A deltoid has sides $a$ and $b(a \neq b)$. The sides of different lengths form a right angle. What is the radius of the circle that touches the extensions of all four sides of the deltoid?
Solution. In the kite $ABCD$, $AB \perp BC$ and $AB=AD=a>b=BC=CD$.

The circle tangent to the extensions of the sides touches the line of the side $AB$ at point $E_{1}$, the line of the side $BC$ at point $E_{2}$, the line of the side $CD$ at point $E_{3}$, and the line of the side $AD$ at point $E_{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_255da1da760839c8ba7bg-1.jpg?height=528&width=425&top_left_y=308&top_left_x=839)

Since $BC \perp AB$ and $OE_{1}$ is also perpendicular to the line of $AB$, $OE_{1} \| BC$. Similarly, since $AB \perp CB$ and $OE_{2} \perp CB$, $OE_{2} \| AB$. The quadrilateral $OE_{1}BE_{2}$ is a rectangle, but since $OE_{1}=OE_{2}=r$ is the radius of the tangent circle, it is a square; thus, $E_{1}B=r$. The triangles $ABC$ and $AE_{1}O$ are similar, and the ratio of their corresponding sides is the same:

$$
a:(a+r)=b: r
$$

From this, the radius of the tangent circle is:

$$
r=\frac{ab}{a-b}
$$