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A maths teacher draws a number of circles on a piece of paper. When she shows this piece of paper to the young mathematician, he claims he can see only five circles. The maths teacher agrees. But when she shows the same piece of paper to another young mathematician, he says that there are exactly eight circles. The teacher confirms that this answer is also correct. How is that possible and how many circles did she originally draw on that piece of paper?

The circles \(\sigma_1\) and \(\sigma_2\) intersect at points \(A\) and \(B\). At the point \(A\) to \(\sigma_1\) and \(\sigma_2\), respectively, the tangents \(l_1\) and \(l_2\) are drawn. The points \(T_1\) and \(T_2\) are chosen respectively on the circles \(\sigma_1\) and \(\sigma_2\) so that the angular measures of the arcs \(T_1A\) and \(AT_2\) are equal (the arc value of the circle is considered in the clockwise direction). The tangent \(t_1\) at the point \(T_1\) to the circle \(\sigma_1\) intersects \(l_2\) at the point \(M_1\). Similarly, the tangent \(t_2\) at the point \(T_2\) to the circle \(\sigma_2\) intersects \(l_1\) at the point \(M_2\). Prove that the midpoints of the segments \(M_1M_2\) are on the same line, independent of the positions of the points \(T_1, T_2\).

A circle is divided up by the points \(A, B, C, D\) so that \({\smile}{AB}:{\smile}{BC}:{\smile}{CD}:{\smile}{DA} = 2: 3: 5: 6\). The chords \(AC\) and \(BD\) intersect at point \(M\). Find the angle \(AMB\).

A circle is divided up by the points \(A\), \(B\), \(C\), \(D\) so that \({\smile}{AB}:{\smile}{BC}:{\smile}{CD}:{\smile}{DA} = 3: 2: 13: 7\). The chords \(AD\) and \(BC\) are continued until their intersection at point \(M\). Find the angle \(AMB\).

The bisector of the outer corner at the vertex \(C\) of the triangle \(ABC\) intersects the circumscribed circle at the point \(D\). Prove that \(AD = BD\).

The vertex \(A\) of the acute-angled triangle \(ABC\) is connected by a segment with the center \(O\) of the circumscribed circle. The height \(AH\) is drawn from the vertex \(A\). Prove that \(\angle BAH = \angle OAC\).

The vertex \(A\) of the acute-angled triangle \(ABC\) is connected by a segment with the center \(O\) of the circumscribed circle. The height \(AH\) is drawn from the vertex \(A\). Prove that \(\angle BAH = \angle OAC\).

From an arbitrary point \(M\) lying within a given angle with vertex \(A\), the perpendiculars \(MP\) and \(MQ\) are dropped to the sides of the angle. From point \(A\), the perpendicular \(AK\) is dropped to the segment \(PQ\). Prove that \(\angle PAK = \angle MAQ\).