## Task $6 / 83$

In every rectangle, the angle bisectors intersect at four points that span a square (if the rectangle is a square, these four points coincide).

The area $A_{Q}$ of this square is to be represented as a function of the side ratio $x=a: b$ (where $a>b, b$ is constant). For which side ratio is the square area $A_{Q}$ equal to the rectangle area $A_{R}$?
Let $A B C D$ be the rectangle, and the sides be $A B=C D=a, B C=D A=b$. Furthermore, let $A^{\prime}$ be the intersection of the angle bisectors through $A$ and $B$; similarly, $B^{\prime}, C^{\prime}$, and $D^{\prime}$ are the intersections of the angle bisectors through $B$ and $C$, $C$ and $D$, and $D$ and $A$, respectively. Then we have

$$
A A^{\prime}=B A^{\prime}=C C^{\prime}=D C^{\prime}=\frac{a}{2} \sqrt{2} \quad \text { and } \quad A D^{\prime}=D D^{\prime}=B B^{\prime}=C B^{\prime}=\frac{b}{2} \sqrt{2}
$$

thus

$$
A^{\prime} B^{\prime}=A A^{\prime}-C B^{\prime}=\frac{a}{2} \sqrt{2}-\frac{b}{2} \sqrt{2}=\frac{1}{2} \sqrt{2}(a-b)
$$

If we set $a=b x$ with $x>1$, then for the area $A_{Q}$ of the square we have

$$
A_{Q}=\frac{1}{2} b^{2}(x-1)^{2}=f(x)
$$

For the area $A_{R}$ of the rectangle, we have $A_{R}=a b=b^{2} x$. By equating, we obtain (among other things because $b \neq 0$)

$$
b^{2} x=\frac{1}{2} b^{2}(x-1)^{2} \rightarrow 0=x^{2}-4 x+1 \rightarrow x_{1 ; 2}=2 \pm \sqrt{3}
$$

Since $x>1$, $x_{2}$ is discarded. Equality thus only occurs for $x=a: b=2+\sqrt{3} \approx 3.73$.