AMC 10 2002

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westin

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Hi,

This is an AMC 10 problem. I understand the first solution, however, I got a hard time understanding the second solution approach. I do not understand why you can just assume any number in "a" and can get the answer. What is the constraint using this assumption? and how can I see this in other similar question like that. and why when the answer is constant, we can assume this trick works.

Thanks
 

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I think Solution 2 makes sense only because the problem is a multiple choice one.
 
thank you for the reply. can you elaborate why if it is a multiple choice then we can use this assumption. I still not quite clear the reason behind the assumption. thanks.
 
This would be a question to whoever posted this problem to you. I.e., am I allowed to assume that the answer does not depend on [imath]a[/imath], or do I have to prove it. My interpretation of this multiple choice problem is that one has to pick one of the answers assuming that one of them is correct, which would imply that it does not depend on [imath]a[/imath]. But check with your teacher about what you are expected to do.
 
AMC 10 is a math contest problem and I don't have a teacher. I am just practicing the past test problems. I know that multiple choice can be done by elimination and one of the answer has to be correct. However, I like to understand why it does not depend on "a" if it is a multiple choice problem so that I can apply this "rule" if I see similar test problem in the future as it is much faster. Dr. Peterson, can you enlighten me when you have time? =)
 
blamocur said:
This would be a question to whoever posted this problem to you. I.e., am I allowed to assume that the answer does not depend on [imath]a[/imath], or do I have to prove it. My interpretation of this multiple choice problem is that one has to pick one of the answers assuming that one of them is correct, which would imply that it does not depend on [imath]a[/imath]. But check with your teacher about what you are expected to do.
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Before asking the teacher, I would run a test with a=0. See what do you get for b and and whether you can satisfy the given constraint.
 
westin said:
What is the constraint using this assumption?
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I think it works because you have a system of 2 equations and three variables. By assuming a=1, you reduce the system to two variables, while it does not violate the condition that a,b,c are real numbers. I think it would work for any assumption b=1, or c=1, or any other real value.
 
AvgStudent said:
I think it works because you have a system of 2 equations and three variables. By assuming a=1, you reduce the system to two variables, while it does not violate the condition that a,b,c are real numbers. I think it would work for any assumption b=1, or c=1, or any other real value.
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try a = 0
 
AvgStudent said:
I think it works because you have a system of 2 equations and three variables. By assuming a=1, you reduce the system to two variables, while it does not violate the condition that a,b,c are real numbers. I think it would work for any assumption b=1, or c=1, or any other real value.
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To expand on what I mean, the problem presented a system of 2 equations and 3 variables, which is unsolvable. Assuming that a=1, the system reduces to 2 equations and 2 variables, which is now solvable. Of course, the assumption a=1 is arbitrary, it can be any real number, and it doesn't have to be a. Obviously, making a different arbitrary assumption for a will yield different results for b and c, but we only care what their relationships [imath]a^2-b^2+c^2[/imath], while still satisfying the 2 equations and the real number value condition.
In short, you can make any assumption you want, as long as it does not violate any conditions.
 
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Yes, I got it now. Thanks avgstudent. 2 equation 3 unknown is unsolvable. But asking for a relationship between 3 unknown can use an arbitrary number for one of the unknowns to calculate the relationship. This will be a great method for future similar problem as it is much faster to expand it. Thank you all for the help. Have a great day!!!
 
westin said:
Yes, I got it now. Thanks avgstudent. 2 equation 3 unknown is unsolvable. But asking for a relationship between 3 unknown can use an arbitrary number for one of the unknowns to calculate the relationship. This will be a great method for future similar problem as it is much faster to expand it. Thank you all for the help. Have a great day!!!
Click to expand...
No problem, first time I'm seeing this method. I'm learning as well :).
 
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