Question

Let $S$ be a square with unit area. Consider any quadrilateral which has one vertex on each side of $S$. If $a,b,c,d$ denotes the lengths of the sides of the quadrilateral, then $\alpha \le a^2+b^2+c^2+d^2\le \beta$ where

• A. $\alpha =1$
• B. $\beta =2$
• C. $\alpha =2$
• D. $\beta =4$

Answer 1

Answer: (C), (D)

The correct options are
C $\alpha =2$
D $\beta =4$

Let the squares of unit area be bounded by the line $x=\pm \frac{1}{2},y=\pm \frac{1}{2}$
$A(x_1,1/2),B(1/2,y_1),C(-x_1,-1/2),D(-1/2,y_2)$
$a^2={(x_1-1/2)}^2+{(1/2-y_1)}^2=x_1^2+y_1^2-x_1-y_1+1/2$
$b^2=x_2^2+y_1^2-x_2-y_1+1/2$
$b^2=x_2^2+y_1^2-x_2-y_1+1/2$
$d^2=x_1^2+y_2^2-x_1-y_2+1/2$
$\Rightarrow a^2+b^2+c^2+d^2=2(x_1^2+x_2^2+y_1^2+y_2^2)+2$
As $0\le x_1^2,x_2^2,y_1^2,y_2^2\le \frac{1}{4}$
$0\le x_1^2+x_2^2+y_1^2+y_2^2\le 1$
Thus $2\le a^2+b^2+c^2+d^2\le 4so\alpha =2and\beta =4$