2002 AMC 10B Problems/Problem 20

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.

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.

Page

Toolbox

Search

2002 AMC 10B Problems/Problem 20

Contents

Problem

Solution

Solution 1

Solution 2

See Also