From the theorem proved in problem 989, it follows that if the cyclic hexagon $A B C D E F=H$ is symmetric with respect to the diagonal $A D$, and a circle can be inscribed in it, then $A B+C D=A D$. What relationship holds between the radii of the circles inscribed in and circumscribed around $H$ and the distance between their centers in this case?[^0]


[^0]:    ${ }^{1}$ See K. M. L. (1960) p. 109.
$H$ is together with the diagonal $A D$ the axis of symmetry for both the circumscribed and the inscribed circle, so the centers $O$ and $M$ are on $A D$. Let the radii of the circles be $R$ and $r$, and $O M=c$. We can assume that the vertices are labeled such that $M$ is on the segment $A O$ - possibly exactly at $O$. Thus, $A M=R-c, D M=R+c$. Let the points of tangency of the inscribed circle with $A B$ and $C D$ be $P$ and $Q$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_efff9ce74d996eb5be53g-1.jpg?height=802&width=816&top_left_y=336&top_left_x=646)

From the similarity of the right triangles $A B D$ and $A P M$, as well as $D C A$ and $D Q M$, and from the right triangles $A P M$ and $D Q M$, we get:

$$
\frac{A B}{A D}=\frac{A P}{A M}=\frac{\sqrt{(R-c)^{2}-r^{2}}}{R-c}, \quad \frac{C D}{A D}=\frac{Q D}{M D}=\frac{\sqrt{(R+c)^{2}-r^{2}}}{R+c}
$$

Substituting these into the form of the given relationship divided by $A D$, we obtain the desired relationship:

$$
\frac{\sqrt{(R-c)^{2}-r^{2}}}{R-c}+\frac{\sqrt{(R+c)^{2}-r^{2}}}{R+c}=1
$$

We get a form of the relationship free of square roots if we square it twice - after intermediate rearrangements:

$$
3\left(R^{2}-c^{2}\right)^{4}-4\left(R^{2}-c^{2}\right)^{2}\left(R^{2}+c^{2}\right) r^{2}-16 R^{2} c^{2} r^{4}=0
$$

Zoltán Demendy (Budapest, XVIII. District Hengersor úti Gymnasium, 3rd grade)