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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>.
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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 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.
  
 
==See Also==
 
==See Also==

Revision as of 19:59, 10 August 2022

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 solve 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. 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

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