Sum of squared distances from a point on the diagonal

Text

Let $ABCD$ be a square. Extend the diagonal $AC$, on the side of $A$, by a segment $AP$ of length $2$. Determine the side of the square so that $PA^2+PB^2+PC^2+PD^2=20.$

Solution

Place $A(0;0)$, $B(\ell;0)$, $C(\ell;\ell)$, $D(0;\ell)$, with side $\ell$. The diagonal $AC$ has direction $(1;1)/\sqrt2$; $P$ is beyond $A$, opposite to $C$, at distance $2$: $P=\left(-\sqrt2;\,-\sqrt2\right)$. The four squares: $PA^2=2,\quad PB^2=(\ell+\sqrt2)^2+2,\quad PD^2=2+(\ell+\sqrt2)^2,\quad PC^2=(2+\ell\sqrt2)^2.$ Summing: $PA^2+PB^2+PC^2+PD^2=16+4\ell^2+8\sqrt2\,\ell.$ Imposing $=20$: $4\ell^2+8\sqrt2\,\ell-4=0\ \Longrightarrow\ \ell^2+2\sqrt2\,\ell-1=0\ \Longrightarrow\ \ell=\sqrt3-\sqrt2.$ (the negative root is discarded). Hence $\boxed{\ell=\sqrt3-\sqrt2\approx0{,}318.}$