Let $S$ be a square of unit area. Consider any quadrilateral which has one vertex on each side of $S

Question

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 the sides of the quadrilateral, then

Professor PC Thomas Classes · Answer

Let $A B C D$ be the required quadrilateral. Let $A, B, C, D$ are the co-ordinates $(p, 0)$, $(1, q),(r, 1)$ and $(0, s)$ respectively.\ In $	riangle O A D$, we have\ $O A^{2}+O D^{2}=A D^{2}$\ $\Rightarrow p^{2}+s^{2}=a^{2}$\ Similarly,\ $b^{2}=q^{2}+(1-p)^{2}$\ $c^{2}=(1-q)^{2}+(1-r)^{2}$\ $d^{2}=r^{2}+(1-s)^{2}$\ $	herefore \quad a^{2}+b^{2}+c^{2}+d^{2}$ $$ \begin{aligned} & p^2+(1-p)^{2}+q^{2}+(1-q)^{2}+r^{2} \ & +(1-r)^{2}+s^{2}+(1-s)^{2} \end{aligned} $$ At $x=1, f^{\prime}(x)$ is minimum Consider the function\ $f(x)=t^{2}+(1-t)^{2}, 0 \leq t \leq t$\ Obviously $f(x)$ has maximum value 1 at $t=0$, 1. Also $f(x)$ has minimum value $1 / 2$ at $t=1 / 2$\ $	herefore$ By (i) $a^{2}+b^{2}+c^{2}+d^{2} \leq 1+1+1+1=4$\ Also $\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2} \leq a^{2}+b^{2}+c^{2}+d^{2}$\ Combining these two, we have\ $2 \leq a^{2}+b^{2}+c^{2}+d^{2} \leq 4$\