Let $ABCD$ be the given parallelogram, $O_1, O_2, O_3$, and $O_4$ be the centers of the circumcircles of triangles $AOB, BOC, COD$, and $AOD$ respectively. Let $\angle AOB = \varphi (0 < \varphi < \pi)$. Since $O_1O_2, O_2O_3, O_3O_4$, and $O_1O_4$ are the perpendicular bisectors of segments $BO, CO, DO$, and $AO$ respectively, $O_1O_2O_3O_4$ is a parallelogram, and $\angle O_1O_4O_3 = \angle AOB = \varphi$. Let $E$ be the projection of point $O_3$ onto line $O_1O_4$. Since $AO = OC$, then $O_3E = AO$. Therefore, $AO = O_3O_4 \cdot \sin \varphi = 16 \sin \varphi$. By the Law of Sines for triangle $ABO$, where the circumradius $R = 5$, we get $AB = 2R \cdot \sin \varphi = 10 \sin \varphi$. In the same triangle, $BO = \frac{1}{2} BD$, so by the Law of Cosines,

$$
AB^2 = AO^2 + BO^2 - 2 \cdot AO \cdot BO \cdot \cos \varphi, \text{ i.e., } 100 \sin^2 \varphi = 256 \sin^2 \varphi + 36 - 192 \sin \varphi \cos \varphi
$$

or

$$
3 \operatorname{ctg}^2 \varphi - 16 \operatorname{ctg} \varphi + 16 = 0
$$

This equation has two solutions: $\operatorname{ctg} \varphi = \frac{4}{3}$ or $\operatorname{ctg} \varphi = 4$. Therefore, $\sin \varphi = \frac{1}{\sqrt{17}}$ or $\sin \varphi = \frac{3}{5}$. If $S$ is the area of parallelogram $ABCD$, then

$$
S = 4 \cdot S_{\triangle AOB} = 4 \cdot \frac{1}{2} \cdot AO \cdot BO \cdot \sin \varphi = 192 \cdot \sin^2 \varphi
$$

In the first case, $S = \frac{192}{17}$, in the second case $S = \frac{1728}{25} \cdot \Gamma$. Two different values of the area of the parallelogram are obtained because the centers of the circumcircles $O_1, O_2$ and $O_3, O_4$ can lie either on opposite sides (first value in the answer) or on the same side (second value) of line $BD$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-39.jpg?height=843&width=869&top_left_y=1425&top_left_x=595)

Answer

$\frac{192}{17}$ or $\frac{1728}{25} = 69.12$.
[Inscribed, circumscribed, and exscribed circles; their radii]

In parallelogram $A B C D$, the diagonals intersect at point $O$, the length of diagonal $B D$ is 12. The distance between the centers of the circles circumscribed around triangles $A O D$ and $C O D$ is 16. The radius of the circle circumscribed around triangle $A O B$ is 5. Find the area of parallelogram $A B C D$.