Orthodiagonal quadrilateral

Question

Is it always possible to construct an orthodiagonal quadrilateral such that the diagonals and perimeter are fixed? More specifically, given 2 fixed diagonals, how is the perimeter of the quadrilateral bounded?

Answer 1

In the above configuration, let $OA=a,OB=b,OC=c,OD=d$ and assume $b>d$.
We may show that if both $b$ and $d$ are replaced by $\frac{b+d}{2}$ (that brings $B$ to $B'$ and $D$ to $D'$), the perimeter of $ABCD$ decreases. This is a consequence of the inequalities $$ \sqrt{a^2+b^2}+\sqrt{a^2+d^2}\geq 2\sqrt{a^2+\left(\frac{b+d}{2}\right)^2} $$ $$ \sqrt{c^2+b^2}+\sqrt{c^2+d^2}\geq 2\sqrt{c^2+\left(\frac{b+d}{2}\right)^2} $$ that follow from the convexity of the functions $g_a(x)=\sqrt{a^2+x^2}$ and $g_c(x)=\sqrt{c^2+x^2}$.
The minimum perimeter is so achieved by the rhombus and the maximum perimeter is achieved by a degenerate orthodiagonal quadrilateral that is a right triangle. If we call $d_1$ and $d_2$ the lengths of the diagonals, $$\boxed{ 2\sqrt{d_1^2+d_2^2}\leq p(ABCD)\leq d_1+d_2+\sqrt{d_1^2+d_2^2} } $$ follows from the Pythagorean theorem.
By continuity, any perimeter belonging to such interval is achievable.