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

drawn
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#