Difference between revisions of "2002 AMC 12P Problems/Problem 18"
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== Solution 1== | == Solution 1== | ||
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> | 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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== 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}} | ||
Revision as of 01:40, 31 December 2023
Problem
If 
are real numbers such that 
, 
and 
, find 
Solution 1
Adding all of the equations gives 
Adding 14 on both sides gives 
Notice that 14 can split into 
and 
which coincidentally makes 
and 
into perfect squares. Therefore, 
An easy solution to this equation is 
and 
Plugging in that solution, we get 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
