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 da,b,candd denote the lengths of sides of the quadrilateral, prove that 2<=a^2+b^2+c^2+d^2<=42a2+b2+c2+d24?

1 Answer
Jul 15, 2018

drawndrawn
Let ABCDABCD be a square of unit area.

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

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

Applying Pythagoras thorem we can write

a^2+b^2+c^2+d^2a2+b2+c2+d2

=x^2+y^2+(1-x)^2+(1-w)^2+w^2+(1-z)^2+z^2+(1-y)^2=x2+y2+(1x)2+(1w)2+w2+(1z)2+z2+(1y)2

=4+2(x^2+y^2+z^2+w^2-x-y-z-w)=4+2(x2+y2+z2+w2xyzw)

=2+2(1+x^2+y^2+z^2+w^2-x-y-z-w)=2+2(1+x2+y2+z2+w2xyzw)

=2+2((x-1/2)^2+(y-1/2)^2+(z-1/2)^2+(w-1/2)^2)=2+2((x12)2+(y12)2+(z12)2+(w12)2)

Now by the problem we have

0<= x<=1=> 0<=(x-1/2)^2<=1/40x10(x12)214

0<= y<=1=> 0<=(y-1/2)^2<=1/40y10(y12)214

0<=z<=1=> 0<=(z-1/2)^2<=1/40z10(z12)214

0<= w<=1=> 0<=(w-1/2)^2<=1/40w10(w12)214

Hence

2<=a^2+b^2+c^2+d^2<=42a2+b2+c2+d24