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

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== Problem ==
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==Problem==
 
 
 
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 ==
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==Solution==
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===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 ==
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===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 19:13, 1 February 2020

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

Solution 1

Rearranging, we get $a+8c=7b+4$ 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 $a^2-b^2+c^2=\boxed{(\text{B})1}$.

Solution 2

The easiest way is to assume a value for $a$ and then solving the system of equations. For $a = 1$, we get the equations

$-7b + 8c = 3$ and

$4b - c = -1$

Multiplying the second equation by $8$, we have

$32b - 8c = -8$

Adding up the two equations yields

$25b = -5$, so $b = -\frac{1}{5}$

We obtain $c = \frac{1}{5}$ after plugging in the value for $b$.

Therefore, $a^2-b^2+c^2 = 1-\frac{1}{25}+\frac{1}{25}=\boxed{1}$ which corresponds to $\text{(B)}$.

This time-saving trick works only because we know that for any value of $a$, $a^2-b^2+c^2$ will always be constant (it's a contest), so any value of $a$ will work.

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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