The equation of a circle is $x^{2}+y^{2}=100$; at the point $M_{1}(6, y>0)$ on the circumference of this circle, we draw a tangent. From the two endpoints of the diameter $EC$ lying on the x-axis, we draw perpendiculars to the tangent, intersecting it at points $A$ and $B$. What is the area of the triangle whose base is $AB$ and whose vertex is the origin $O$ of the coordinate system?
The ordinate drawn at point $M_{1}$ is $y=8$. Let's draw a line from $C$ parallel to $M_{1} B$, which intersects $O M_{1}$ at $S$. Since

$$
O S C \triangle \sim O M_{1} D \triangle
$$

it follows that

$$
r: y=r: M_{1} B
$$

and thus

$$
M_{1} B=y=8
$$

Since $A O B$ is an isosceles triangle, its area is:

$$
t=M_{1} B \cdot r=8 \cdot 10=80
$$

(Géza Sasvári.)

The problem was also solved by: Bella I., Juvancz I., Kohn B., Koós A., Krausz B., Krisztián Gy., Lukhaub Gy., Obláth R., Prohászka J.