88. Quadrilateral $ABCD$ is circumscribed around a circle. The point of tangency of the circle with side $AB$ divides this side into segments $a$ and $b$, and the point of tangency of the circle with side $AD$ divides it into segments $a$ and $c$. Within what limits can the radius of the circle vary?
88. The radius of the inscribed circle is bounded by the magnitudes of the radii of two limiting cases. It cannot be less than

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-085.jpg?height=371&width=354&top_left_y=932&top_left_x=167)

Fig. 20. the radius of the circle inscribed in a triangle with sides \(a+b, b+c, c+a\), which is equal to \(S / p\), where \(S\) is the area, \(p\) is the semiperimeter of the triangle; thus,

\[
\begin{aligned}
& r>\frac{S}{p}=\frac{\sqrt{(a+b+c) a b c}}{a+b+c}= \\
& =\sqrt{\frac{a b c}{a+b+c}}
\end{aligned}
\]

On the other hand, the radius is less than the radius of the circle depicted in Fig. 20 (on this figure, opposite tangents are parallel, point \(C\) "escapes" to infinity). Since for the angles \(\alpha, \beta\), and \(\gamma\) marked on the figure, the equality \(\alpha+\beta+\gamma=\frac{\pi}{2}\) holds and \(\operatorname{tg} \alpha=\frac{c}{\rho}, \operatorname{tg} \beta=\frac{a}{\rho}, \operatorname{tg} \gamma=\frac{b}{\rho}\), where \(\rho\) is the radius of the depicted circle, we have \(\operatorname{tg}(\alpha+\beta)=\frac{1}{\operatorname{tg} \gamma}\), or \(\frac{\frac{c}{\rho}+\frac{a}{\rho}}{1-\frac{a c}{\rho^{2}}}=\frac{\rho}{b}\), from which \(\rho=\sqrt{a b+b c+c a}\). Thus, \(\sqrt{\frac{a b c}{a+b+c}}<r<\sqrt{a b+b c+c a}\).