2002 AMC 10B Problem 20

Below is the professionally curated solution for Problem 20 of the 2002 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

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Concepts:algebraic manipulation system of equations

Difficulty rating: 1790

20.

Let $a,$ $b,$ and $c$ be real numbers such that $a - 7b + 8c = 4$ and $8a + 4b - c = 7.$ What is $a^2 - b^2 + c^2?$

$0$

$1$

$4$

$7$

$8$

Solution:

Rewrite the equations as $a + 8c = 4 + 7b$ and $8a - c = 7 - 4b.$ Squaring both and adding, $(a + 8c)^2 + (8a - c)^2 = (4 + 7b)^2 + (7 - 4b)^2.$

The left side expands to $65a^2 + 65c^2$ (the $ac$ terms cancel), and the right side expands to $65 + 65b^2$ (the $b$ terms cancel). So $65(a^2 + c^2) = 65(1 + b^2),$ giving $a^2 - b^2 + c^2 = 1.$

Thus, the correct answer is B.

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