A rhombus has diagonals of length $2a$ and $2b$. The tangents of the inscribed circle parallel to the diagonals and the sides of the rhombus form an octagon. Express the area of the octagon in terms of $a$ and $b$.
I. solution. We will calculate the area $t$ of the octagon by subtracting the area of the four triangles cut off by the tangents from the area $T=2 a b$ of the rhombus. Let's use the notations of the diagram. Let the area of the triangle $A E B$ be $t_{1}$, and the area of the triangle $C F D$ be $t_{2}$. It is easy to see that $t_{1}=\frac{(b-x)(a-r)}{2}$ and $t_{2}=\frac{(a-y)(b-r)}{2}$. By the theorem of parallel chords, $\frac{x}{r}=\frac{b}{a}$, from which $x=\frac{b r}{a}$, and similarly $y=\frac{a r}{b}$. Considering that the side of the rhombus is $\sqrt{a^{2}+b^{2}}$, the double area of the triangle $A O D$ can be calculated in two ways: $a b=r \sqrt{a^{2}+b^{2}}$, so $r=\frac{a b}{\sqrt{a^{2}+b^{2}}}$. With these values, the sum of the areas of the four triangles cut off by the tangents is:

$$
\begin{gathered}
4 t_{1}+4 t_{2}=2\left[(a-r)\left(b-\frac{b r}{a}\right)+(b-r)\left(a-\frac{a r}{b}\right)\right]= \\
=2\left(2 a b-2 a r-2 b r+r^{2} \frac{a^{2}+b^{2}}{a b}\right)= \\
=2\left(2 a b-2(a+b) \frac{a b}{\sqrt{a^{2}+b^{2}}}+\frac{a^{2} b^{2}}{a^{2}+b^{2}} \cdot \frac{a^{2}+b^{2}}{a b}\right)=2 \cdot\left(3 a b-2 a b \frac{a+b}{\sqrt{a^{2}+b^{2}}}\right)
\end{gathered}
$$

The area of the octagon:

$$
t=T-\left(4 t_{1}+4 t_{2}\right)=2 a b-2 \cdot\left(3 a b-2 a b \frac{a+b}{\sqrt{a^{2}+b^{2}}}\right)=4 a b\left(\frac{a+b}{\sqrt{a^{2}+b^{2}}}-1\right)
$$

Előd Csirmaz (Fazekas M. Főv. Gyak. Gimn., 9th grade) Dávid Siklósi (Fazekas M. Főv. Gyak. Gimn., 9th grade)

II. solution. By combining the opposite triangles cut off by the tangents, we obtain rhombuses similar to the original rhombus. It is known that the ratio of the areas of similar figures is equal to the square of the similarity ratio. Based on this, the area $4 t_{1}$ of the rhombus formed by the triangle $A K B$ and its reflection about $O D$ can be obtained as follows: $\frac{4 t_{1}}{T}=\left(\frac{a-r}{a}\right)^{2}$, from which $4 t_{1}=2 a b\left(\frac{a-r}{a}\right)^{2}$, using the fact that $T=2 a b$. Similarly, we get: $4 t_{2}=2 a b\left(\frac{b-r}{b}\right)^{2}$. The area of the octagon: $t=2 a b\left[1-\left(\frac{a-r}{a}\right)^{2}-\left(\frac{b-r}{b}\right)^{2}\right]$, from which, after substituting $r=\frac{a b}{\sqrt{a^{2}+b^{2}}}$ and some calculations, we obtain the result found in the first solution.

![](https://cdn.mathpix.com/cropped/2024_05_02_62df3c13b7cf95dcdef9g-1.jpg?height=721&width=531&top_left_y=1488&top_left_x=797)