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Difference between revisions of "2002 AMC 12P Problems/Problem 18"

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== Solution 1==
 
== Solution 1==
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Adding all of the equations gives <math>a^2 + b^2 +c^2 + 6a + 2b + 4c=-14.</math> Adding 14 on both sides gives <math>a^2 + b^2 +c^2 + 6a + 2b + 4c+14=0.</math> Notice that 14 can split into <math>9, 1,</math> and <math>4,</math> which coincidentally makes <math>a^2 +6a, b^2+2b,</math> and <math>c^2+4c</math> into perfect squares. Therefore, <math>(a+3)^2 + (b+1)^2 + (c+2) ^2 =0.</math> An easy solution to this equation is <math>a=-3, b=-1,</math> and <math>c=-2.</math> Plugging in that solution, we get <math>a^2+b^2+c^2=-3^2+-1^2+-2^2=\boxed{\textbf{(B) } 14}.</math>
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Adding all of the equations gives <math>a^2 + b^2 +c^2 + 6a + 2b + 4c=-14.</math> Adding 14 on both sides gives <math>a^2 + b^2 +c^2 + 6a + 2b + 4c+14=0.</math> Notice that 14 can split into <math>9, 1,</math> and <math>4,</math> which coincidentally makes <math>a^2 +6a, b^2+2b,</math> and <math>c^2+4c</math> into perfect squares. Therefore, <math>(a+3)^2 + (b+1)^2 + (c+2) ^2 =0.</math> An easy solution to this equation is <math>a=-3, b=-1,</math> and <math>c=-2.</math> Plugging in that solution, we get <math>a^2+b^2+c^2=-3^2+-1^2+-2^2=\boxed{\textbf{(A) } 14}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=17|num-a=19}}
 
{{AMC12 box|year=2002|ab=P|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:40, 31 December 2023

Problem

If $a,b,c$ are real numbers such that $a^2 + 2b =7$, $b^2 + 4c= -7,$ and $c^2 + 6a= -14$, find $a^2 + b^2 + c^2.$

$\text{(A) }14 \qquad \text{(B) }21 \qquad \text{(C) }28 \qquad \text{(D) }35 \qquad \text{(E) }49$

Solution 1

Adding all of the equations gives $a^2 + b^2 +c^2 + 6a + 2b + 4c=-14.$ Adding 14 on both sides gives $a^2 + b^2 +c^2 + 6a + 2b + 4c+14=0.$ Notice that 14 can split into $9, 1,$ and $4,$ which coincidentally makes $a^2 +6a, b^2+2b,$ and $c^2+4c$ into perfect squares. Therefore, $(a+3)^2 + (b+1)^2 + (c+2) ^2 =0.$ An easy solution to this equation is $a=-3, b=-1,$ and $c=-2.$ Plugging in that solution, we get $a^2+b^2+c^2=-3^2+-1^2+-2^2=\boxed{\textbf{(A) } 14}.$

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 12 Problems and Solutions

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