2002 AMC 10B Problem 20

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 $\boxed{\textbf{(B)}\ 1}$.

The identity $(pa + qc)^2 + (qa - pc)^2 = (p^2 + q^2)(a^2 + c^2)$ turns two linear equations with swapped, sign-flipped coefficients into a statement about $a^2 + c^2$. The problem was built so that the same identity works on the right side with $(4, 7)$ and $(7, -4)$. When coefficients appear in this mirrored pattern, square and add.

Since the answer choices are constants, the expression must be the same for every solution. Take $b = 0$: then $a + 8c = 4$ and $8a - c = 7$, so $c = 8a - 7$ and $65a = 60$, giving $a = \frac{12}{13}$, $c = \frac{5}{13}$. Then $a^2 + c^2 = \frac{144 + 25}{169} = 1$.

• Trying to solve the two equations for a, b, c individually; the system has infinitely many solutions, and the target expression is the same for all of them.
• Squaring and adding the original equations as written (with $b$ still mixed in), where the cross terms do not cancel.