Quadrilateral $A B C D$ has $\angle B C D=\angle D A B=90^{\circ}$. The perimeter of $A B C D$ is 224 and its area is 2205. One side of $A B C D$ has length 7. The remaining three sides have integer lengths. The sum of the squares of the side lengths of $A B C D$ is $S$. What is the integer formed by the rightmost two digits of $S$ ?
Solution 1

Suppose that $A B=x, B C=y, C D=z$, and $D A=7$. (It does not matter to which side length we assign the fixed length of 7.)

We are told that $x, y$ and $z$ are integers.

Since the perimeter of $A B C D$ is 224 , we have $x+y+z+7=224$ or $x+y+z=217$.

Join $B$ to $D$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-010.jpg?height=260&width=287&top_left_y=518&top_left_x=992)

The area of $A B C D$ is equal to the sum of the areas of $\triangle D A B$ and $\triangle B C D$.

Since these triangles are right-angled, then $2205=\frac{1}{2} \cdot D A \cdot A B+\frac{1}{2} \cdot B C \cdot C D$.

Multiplying by 2 , we obtain $4410=7 x+y z$.

Finally, we also note that, using the Pythagorean Theorem twice, we obtain

$$
D A^{2}+A B^{2}=D B^{2}=B C^{2}+C D^{2}
$$

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 $4410=7 x+y z$ and proceeding algebraically, we obtain successively

$$
\begin{aligned}
& 4410=7 x+y z \\
& 4410=7(217-y-z)+y z \\
& 4410=1519-7 y-7 z+y z \\
& 2891=y z-7 y-7 z \\
& 2891=y(z-7)-7 z \\
& 2891=y(z-7)-7 z+49-49 \\
& 2940=y(z-7)-7(z-7) \\
& 2940=(y-7)(z-7)
\end{aligned}
$$

Therefore, $y-7$ and $z-7$ form a positive divisor pair of 2940 . (Since their product is postiive, they are either both positive or both negative. Since $y$ and $z$ are positive, if both of $y-7$ and $z-7$ are negative, we would have $0<y<7$ and $0<z<7$ which could not be large enough to allow for a feasible value of $x$.)

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 Pythaogrean condition. We make a table:

| $y-7$ | $z-7$ | $y$ | $z$ | $x=217-y-z$ | $y^{2}+z^{2}-x^{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 20 | 147 | 27 | 154 | 36 | 23149 |
| 21 | 140 | 28 | 147 | 42 | 20629 |
| 28 | 105 | 35 | 112 | 70 | 8869 |
| 30 | 98 | 37 | 105 | 75 | 6769 |
| 35 | 84 | 42 | 91 | 84 | 2989 |
| 42 | 70 | 49 | 77 | 91 | 49 |
| 49 | 60 | 56 | 67 | 94 | -1211 |

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 .

## Solution 2

As in Solution 1, we have $x+y+z=217$ and $4410=7 x+y z$ and $x^{2}+49=y^{2}+z^{2}$.

Re-arranging and squaring the first equation and using the second and third equations, we obtain

$$
\begin{aligned}
y+z & =217-x \\
y^{2}+z^{2}+2 y z & =x^{2}-434 x+217^{2} \\
\left(x^{2}+49\right)+2(4410-7 x) & =x^{2}-434 x+217^{2} \\
49+8820-14 x & =-434 x+217^{2} \\
420 x & =217^{2}-8820-49 \\
420 x & =38220 \\
x & =91
\end{aligned}
$$

Thus, $y+z=217-91=126$ and $y z=4410-7 \cdot 91=3773$.

This gives, $y(126-y)=3773$ and so $y^{2}-126 y+3773=0$ or $(y-49)(y-77)=0$.

Therefore, $y=49$ (which means $z=77$ ) or $y=77$ (which means $z=49$ ).

We note that $y^{2}+z^{2}=49^{2}+77^{2}=8330=91^{2}+7^{2}=x^{2}+7^{2}$ which verifies the remaining equation.

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 .

ANSWER: 60