A circle $K$ internally touches a circle $k$ with half its radius. Construct a chord in $K$ that is perpendicular to the line connecting the centers of the two circles, and which is divided into three equal parts by its intersection points with $k$.

 untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 

Note: The last sentence is in Chinese and is not part of the original problem statement. It is a note about the translation request and should not be included in the translated text. Here is the final translation without the note:

A circle $K$ internally touches a circle $k$ with half its radius. Construct a chord in $K$ that is perpendicular to the line connecting the centers of the two circles, and which is divided into three equal parts by its intersection points with $k$.
I. solution. Let's use the notations of Figure 1. Let the length of the chord that meets the conditions be $3 x = A E$, and its distance from the center $O$ of the larger circle be $y = C O$. Let the radius of the large circle be $r$, and the radius of the small circle be $\frac{1}{2} r$.

![](https://cdn.mathpix.com/cropped/2024_05_02_755fdae782f22a6e1235g-1.jpg?height=447&width=458&top_left_y=302&top_left_x=823)

Figure 1

We know that $A B = B D = D E$, $B C = \frac{1}{2} B D$, and thus $A C = 3 B C$. Applying the Pythagorean theorem to the triangle $B K C$:

$$
\begin{aligned}
B C^{2} & = B K^{2} - C K^{2} \\
B C & = \sqrt{\left(\frac{1}{2} r\right)^{2} - \left(\frac{1}{2} r - y\right)^{2}}
\end{aligned}
$$

Applying the Pythagorean theorem to the triangle $A O C$:

$$
A C^{2} = A O^{2} - C O^{2}, \quad A C = \sqrt{r^{2} - y^{2}}
$$

Substituting the right-hand sides of (1) and (2) into the equation $A C = 3 B C$ and squaring it:

$$
r^{2} - y^{2} = 9\left[\frac{1}{4} r^{2} - \left(\frac{1}{4} r^{2} - r y + y^{2}\right)\right], \quad \text { hence } \quad -8 y^{2} + 9 r y - r^{2} = 0
$$

Solving for $y$:

$$
y_{1,2} = \frac{-9 r \pm 7 r}{-16}, \quad y_{1} = \frac{1}{8} r, \quad y_{2} = r
$$

$y_{2}$ is not a valid solution, as in this case the "chord" would be the point of tangency of the two circles.

Thus, the desired chord is the perpendicular to the segment $P O$ at the point closer to $O$ that divides it into an eighth.

II. solution. Let the length of the equal segments cut by the circles from the desired line be $2 m$.

The triangles $A B C$ and $B D E$ on Figure 2 are right-angled by Thales' theorem. The altitude to the hypotenuse of a right-angled triangle divides it into similar triangles, and thus the square of the altitude is equal to the product of the two segments of the base. Applying this to the two triangles above, we get $x(r - x) = m^{2}$ and $(r + x)(r - x) = (3 m)^{2}$. Substituting 9 times the first equation into the right-hand side of the second:

$$
r^{2} - x^{2} = 9 x(r - x), \quad \text { hence } \quad 8 x^{2} - 9 r x + r^{2} = 0
$$

The solutions to this are $x = r$ and $x = \frac{r}{8}$, but only the latter is valid.

![](https://cdn.mathpix.com/cropped/2024_05_02_755fdae782f22a6e1235g-1.jpg?height=445&width=469&top_left_y=2171&top_left_x=817)

Figure 2

Thus, the line must be drawn at a distance of $\frac{1}{8}$ of the radius from the center of the large circle. The eighth can be constructed using segment bisections.

Remark. The theorem used is the so-called altitude theorem. The same equations must be used if someone uses the theorem about the segments of a chord from an internal point of a circle.

III. solution. Let the radius of the circle $K$ be $R$, and the radius of the circle $k$ be $r$. Place the circles in a coordinate system as shown in Figure 3, where $R = 2$. The center of $K$ is the origin, the center of $k$ is the point $(1; 0)$, and the line through the centers of the circles is the $x$-axis. The desired chord is perpendicular to this, so its equation is of the form $x = x_{0}$, where $0 < x_{0} < 2$. Let the endpoints of the chord be $P_{1}$ and $P_{2}$, the intersection points with the circle $k$ be $Q_{1}$ and $Q_{2}$, and the intersection point with the $x$-axis be $T$. We know that $Q_{1}$ and $Q_{2}$ are the trisection points of the segment $P_{1} P_{2}$, and $T$ bisects the segment $Q_{1} Q_{2}$, so $3 T Q_{1} = T P_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_755fdae782f22a6e1235g-2.jpg?height=496&width=526&top_left_y=619&top_left_x=797)

Figure 3

Write the equations of the two circles, and using the above equality, calculate the common abscissa of $Q_{1}$ and $P_{1}$, which we know is positive.

The equation of $K$ is $x^{2} + y^{2} = 4$, from which $y = \sqrt{4 - x^{2}}$.

The equation of $k$ is $(x - 1)^{2} + y^{2} = 1$, from which $y = \sqrt{-x^{2} + 2 x}$.

We are looking for the number $x_{0}$ such that $3 \cdot \sqrt{-x_{0}^{2} + 2 x_{0}} = \sqrt{4 - x_{0}^{2}}$.

Squaring and rearranging:

$$
4 x_{0}^{2} - 9 x_{0} + 2 = 0
$$

The solutions to the equation are 2 and $\frac{1}{4}$, but only the latter is valid. Thus, $x_{0} = \frac{1}{4}$.

Remark. There is another, quite surprising construction, as shown in Figure 4.

![](https://cdn.mathpix.com/cropped/2024_05_02_755fdae782f22a6e1235g-2.jpg?height=580&width=631&top_left_y=1789&top_left_x=736)

The circle with radius $3 r$ whose center is the intersection of the line connecting the centers of the two given circles and $K$ cuts the two endpoints of the desired chord from $K$. Indeed: the equation of the circle in the previous coordinate system is: $(x + 2)^{2} + y^{2} = 9$, and from the equation of the circle $K$, $x^{2} + y^{2} = 4$:

$$
P_{1}\left(\frac{1}{4}, \sqrt{4 - \frac{1}{16}}\right) = P_{1}\left(\frac{1}{4}, \sqrt{\frac{63}{16}}\right)
$$

which satisfies the equation of the circle (1).