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=1997|num-b=2|num-a=4}} | {{AHSME box|year=1997|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:12, 5 July 2013
Problem 3
If 
, 
, and 
are real numbers such that
,then 
Solution
If the sum of three squared expressions is zero, then each expression itself must be zero, since 
with the equality iff 
.
In this case, 
, 
, and 
. Adding the three equations and moving the constant to the right gives 
, and the answer is 
.
See also
| 1997 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
