Difference between revisions of "2002 AMC 10B Problems/Problem 20"
(→Solution 2) |
m (→Solution 2: added WLOG addendum) |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 15: | Line 15: | ||
===Solution 2=== | ===Solution 2=== | ||
− | The easiest way is to assume a value for <math>a</math> and then | + | The easiest way is to assume a value for <math>a</math> and then solve the system of equations. For <math>a = 1</math>, we get the equations |
− | |||
<math>-7b + 8c = 3</math> and | <math>-7b + 8c = 3</math> and | ||
− | |||
<math>4b - c = -1</math> | <math>4b - c = -1</math> | ||
− | |||
Multiplying the second equation by <math>8</math>, we have | Multiplying the second equation by <math>8</math>, we have | ||
− | |||
<math>32b - 8c = -8</math> | <math>32b - 8c = -8</math> | ||
− | |||
Adding up the two equations yields | Adding up the two equations yields | ||
− | |||
<math>25b = -5</math>, so <math>b = -\frac{1}{5}</math> | <math>25b = -5</math>, so <math>b = -\frac{1}{5}</math> | ||
− | |||
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. This is also called [[without loss of generality]] or WLOG. | |
− | 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 20:59, 10 August 2022
Problem
Let a, b, and c be real numbers such that and . Then is
Solution
Solution 1
Rearranging, we get and
Squaring both, and are obtained.
Adding the two equations and dividing by gives , so .
Solution 2
The easiest way is to assume a value for and then solve 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. This is also called without loss of generality or WLOG.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.