1997 AHSME Problems/Problem 3

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Problem 3

If $x$, $y$, and $z$ are real numbers such that

$(x-3)^2 + (y-4)^2 + (z-5)^2 = 0$,

then $x + y + z =$

$\mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \  } 8 \qquad \mathrm{(D) \  } 12 \qquad \mathrm{(E) \  }50$

Solution

If the sum of three squared expressions is zero, then each expression itself must be zero, since $a^2 \ge 0$ with the equality iff $a=0$.

In this case, $x-3=0$, $y-4=0$, and $z-5=0$. Adding the three equations and moving the constant to the right gives $x + y + z = 12$, and the answer is $\boxed{D}$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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