Again, minimum value
x 2 + y 2 − x y + ( 1 − x ) 2 + ( 2 − y ) 2 − ( 1 − x ) ( 2 − y )
If 0 < x < 1 and 0 < y < 2 , then the minimum value of the expression above is s . What is ⌊ 1 0 0 0 s ⌋ ?
The answer is 1732.
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3 solutions
How did you obtained the expression from the geometry?
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it is usually linked to Cosine Rule if we have x 2 + y 2 − x y , as we have the formula z 2 = x 2 + y 2 − 2 x y cos θ .
Awesome solution sir ,(+1)!
I found an elegant algebraic proof. We notice the function is symmetrical across (1-x) and (2-y), in such f(1-x,2-y)=f(x,y). Hence a minimum must be obtained at the point which is distinctly reached once. So 1-x=x, 2-y=y. Now solve and plug them in.
Good point. Thanks for sharing.
We get x=1/2 and y =1, which implies the value of the above expression √2
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Check your computation.
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Are my values of x and y right?
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@Puneet Pinku – Yes. Plug in 1/2 and 1. You get 2*sqrt(3/4), which is sqrt(3). Where is sqrt(2) coming from?
We notice the function is symmetrical ... such f(1-x,2-y)=f(x,y). Hence a minimum must be obtained at the point which is distinctly reached once.
That statement is not true. For example, the function f ( x ) = x 2 ( 1 − x ) 2 is symmetrical such that f ( 1 − x ) = f ( x ) . However, it doesn't have a "distinctly reached minimum".
Please avoid making the mistake of "Oh this is an inequality so the conditions must be that all my nice variables are equal". Check out inequalities with strange equality conditions .
x^2 + y^2 - xy = (x - y/2)^2 + 3y^2/4
>= 3y^2/4, equality happens when y = 2x,
Likewise,
(1-x)^2+(2-y)^2 - (1-x)(2-y)
=x^2 + y^2 - xy - 3y + 3
=(x - y/2)^2 + 3(1 - y + y^2/4)
=(x - y/2)^2 + (3/4)(2 - y)^2
>= (3/4)(2 - y)^2, the equality happens in cases when y = 2x, therefore, the expression in question,
>= √{(3/4)(y^2)} + √{(3/4)(2 - y)^2},
>= (√3/2)(y + 2 - y) = √3,
Hence, the minimum value of expression is
√3 and that occurs when y = 2x.
As shown in the above figure, let A B = 1 , C D = 2 make a 6 0 ∘ and intersect at E . Let E B = x and E D = y . Then A C + B D = x 2 + y 2 − x y + ( 1 − x ) 2 + ( 2 − y ) 2 − ( 1 − x ) ( 2 − y )
Now construct B F such that B F = D C . This means that C F = D B and so A C + B D = A C + C F ≤ A F = A B 2 + B F 2 − 2 ( A B ) ( B F ) cos 6 0 ∘ = 3
Hence ⌊ 1 0 0 0 s ⌋ = ⌊ 1 0 0 0 3 ⌋ = 1 7 3 2 .
Any algebraic proof?