These numbers... where have I seen them?

Algebra Level 5

Given that $0<x<2,\quad 0<y<2$ ,

Let the minimum value of $\sqrt { 2{ x }^{ 2 }+2{ y }^{ 2 } } +\sqrt { { y }^{ 2 }+{ x }^{ 2 }-4y+4 } +\sqrt { { x }^{ 2 }+{ y }^{ 2 }-4y-4x+8 }$ be $a\sqrt { b }$ such that $b$ is square free. Find $a+b$ .

The answer is 7.

1 solution

Joel Tan
Sep 17, 2014

Let $A$ be the distance from $(0, 0)$ to $(x+y, y-x)$ on the coordinate plane. Let $B$ be the distance from $(x+y, y-x)$ to $(x+2, y)$ . Let $C$ be the distance from $(x+2, y)$ to $(4, 2)$ . Then $A+B+C$ , by triangle inequality, is not less than the length from $(0, 0)$ to $(4, 2)$ which is just $\sqrt {20}=2\sqrt {5}$ Equality can occur when the points $(0, 0), (x+y, y-x), (x+2, y), (4, 2)$ are in a straight line, or when $(x, y)=(\frac {2}{5}, \frac {6}{5})$ .

could u pls tell about what u wrote about A+B+C triangle inequality in a more detail!! @Joel Tan

A Former Brilliant Member - 6 years, 8 months ago

A+B+C is just the huge expression in the question. This is by pythagoras theorem (distance between two points = [(difference in x coordinates)^2+(difference in y coordinates)^2]^0.5) The shortest path from the origin to the final destination is a straight line as this is a plane.

Joel Tan - 6 years, 8 months ago

These numbers... where have I seen them?

The answer is 7.

1 solution

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