Difference between revisions of "1997 AHSME Problems/Problem 3"

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==Problem 3==
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If <math>x</math>, <math>y</math>, and <math>z</math> are real numbers such that
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<center><math>(x-3)^2 + (y-4)^2 + (z-5)^2 = 0</math>,</center>
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then <math>x + y + z =</math>
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<math> \mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \  } 8 \qquad \mathrm{(D) \  } 12 \qquad \mathrm{(E) \  }50  </math>
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==Solution==
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If the sum of three squared expressions is zero, then each expression itself must be zero, since <math>a^2 \ge 0</math> with the equality iff <math>a=0</math>.
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In this case, <math>x-3=0</math>, <math>y-4=0</math>, and <math>z-5=0</math>.  Adding the three equations and moving the constant to the right gives <math>x + y + z = 12</math>, and the answer is <math>\boxed{D}</math>.
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== See also ==
 
== See also ==
{{AHSME box|year=1998|num-b=2|num-a=4}}
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{{AHSME box|year=1997|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 14:12, 5 July 2013

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
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 26 27 28 29 30
All AHSME Problems and Solutions

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