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

$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$

Solutions

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)

Solution 4 (bashy w/ substitution)

Notice that for both given equations, the coefficients for $a$ and $c$ are $1$ and $-1$ respectively, so we isolate $a$ and $c$ in their respective equations. We now have $a=7b-8c+4$ (let this be equation A) and $c=8a+4b-7$ (let this be equation B).

We can now substitute equation B for the value of $c$ in equation A, yielding the result $a=7b-64a-32b+60$ . Simplifying and separating the constant from the variables, this turns into $65a+26b=60$ .

Now, we subtract 8 times $8a+4b-c=7$ (the second given equation) from $65a+26b=60$ , yielding the result $a-6b+8c=4$ (let this be equation C).

We can then subtract (the first given equation) from equation C, yielding $b=0$ , which we can then substitute into $65a+26b=60$ , getting $65a=60$ and $a=(12/13)$ , which we then substitute again into $c=8a+4b-7$ , yielding $c=(5/13)$ . We can now plug in all of the values we found for $a$ , $b$ , and $c$ into $a^2=b^2+c^2$ , obtaining the solution $B$ .

please reformat my LaTex and make it less wordy - a2m

Video Solution

https://www.youtube.com/watch?v=3Oq21r5OezA ~David

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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