Problem 111

Suppose that $AB=x, BC=y, CD=z$, and $DA=7$. Since the perimeter of $ABCD$ is 224, we have $x+y+z+7=224$ or $x+y+z=217$. Join $B$ to $D$. The area of $ABCD$ is equal to the sum of the areas of $\triangle DAB$ and $\triangle BCD$. Since these triangles are right-angled, then $2205=\frac{1}{2} \cdot DA \cdot AB+\frac{1}{2} \cdot BC \cdot CD$. Multiplying by 2, we obtain $4410=7x+yz$. Finally, we also note that, using the Pythagorean Theorem twice, we obtain $DA^{2}+AB^{2}=DB^{2}=BC^{2}+CD^{2}$ and so $49+x^{2}=y^{2}+z^{2}$. We need to determine the value of $S=x^{2}+y^{2}+z^{2}+7^{2}$. Since $x+y+z=217$, then $x=217-y-z$. Substituting into $4410=7x+yz$ and proceeding algebraically, we obtain successively $4410=7x+yz$ $4410=7(217-y-z)+yz$ $4410=1519-7y-7z+yz$ $2891=yz-7y-7z$ $2891=y(z-7)-7z$ $2891=y(z-7)-7z+49-49$ $2940=y(z-7)-7(z-7)$ $2940=(y-7)(z-7)$. Therefore, $y-7$ and $z-7$ form a positive divisor pair of 2940. We note that $y+z=217-x$ and so $y+z<217$ which means that $(y-7)+(z-7)<203$. Since $2940=20 \cdot 147=2^{2} \cdot 5 \cdot 3 \cdot 7^{2}$ then the divisors of 2940 are the positive integers of the form $2^{r} \cdot 3^{s} \cdot 5^{t} \cdot 7^{u}$ where $0 \leq r \leq 2$ and $0 \leq s \leq 1$ and $0 \leq t \leq 1$ and $0 \leq u \leq 2$. Thus, these divisors are $1,2,3,4,5,6,7,10,12,14,15,20,21,28,30,35,42,49$. We can remove divisor pairs from this list whose sum is greater than 203. This gets us to the shorter list $20,21,28,30,35,42,49,60,70,84,98,105,140,147$. This means that there are 7 divisor pairs remaining to consider. We can assume that $y<z$. Using the fact that $x+y+z=217$, we can solve for $x$ in each case. These values of $x, y$ and $z$ will satisfy the perimeter and area conditions, but we need to check the Pythagorean condition. We make a table: Since we need $y^{2}+z^{2}-x^{2}=49$, then we must have $y=49$ and $z=77$ and $x=91$. This means that $S=x^{2}+y^{2}+z^{2}+7^{2}=91^{2}+49^{2}+77^{2}+7^{2}=16660$. The rightmost two digits of $S$ are 60.