Difference between revisions of "2002 AMC 10B Problems/Problem 20"
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| − | == Problem == | + | ==Problem== |
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Let a, b, and c be real numbers such that <math>a-7b+8c=4</math> and <math>8a+4b-c=7</math>. Then <math>a^2-b^2+c^2</math> is | Let a, b, and c be real numbers such that <math>a-7b+8c=4</math> and <math>8a+4b-c=7</math>. Then <math>a^2-b^2+c^2</math> is | ||
<math> \mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8 </math> | <math> \mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8 </math> | ||
| − | == Solution == | + | ==Solution== |
| + | ===Solution 1=== | ||
Rearranging, we get <math>a+8c=7b+4</math> and <math>8a-c=7-4b</math> | Rearranging, we get <math>a+8c=7b+4</math> and <math>8a-c=7-4b</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>. | 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>. | ||
| − | == Solution 2 == | + | ===Solution 2=== |
The easiest way is to assume a value for <math>a</math> and then solving the system of equations. For <math>a = 1</math>, we get the equations | The easiest way is to assume a value for <math>a</math> and then solving the system of equations. For <math>a = 1</math>, we get the equations | ||
Revision as of 18:13, 1 February 2020
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 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
