2002 AMC 10B Problem 20

Attempt Problem 20 of the 2002 AMC 10B below, then check your answer against the professionally curated solution 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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20.

Let a,a, b,b, and cc be real numbers such that a7b+8c=4a - 7b + 8c = 4 and 8a+4bc=7.8a + 4b - c = 7. What is a2b2+c2?a^2 - b^2 + c^2?

00

11

44

77

88

Answer: B
Concepts:algebraic manipulationsystem of equations
Difficulty rating: 1790
Solution:

Rewrite the equations as a+8c=4+7ba + 8c = 4 + 7b and 8ac=74b.8a - c = 7 - 4b. Squaring both and adding, (a+8c)2+(8ac)2=(4+7b)2+(74b)2. \begin{aligned} &(a + 8c)^2 + (8a - c)^2 \\ &= (4 + 7b)^2 \\ &\quad {}+ (7 - 4b)^2. \end{aligned}

The left side expands to 65a2+65c265a^2 + 65c^2 (the acac terms cancel), and the right side expands to 65+65b265 + 65b^2 (the bb terms cancel). So 65(a2+c2)=65(1+b2),65(a^2 + c^2) = 65(1 + b^2), giving a2b2+c2=1.a^2 - b^2 + c^2 = 1.

Thus, the correct answer is B.

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