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>. | + | 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 and . Then is
Solution
and
Squaring both, and are obtained.
Adding the two equations and dividing by gives , so .
See Also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
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 |
[[Category:]] The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.