Team Selection Test for IMO

Let the quadrilateral $EFGH$ be inscribed in the triangle $ABC$ so that $E$ and $F$ lie on $BC$ and $G$ lies on $AC$ and $H$ lies on $AB$. Let us denote the side-lengths of the triangle $ABC$ by $a$, $b$ and $c$ and let $h$ denote the length of the height drawn from $A$ to $BC$. We put $\overline{AH}=x$, $\overline{EF}=u$ and $\overline{FG}=v$.

From $△AHG\sim △ABC$ we have $u=\frac{ax}{c}$, from $△BEH\sim △BVA$ we get $v=\frac{h(c-x)}{c}$. If $l$ denotes the length of the diagonal of $EFGH$, we get $$l^2=\frac{a^2{x}^2}{c^2}+\frac{h^2(c-x{)}^2}{c^2}.$$ The smallest value of the parabola $$f(x)=\frac{a^2{x}^2}{c^2}+\frac{h^2(c-x{)}^2}{c^2}$$ is $\frac{a^2{h}^2}{a^2+h^2}$ and it is attained when $x=\frac{h^{2}c}{a^2+h^2}$. Let us note that $$\frac{a^2{h}^2}{a^2+h^2}=\frac{4P^2}{a^2+\frac{4P^2}{a^2}},$$ where $P$ is the area of the triangle. Now if we do the same when the rectangle has two neighbouring vertices lying on the side $AC$, we get $\frac{4P^2}{b^2+\frac{4P^2}{b^2}}$ for the smallest possible length of the diagonal.

We have $$\frac{a^2+\frac{4P^2}{a^2}-b^2-\frac{4P^2}{b^2}}{a^2+\frac{4P^2}{b^2}}=(a^2-b^2)\left(1-\frac{4P^2}{a^2{b}^2}\right).$$ Since $ab\ge 2P$ the last expression is greater or equal to $0$ if and only if $a\ge b$. In other words, the smallest value of $l$ is obtained when the rectangle has two neighbouring vertices lying on the longest side of the triangle. Let $a$ be the longest side. We already saw that the smallest value of $l$ is $$\frac{2P}{\sqrt{a^2+\frac{4P^2}{a^2}}}$$ and it is obtained when $\overline{AH}=x=\frac{h^{2}c}{a^2+h^2}=\frac{4P^{2}c}{a^4+4P^2}$. Let us note that the area is known and it can be expressed through the side-lengths of the triangle by e.g. Heron's formula.