Let S be a square of unit area. Consider any quadrilateral which has one vertex on each side of S. If a,b,c and d denote the lengths of sides of the quadrilateral, prove that 2<=a^2+b^2+c^2+d^2<=4?

1 Answer
Jul 15, 2018

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Let ABCD be a square of unit area.

So AB=BC=CD=DA=1 unit.

Let PQRS be a quadrilateral which has one vertex on each side of the square. Here let PQ=b,QR=c,RS=dandSP=a

Applying Pythagoras thorem we can write

a^2+b^2+c^2+d^2

=x^2+y^2+(1-x)^2+(1-w)^2+w^2+(1-z)^2+z^2+(1-y)^2

=4+2(x^2+y^2+z^2+w^2-x-y-z-w)

=2+2(1+x^2+y^2+z^2+w^2-x-y-z-w)

=2+2((x-1/2)^2+(y-1/2)^2+(z-1/2)^2+(w-1/2)^2)

Now by the problem we have

0<= x<=1=> 0<=(x-1/2)^2<=1/4

0<= y<=1=> 0<=(y-1/2)^2<=1/4

0<=z<=1=> 0<=(z-1/2)^2<=1/4

0<= w<=1=> 0<=(w-1/2)^2<=1/4

Hence

2<=a^2+b^2+c^2+d^2<=4