## Task 3 - 120913

A diameter $AB$ of a circle is intersected by a chord $CD$ at a point $E$ that divides $AB$ internally in the ratio 2:5. The chord $CD$ intersects the diameter $AB$ at an angle of $30^{\circ}$.

Determine the distance of the chord from the center $M$ of the circle, given the length $d$ of the diameter!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0952.jpg?height=411&width=488&top_left_y=1639&top_left_x=287)
For a suitable choice of labels $A, B$, it follows from the assumption that

$$
\begin{aligned}
& |A E|=\frac{2}{7} d \quad \text { and } \\
& |E B|=\frac{5}{7} d, \quad \text { thus } \\
& |E M|=\frac{1}{2} d-\frac{2}{7} d=\frac{3}{14} d
\end{aligned}
$$

Let $F$ be the foot of the perpendicular from $M$ to $C D$. Then $\triangle E M F$ is a right triangle with $|\angle M E F|=30^{\circ}$ (by assumption), so

$$
|\angle F M E|=60^{\circ}
$$

Now reflect $\triangle M E F$ across $C D$. For the resulting right triangle $\triangle E F M^{\prime}$, we have

$$
\left|\angle F E M^{\prime}\right|=30^{\circ} \text { and }\left|\angle E M^{\prime} F\right|=60^{\circ}
$$

From (4) and (5), the triangle $\triangle E M M^{\prime}$ is equilateral, and thus the foot of the altitude $F$ is also the midpoint of $M M^{\prime}$. Considering (3), it follows that $|M F|=\frac{3}{28} d$. Therefore, the distance of the chord $C D$ from the center of the circle is $\frac{3}{28} d$, where $d$ is the diameter of the circle.