500 followers problem-3

Algebra Level 4

Let a a and b b be positive real numbers. Find the minimum possible value of a 2 + b 2 + ( a − 1 ) 2 + b 2 + a 2 + ( b − 1 ) 2 + ( a − 1 ) 2 + ( b − 1 ) 2 . \sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2} .

Enter your answer to 2 decimal places.


The answer is 2.82842712474619.

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1 solution

Let ABCD be a square with A = ( 0 , 0 ) , B = ( 1 , 0 ) , C = ( 1 , 1 ) , D = ( 0 , 1 ) A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1) , and P be a point in the same plane as ABCD. Then the desired expression is equivalent to A P + B P + C P + D P AP + BP + CP + DP .

By the triangle inequality, A P + C P ≥ A C AP + CP \ge AC and B P + D P ≥ B D BP + DP ≥ BD , adding the two inequalities we get the minimum possible value of the expression as A C + B D = 2 2 ≈ 2.82 AC + BD = 2\sqrt{2} \approx 2.82 .

Equality holds if and only if P = ( 1 2 , 1 2 P=(\frac{1}{2},\frac{1}{2} ) or a = b = 1 2 a=b=\frac{1}{2} .

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And furthemore this equality is reached at P = ( 1 2 , 1 2 ) P =(\frac{1}{2}, \frac{1}{2})

Guillermo Templado - 5 years, 3 months ago

did it the same way

aryan goyat - 5 years, 3 months ago

Did the same way

I Gede Arya Raditya Parameswara - 4 years, 1 month ago

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