## Task 32/62

The diameter $d$ of a circle is intersected by a chord at an angle of $30^{\circ}$, dividing it in the ratio $\frac{a}{b}=\frac{1}{3}$.

a) How long is the chord?

b) What is the distance from the chord to the center of the circle?
Since the length $d$ of the diameter is not given in the problem, it is arbitrarily assumed to be $d=2r=2$ (length units). If $a$ and $b$ are the two segments of the diameter, then $\frac{a}{b}=\frac{1}{3}$ and $a+b=d=2$. This implies $b=3a=1.5$ and $a=0.5$.

We introduce further relationships according to the diagram.

Let $FM$ be the perpendicular from $M$ to $A_{1}A_{2}$. From the diagram, we can see:

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-056.jpg?height=411&width=534&top_left_y=1959&top_left_x=276)

1. Since the angle $MDF=30^{\circ}$, the angle $DMF$ is $60^{\circ}$ due to the right-angled triangle. Reflecting the triangle $MDF$ over $A_{1}A_{2}$ results in an equilateral triangle $MDM'$, where $M'F=MF$ and $M'D=DM=0.5$.

Thus, $MF=0.5DM=0.25$ is the distance from the chord $A_{1}A_{2}$ to the center $M$ of the circle.

2. From the right-angled triangle $MFA_{2}$, we have

$$
FA_{2}^{2}=MA_{2}^{2}-MF^{2}=r^{2}-0.25^{2}=1-\frac{1}{16}=\frac{15}{16} \rightarrow FA_{2}=\sqrt{\frac{15}{16}}=\frac{1}{4}\sqrt{15}
$$

Therefore, $A_{1}A_{2}=2FA_{2}=\frac{1}{2}\sqrt{15} \approx 1.936$.