IMO 1996 SL G6

Origin: ARM | Category: Geometry

Problem

Let the sides of two rectangles be ${a, b}$ and ${c, d}$ with

$a < c \leq d < b$ and $ab < cd$. Prove that the first rectangle can be placed

within the second one if and only if

$(b^2 - a^2)^2 \leq (bd - ac)^2 + (bc - ad)^2.$

Solution

Denote by $ABCD$ and $EFGH$ the two rectangles, where $AB = a$, $BC =

b$, $EF = c$, and $FG = d$. Obviously, the first rectangle can be placed

within the second one with the angle $\alpha$ between $AB$ and $EF$ if and only

if

$$\begin{cases} a \cos \alpha + b \sin \alpha \leq c,\ a \sin \alpha + b \cos \alpha \leq d. \end{cases} \tag{1}$$

Hence $ABCD$ can be placed within $EFGH$ if and only if there is an

$\alpha \in [0, \pi/2]$ for which \eqref{1} holds.

The lines $l_1: ax + by = c$ and $l_2: bx + ay = d$ and the axes $x$ and $y$ bound

a region $R$. By \eqref{1}, the desired placement of the rectangles is possible if

and only if $R$ contains some point $(\cos \alpha, \sin \alpha)$ of the unit circle centered

at the origin $(0, 0)$. This in turn holds if and only if the intersection point

$L$ of $l_1$ and $l_2$ lies outside the unit circle. It is easily computed that $L$ has

coordinates

$$\left( \frac{bd - ac}{b^2 - a^2}, \frac{bc - ad}{b^2 - a^2} \right).$$

Now $L$ being outside the unit circle is exactly

equivalent to the inequality we want to prove.

Remark. If equality holds, there is exactly one way of placing. This happens,

for example, when $(a, b) = (5, 20)$ and $(c, d) = (16, 19)$.

Second remark. This problem is essentially very similar to (SL89-2).