Difference between revisions of "2002 AMC 10B Problems/Problem 20"
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We obtain <math>c = \frac{1}{5}</math> after plugging in the value for <math>b</math>. | We obtain <math>c = \frac{1}{5}</math> after plugging in the value for <math>b</math>. | ||
| − | Therefore, <math>a^2-b^2+c^2 = 1-\frac{1}{25}+\frac{1}{25}=\boxed{1}</math> which corresponds to <math>\text{(B)}</math> | + | Therefore, <math>a^2-b^2+c^2 = 1-\frac{1}{25}+\frac{1}{25}=\boxed{1}</math> which corresponds to <math>\text{(B)}</math>. |
| + | |||
| + | This time-saving trick works only because we know that for any value of <math>a</math>, <math>a^2-b^2+c^2</math> will always be constant (it's a contest), so any value of <math>a</math> will work. | ||
| + | . | ||
==See Also== | ==See Also== | ||
Revision as of 12:31, 30 November 2019
Contents
Problem
Let a, b, and c be real numbers such that 
and 
. Then 
is
Solution
Rearranging, we get 
and 
Squaring both, 
and 
are obtained.
Adding the two equations and dividing by 
gives 
, so 
.
Easiest Solution
The easiest way is to assume a value for 
and then solving the system of equations. For 
, we get the equations
and
Multiplying the second equation by 
, we have
Adding up the two equations yields
, so 
We obtain 
after plugging in the value for 
.
Therefore, 
which corresponds to 
.
This time-saving trick works only because we know that for any value of , will always be constant (it's a contest), so any value of will work.
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
