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Geometry Level 4

A = 2 x 2 − 12 x + 20 + 2 x 2 − 8 x + 10 \large A = \sqrt {2x^{2} - 12x + 20} + \sqrt {2x^{2} - 8x +10}

For positive real number x x , let the minimum value of A A be y y . Find y 2 y^{2} .


The answer is 10.000.

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Harsh Shrivastava by Harsh Shrivastava , 20, India

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1 solution

Rajen Kapur
May 11, 2015

A = 2 x 2 − 12 x + 20 + 2 x 2 − 8 x + 10 = ( x − 2 ) 2 + ( x − 4 ) 2 + ( x − 1 ) 2 + ( x − 3 ) 2 \large\displaystyle A = \sqrt{2x^2 - 12x +20} + \sqrt{2x^2 - 8 x + 10}\\= \sqrt{(x - 2)^2 + (x - 4)^2} + \sqrt{(x - 1)^2 + (x - 3)^2} A geometrical interpretation of A could be the sum of distances from a point on y = x from (2,4) and (3, 1). The two points are on the opposite sides of the line y = x. As a straight line is the minimum distance which when squared equals (2 - 3)^2 + (4 - 1)^2 = 10. Please see my question The minimum you can do.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Extremely well approached!!

Aran Pasupathy - 6 years ago

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