107. What dependence should exist between the radii $R$ and $r$ and the distance $d$ between the centers of two circles, so that a triangle can be constructed that is inscribed in the first circle and circumscribed around the second circle
107. Let $O$ and $o$ be the centers of the circles with radii $R$ and $r$, respectively, and $Oo = d$. Let $ABC$ be a triangle inscribed in the first circle and circumscribed around the second circle (Fig. 235). Extend $Oo$ to intersect the circumcircle of triangle $ABC$ at points $M$ and $N$. Then $OM = R + d$, $ON = R - d$. Let $D$ be the second point of intersection of line $Co$ with the circumcircle; in this case, we have

$$
\begin{gathered}
oC \cdot oD = oM \cdot oN = \\
= (R + d)(R - d) = \\
= R^2 - d^2
\end{gathered}
$$

Now let's find another expression for $oC \cdot oD$. Draw the diameter $DE$ of the circumcircle and the segments $DA$, $DB$, and $oA$. Since $CD$ is the angle bisector of angle $C$ of triangle $ABC$, we have

$$
DA = DB
$$

Furthermore, triangle $AoD$ is isosceles; indeed, since $\angle DAB = \angle DCB = \frac{1}{2} \angle C$ and $\angle oAB = \frac{1}{2} \angle A$, we have

$$
\angle oAD = \angle oAB + \angle BAD = \frac{\angle A}{2} + \frac{\angle C}{2}
$$

On the other hand,

$$
\angle AoD = \angle oAC + \angle oCA = \frac{\angle A}{2} + \frac{\angle C}{2}
$$

(as the external angle of triangle $AoC$). Therefore, $AD = oD$, and hence $BD = oD$.

Now drop a perpendicular $oP$ from point $o$ to side $AC$ of triangle $ABC$. In the right triangles $CoP$ and $EBD$, $\angle DEB = \angle DCB = \angle oCP$; therefore, these triangles are similar, and $\frac{Co}{oP} = \frac{DE}{DB}$, so $oC \cdot DB = oP \cdot DE = r \cdot 2R$. Hence,

$$
oC \cdot oD = 2Rr.
$$

Comparing this relation with the one obtained earlier, we have

$$
R^2 - d^2 = 2Rr
$$

or

$$
d^2 = R^2 - 2Rr
$$

or

$$
\frac{1}{R+d} + \frac{1}{R-d} = \frac{1}{r}
$$

Conversely, let the radii $R$ and $r$ of the given circles and the distance between their centers be related by the equation (*). Then $R^2 - 2Rr \geq 0$, from which $R \geq 2r$. Therefore, there exists a triangle $ABC$ such that the circumcircle and the incircle of this triangle have radii $R$ and $r$ (see problem 106b)). But then the distance between the centers of the circumcircle and the incircle of triangle $ABC$ must, by the proven result, equal $\sqrt{R^2 - 2Rr} = d$, from which it follows that there exists a triangle (congruent to $\triangle ABC$) inscribed in one of the given circles and circumscribed around the other.

If the radii of two circles and the distance between their centers do not satisfy the relation $d^2 = R^2 - 2Rr$, then there does not exist a triangle inscribed in one circle and circumscribed around the other. If this relation holds, then there are infinitely many such triangles: if from an arbitrary point $C$ on the larger circle, we draw chords $AC$ and $CB$ of this circle that are tangent to the smaller circle, it can be shown that the chord $AB$ will also be tangent to this circle, and triangle $ABC$ will be inscribed in one circle and circumscribed around the other.