If \(a-b > \sqrt{(b+c-a)^2}\) which of the following must be true?
A. \(a>0\)
B. \(a<b\)
C. \(b=0\)
D. \(c>0\)
E. \(c<0\)
Answer: D
Source: Economist
If \(a-b > \sqrt{(b+c-a)^2}\) which of the following must be true?
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Recognize that the right side is positive, so A must be greater than B.
Unfortunately, that is not one of the answer choices. Too bad.
For a given result on the right side, there are two ways to achieve this.
If B+C>A, then right side equals B+C-A.
For example, B=2, C=4 and A=5. So right side would equal 1.
So, A-B >B+C-A. Unfortunately, this doesn't help us since any manipulation of this won't yield any of the answer choices. Too bad again.
However, right side can also = 1 if B+C-A= -1, since squaring -1=1 and square root of 1=1.
B could=2, C=4 and A=7 for example.
But squaring this and taking the square root loses the relationship between A,B and C.
Fortunately B+C-A can be rewritten as A-B-C to yield +1 in this example.
So, in this situation, A-B>A-B-C
Subtract A and add B to both sides yields 0>-C. This is only true if
C>0 D
Unfortunately, that is not one of the answer choices. Too bad.
For a given result on the right side, there are two ways to achieve this.
If B+C>A, then right side equals B+C-A.
For example, B=2, C=4 and A=5. So right side would equal 1.
So, A-B >B+C-A. Unfortunately, this doesn't help us since any manipulation of this won't yield any of the answer choices. Too bad again.
However, right side can also = 1 if B+C-A= -1, since squaring -1=1 and square root of 1=1.
B could=2, C=4 and A=7 for example.
But squaring this and taking the square root loses the relationship between A,B and C.
Fortunately B+C-A can be rewritten as A-B-C to yield +1 in this example.
So, in this situation, A-B>A-B-C
Subtract A and add B to both sides yields 0>-C. This is only true if
C>0 D
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- Master | Next Rank: 500 Posts
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- Joined: Thu Oct 15, 2009 11:52 am
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Another way, probably better.
Rearrange B+C-A to be C-(A-B)
Call (A-B)= X
Now X > $$\sqrt{\left(C-X\right)}^2$$
Square both sides
X $$^2$$ > $$^{\left(C-X\right)^2}$$
So $$^{\left(X^2\right)}$$ > $$^{\left(C^2\right)}$$ -2XC+ $$^{\left(X^2\right)}$$
Subtract $$^{\left(X^2\right)}$$ from both sides
0>C(C-2X)
This can only be true if either
C>0 and (C-2X)<0
Or
C<0 and (C-2X)>0
Let's examine the second choice first
For C-2X >0 C must be> 2X
2X=2(A-B) from our substitution.
But A is greater than B because the right side of the problem statement is positive always, so 2(A-B) is positive.
But for C to be greater than 2(A-B) means therefore that C>0.
But C can't be >0 per the condition, creating a contradiction, so the second choice is eliminated, leaving only the first choice C>0 D
Rearrange B+C-A to be C-(A-B)
Call (A-B)= X
Now X > $$\sqrt{\left(C-X\right)}^2$$
Square both sides
X $$^2$$ > $$^{\left(C-X\right)^2}$$
So $$^{\left(X^2\right)}$$ > $$^{\left(C^2\right)}$$ -2XC+ $$^{\left(X^2\right)}$$
Subtract $$^{\left(X^2\right)}$$ from both sides
0>C(C-2X)
This can only be true if either
C>0 and (C-2X)<0
Or
C<0 and (C-2X)>0
Let's examine the second choice first
For C-2X >0 C must be> 2X
2X=2(A-B) from our substitution.
But A is greater than B because the right side of the problem statement is positive always, so 2(A-B) is positive.
But for C to be greater than 2(A-B) means therefore that C>0.
But C can't be >0 per the condition, creating a contradiction, so the second choice is eliminated, leaving only the first choice C>0 D