2002 AMC 10B Problem 20

We want $a^2 + c^2 - b^2$, which has $a$ and $c$ together and $b$ on its own. So put $a$ and $c$ on the left and $b$ on the right.

1. Rearrange the two equations:
$$a + 8c = 4 + 7b, \qquad 8a - c = 7 - 4b.$$
2. Squares appear in the target, so square both equations. Left sides:
$$(a + 8c)^2 = a^2 + 16ac + 64c^2, \qquad (8a - c)^2 = 64a^2 - 16ac + c^2.$$
Right sides:
$$(4 + 7b)^2 = 16 + 56b + 49b^2, \qquad (7 - 4b)^2 = 49 - 56b + 16b^2.$$
3. Notice the mixed terms: $+16ac$ and $-16ac$, and $+56b$ and $-56b$. Adding the two squared equations makes them cancel:
$$65a^2 + 65c^2 = 65 + 65b^2.$$
4. Divide by $65$: $a^2 + c^2 = 1 + b^2$, so $a^2 - b^2 + c^2 = 1$.

The answer is $\boxed{\textbf{(B)}\ 1}$.

The numbers were chosen in a mirrored pattern: $1, 8$ in one equation and $8, -1$ in the other (and $4, 7$ with $7, -4$ on the right). With that pattern, squaring and adding always cancels the mixed terms and leaves a multiple of $a^2 + c^2$. When you see coefficients that swap places and flip a sign, try squaring and adding.

The answer choices are fixed numbers, so the expression must have the same value for every solution. Pick an easy one by setting $b = 0$. Then $a + 8c = 4$ and $8a - c = 7$. From the second, $c = 8a - 7$; putting that in the first gives $a + 64a - 56 = 4$, so $a = \frac{60}{65} = \frac{12}{13}$ and $c = \frac{96}{13} - \frac{91}{13} = \frac{5}{13}$. Then $a^2 + c^2 - 0 = \frac{144 + 25}{169} = 1$.

The target $a^2 + c^2 - b^2$ suggests keeping $a, c$ on one side and $b$ on the other:
$$a + 8c = 4 + 7b, \qquad 8a - c = 7 - 4b.$$

The coefficient pairs $(1, 8)$ and $(8, -1)$ are "perpendicular," so squaring both equations and adding makes the $ac$ cross terms cancel on the left. The same happens with the $b$ terms on the right:
$$(a + 8c)^2 + (8a - c)^2 = (a^2 + 16ac + 64c^2) + (64a^2 - 16ac + c^2) = 65(a^2 + c^2),$$
$$(4 + 7b)^2 + (7 - 4b)^2 = (16 + 56b + 49b^2) + (49 - 56b + 16b^2) = 65(1 + b^2).$$

Hence $a^2 + c^2 = 1 + b^2$, that is, $a^2 - b^2 + c^2 = 1$.

The answer is $\boxed{\textbf{(B)}\ 1}$.