# 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. This is also called without loss of generality or WLOG.

## Solution 3 (fakesolve)

Notice that the coefficients of $a$ and $c$ are pretty similar (15s for reading and noticing), so let $b=0$ gives $a+8c=4$, and $8a-c=7$ (10s writing). Since the desired quantity simplifies to $a^2+c^2$, the $ac$ term of the quadratics after squaring gets canceled by adding up the squares of the two equations because they have the same coefficients but opposite sign (15s mind-binom). This simplifies to $65(a^2+c^2)=16+49$, or $a^2-b^2+c^2=\frac{65}{65}=\boxed{1}$(15s writing and addition and fraction simplification and (B) circling and submission)

## Video Solution

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