Difference between revisions of "2002 AMC 10B Problems/Problem 20"

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Squaring both, <math>a^2+16ac+64c^2=49b^2+56b+16</math> and <math>64a^2-16ac+c^2=16b^2-56b+49</math> are obtained.
 
Squaring both, <math>a^2+16ac+64c^2=49b^2+56b+16</math> and <math>64a^2-16ac+c^2=16b^2-56b+49</math> are obtained.
  
Adding the two equations and dividing by <math>65</math> gives <math>a^2+c^2=b^2+1</math>, so <math>a^2-b^2+c^2=\boxed{(B)1}</math>.
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Adding the two equations and dividing by <math>65</math> gives <math>a^2+c^2=b^2+1</math>, so <math>a^2-b^2+c^2=\boxed{(\text{B})1}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 22:26, 17 November 2013

Problem

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

$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$

Solution

$a+8c=7b+4$ and $8a-c=7-4b$

Squaring both, $a^2+16ac+64c^2=49b^2+56b+16$ and $64a^2-16ac+c^2=16b^2-56b+49$ are obtained.

Adding the two equations and dividing by $65$ gives $a^2+c^2=b^2+1$, so $a^2-b^2+c^2=\boxed{(\text{B})1}$.

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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