2205. One side of ABCD has length 7. The remaining three sides have integer length. The sum of the squares of the side length of ABCD is $S$ . What is the integer formed by the rightmost two digits of $S$ ?”

Without loss of generality, we let $a, b, c, d$ be four side lengths of the ABCD as shown above and $d=7$ . Then we have the following system of equations.

$a+b+c+7=224$ …(1)(by perimeter)

$ab+7c=4410$ …(2)(by area)

$a^2+b^2=c^2+7^2$ …(3)(by pythagorean)

Notice that $S=a^2+b^2+c^2+7^2=2(a^2+b^2)$ . This suggests that if we can find the value of $a^2+b^2$ , we’re done. Another notice is about the connections between the expressions $a^2+b^2, a+b$ and $ab$ . If we let $x=ab, y=a+b$ , then $a^2+b^2=(a+b)^2-2ab=y^2-2x$ . Also, from the equation(1), we have $c=217-y$ which can be substituted into the equations (2) and (3) to get a new system about $x$ and $y$ as follows.