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?

Jul 15, 2018

Let $A B C D$ be a square of unit area.

So $A B = B C = C D = D A = 1$ unit.

Let $P Q R S$ be a quadrilateral which has one vertex on each side of the square. Here let $P Q = b , Q R = c , R S = \mathrm{da} n \mathrm{dS} P = a$

Applying Pythagoras thorem we can write

${a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}$

$= {x}^{2} + {y}^{2} + {\left(1 - x\right)}^{2} + {\left(1 - w\right)}^{2} + {w}^{2} + {\left(1 - z\right)}^{2} + {z}^{2} + {\left(1 - y\right)}^{2}$

$= 4 + 2 \left({x}^{2} + {y}^{2} + {z}^{2} + {w}^{2} - x - y - z - w\right)$

$= 2 + 2 \left(1 + {x}^{2} + {y}^{2} + {z}^{2} + {w}^{2} - x - y - z - w\right)$

$= 2 + 2 \left({\left(x - \frac{1}{2}\right)}^{2} + {\left(y - \frac{1}{2}\right)}^{2} + {\left(z - \frac{1}{2}\right)}^{2} + {\left(w - \frac{1}{2}\right)}^{2}\right)$

Now by the problem we have

$0 \le x \le 1 \implies 0 \le {\left(x - \frac{1}{2}\right)}^{2} \le \frac{1}{4}$

$0 \le y \le 1 \implies 0 \le {\left(y - \frac{1}{2}\right)}^{2} \le \frac{1}{4}$

$0 \le z \le 1 \implies 0 \le {\left(z - \frac{1}{2}\right)}^{2} \le \frac{1}{4}$

$0 \le w \le 1 \implies 0 \le {\left(w - \frac{1}{2}\right)}^{2} \le \frac{1}{4}$

Hence

$2 \le {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} \le 4$