Difference between revisions of "2002 AMC 12P Problems/Problem 18"
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| + | {{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #18]] and [[2002 AMC 10P Problems|2002 AMC 10P #19]]}} | ||
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== Problem == | == Problem == | ||
If <math>a,b,c</math> are real numbers such that <math>a^2 + 2b =7</math>, <math>b^2 + 4c= -7,</math> and <math>c^2 + 6a= -14</math>, find <math>a^2 + b^2 + c^2.</math> | If <math>a,b,c</math> are real numbers such that <math>a^2 + 2b =7</math>, <math>b^2 + 4c= -7,</math> and <math>c^2 + 6a= -14</math>, find <math>a^2 + b^2 + c^2.</math> | ||
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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{( | + | 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> |
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== See also == | == See also == | ||
| + | {{AMC10 box|year=2002|ab=P|num-b=18|num-a=20}} | ||
{{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 13:08, 19 July 2024
- The following problem is from both the 2002 AMC 12P #18 and 2002 AMC 10P #19, so both problems redirect to this page.
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 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 10 Problems and Solutions | ||
| 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. 
