In a circle, we drew the chords $A B$ and $C D$ which are perpendicular to each other and different from the diameter. $C D$ divides the chord $A B$ in the ratio $1: 5$, and the longer arc of $A B$ in the ratio $1: 2$. In what ratio does $A B$ divide the chord $C D$?
I. solution. Let the intersection point of the chords be denoted by $M$, and let $A M=1$ unit. Reflect the chord $C D$ over the perpendicular bisector of $A B$ (Figure 1). Clearly, $C D D^{\prime} C^{\prime}$ is a rectangle, and therefore $D D^{\prime}=C C^{\prime}=M M^{\prime}=4$. Since the longer arc $A B$ contains the point $D$, the arc $A D$ is half of the arc $D B$. Due to the reflection, the arcs $A D$ and $B D^{\prime}$ are equal, so from the previous observation, $A D=B D^{\prime}=D D^{\prime}=4$. From the above, it follows that in the quadrilateral $A D C^{\prime} C$, $A D=C C^{\prime}=4$, and this cyclic quadrilateral has the side $D C^{\prime}$ as a diameter of the circle. Therefore, this quadrilateral is symmetric with respect to the perpendicular bisector of $D C^{\prime}$. Then the triangles $A D C$ and $C C^{\prime} A$ are congruent, which means their areas are equal: $\frac{C D \cdot 1}{2}=\frac{4 \cdot C M}{2}$. This implies that $C D=4 \cdot C M$, and thus $M D=3 \cdot C M$, so $C M: M D=1: 3$.

Based on the work of Pap Júlia (Debrecen, Fazekas M. Gimn, I. o.t.)

II. solution. Let $\angle A O D = \alpha$. According to the second condition, $\angle B O D = 2 \alpha$, and thus $\angle A O B = 360^{\circ} - 3 \alpha$ (see Figure 2). Based on the relationship between inscribed and central angles, $\angle A B D = \frac{\alpha}{2}$, $\angle B C D = \alpha$, and $\angle A D B = 180^{\circ} - \frac{3}{2} \alpha$. From the perpendicularity of the chords, it follows that $\angle A B C = \angle A D C = 90^{\circ} - \alpha$. Then we show that the triangle $B C D$ is isosceles. $\angle C D B = 180^{\circ} - \frac{3}{2} \alpha - (90^{\circ} - \alpha) = 90^{\circ} - \frac{\alpha}{2}$, $\angle C B D = \frac{\alpha}{2} + (90^{\circ} - \alpha) = 90^{\circ} - \frac{\alpha}{2}$, which indeed implies that $C B = C D = c + d$. Let $A M = 1$ again. From the similarity of triangles $A M D$ and $C M B$, $\frac{1}{d} = \frac{c}{5}$, so $c = \frac{5}{d}$. Applying the Pythagorean theorem to triangle $C M B$, $c^2 + 5^2 = (c + d)^2$, and using $c = \frac{5}{d}$, we get $d = \sqrt{15}$. The desired ratio is $c: d = \frac{5}{\sqrt{15}}: \sqrt{15} = 1: 3$.

Based on the work of Bakos Péter (Eger, Szilágyi E. Gimn., III. o.t.)

Note: Several others solved the problem using trigonometric knowledge.
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