\section*{Problem 4 - 191044}

Given are two lengths \(a, b\) and an area \(F \leq \frac{1}{2} a b\).

Calculate from these given values \(a, b, F\) all the lengths \(r\) that have the property that a triangle \(A B C\) with \(B C=a, A C=b\), the area \(F\), and the circumradius \(r\) exists!
}

The solution consists of two parts. First, \(r\) is to be expressed in terms of \(a, b\) and \(F\), and second, the existence of a triangle with \(a, b, r\) and \(F\) is to be demonstrated.

I. Suppose there exists a triangle \(A B C\) with \(a = BC, b = AC\), the area \(F\), and the circumradius \(r\). Then, with \(c = AB\) and \(\gamma = \angle ACB\), we have \(\sin \gamma = \frac{c}{2r}\) (see District Olympiad Problem 191034),

\[
F = \frac{1}{2} a b \sin \gamma \quad ; \quad \cos \gamma = \pm \sqrt{1 - \sin^2 \gamma}
\]

and by the Law of Cosines, \(c = \sqrt{a^2 + b^2 - 2ab \cos \gamma}\) (2). Thus, we get

\[
r = \frac{c}{2 \sin \gamma} = \frac{abc}{4F} = \frac{ab}{4F} \sqrt{a^2 + b^2 - 2ab \cos \gamma} = \frac{ab}{4F} \sqrt{a^2 + b^2 \pm 2 \sqrt{a^2 b^2 - 4F^2}}
\]

II. Since \(0 < \frac{\sqrt{a^2 b^2 - 4F^2}}{ab} < 1\), it follows that \(a^2 b^2 - 4F^2 \geq 0\) and

\[
a^2 + b^2 \pm 2 \sqrt{a^2 b^2 - 4F^2} \geq a^2 + b^2 - 2 \sqrt{a^2 b^2 - 4F^2} > a^2 + b^2 - 2ab \geq 0
\]

i.e., each of the \(r\)-values specified in (3) exists. Since \(0 \leq \frac{\sqrt{a^2 b^2 - 4F^2}}{ab} < 1\) and (4) follows

\[
\sin \gamma = \sqrt{1 - \cos^2 \gamma} = \sqrt{1 - \frac{a^2 b^2 - 4F^2}{a^2 b^2}} = 2 \frac{F}{ab}
\]

thus, \(F = \frac{1}{2} ab \sin \gamma\) is the area of the triangle \(A B C\).

From (3), (4), and (2), we get

\[
r = \frac{abc}{4F}
\]

i.e., \(r\) (with the corresponding sign) is the circumradius of the triangle \(A B C\). Therefore, exactly the lengths \(r\) specified in (3) have the required property.