Please don't use just algebra
Let a , b ∈ R be fixed, with a < 0 and b > 0 . Then suppose that f ( α , β ) = ( x − α ) 2 + ( y − β ) 2 and some x , y ∈ R give the minimum possible value of the expression below:
f ( 1 , 1 ) + f ( − 1 , − 1 ) + f ( 1 , − 1 ) + f ( a , b )
Then we find that
x + y = 2 b − a + K a + b
Find K.
The answer is 2.
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1 solution
This assumes that the line segment joining ( 1 , 1 ) and ( − 1 , − 1 ) , and the line segment joining ( a , b ) and ( 1 , − 1 ) intersect. This would not be the case, for example, when ( a , b ) = ( 2 , − 2 ) .
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Yeah, I think Jon Haussmann is right. @Dylan Pentland , your problem statements needs fixing.
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Yup, I forgot to include that a , b was in the second quadrant. (for some reason these recent problems I have posted seem to be full of small errors, I suppose I should proofread more before posting)
Observe that minimizing the expression is equivalent to finding the point ( x , y ) with a minimum sum of distances to the points ( 1 , 1 ) , ( − 1 , − 1 ) , ( 1 , − 1 ) , ( a , b ) . This point is the intersection of the diagonals from ( a , b ) to ( 1 , − 1 ) and from ( 1 , 1 ) to ( − 1 , − 1 ) - we can see that if a point is off either diagonal, it would be better if it were on the diagonal by the triangle inequality.
Hence, we find the intersection of y = x and y − b = a − 1 b + 1 ( x − a ) . We let y = x and solve for x , getting
x = y = b − a + 2 a + b
Evidently, K = 2 .