The unit-length rods $O A$ and $A B$ are connected by a hinge at $A$. To $A B$, we attach at $B$ a rod $B C$ of equal length, such that $C$ lies in the plane $O A B$ and $\angle A B C = 90^{\circ}$. Keeping $O$ at the origin, we move our structure in the coordinate plane such that $B$ moves along the line through $O$ with direction vector $(1 ; 1-\sqrt{2})$. What path does the point $C$ trace as $A$ moves around the unit circle centered at the origin?
Let the given line be $e$. The rod $BC$ can be attached to $AB$ in two different ways at point $B$, depending on whether the directed angle $ABC$ is $+90^{\circ}$ or $-90^{\circ}$. We will now deal with the case where the directed angle is $-90^{\circ}$. The other case can be reduced to this one by reflecting over the line $e$.

Let the angle between the vector $\overrightarrow{OA}$ and the $x$-axis be $\alpha$, and fix the point $A$. The point $B$ can generally take two different positions. Let's calculate the coordinates of point $C$ in both cases!

![](https://cdn.mathpix.com/cropped/2024_05_02_5c0facb06f15585edbbbg-1.jpg?height=559&width=631&top_left_y=379&top_left_x=736)

1. Figure
2. If points $B$ and $O$ coincide, then by rotating $OA$ by $-90^{\circ}$, we get $OC$, so the coordinates of $C$ are

$$
\begin{aligned}
& C_{x}=\cos \left(\alpha-90^{\circ}\right)=\sin \alpha \\
& \quad C_{y}=\sin \left(\alpha-90^{\circ}\right)=-\cos \alpha
\end{aligned}
$$

2. If $B$ and $O$ do not coincide, then the position of $B$ is already unique. Determine the angle between the directed rays $\overrightarrow{AC}$ and $\overrightarrow{OX}$! By rotating $\overrightarrow{OB}$ by $+22.5^{\circ}$, we get $\overrightarrow{OA}$, because the line $e$ forms an angle of $\arctan \frac{1-\sqrt{2}}{1} = -22.5^{\circ}$ with the $x$-axis. The triangle $AOB$ is isosceles, so by rotating $\overrightarrow{OB}$ by $-\left(\alpha+22.5^{\circ}\right)$, we get $\overrightarrow{AB}$. By rotating $\overrightarrow{AB}$ by $+45^{\circ}$, we get $\overrightarrow{AC}$. Thus, by rotating $\overrightarrow{OX}$ by $\left(-22.5^{\circ}-\alpha-22.5^{\circ}+45^{\circ}\right)=-\alpha$, we get $\overrightarrow{AC}$, meaning the angle between them is $-\alpha$. The length of vector $\overrightarrow{AC}$ is $\sqrt{2}$ from the isosceles right triangle $ABC$, so the vector $\overrightarrow{AC}$ is

$$
\overrightarrow{AC}=(\sqrt{2} \cos (-\alpha), \sqrt{2} \sin (-\alpha))
$$

By adding vector $\overrightarrow{AC}$ to vector $\overrightarrow{OA}(\cos \alpha, \sin \alpha)$, we get the coordinates of point $C$:

$$
\begin{aligned}
& C_{x}=\cos \alpha+\sqrt{2} \cos (-\alpha)=(\sqrt{2}+1) \cos \alpha \\
& C_{y}=\sin \alpha+\sqrt{2} \sin (-\alpha)=(\sqrt{2}-1) \sin \alpha
\end{aligned}
$$

Thus, for any $\alpha$, point $C$ lies on the curve described by equations (1) or (2): The curve described by equation (1) is the unit circle. Equation (2) describes an ellipse:

$$
x=(\sqrt{2}+1) \cos \alpha, \quad y=(1-\sqrt{2}) \sin \alpha
$$

With the choice of

$$
\frac{x^{2}}{(\sqrt{2}+1)^{2}}+\frac{y^{2}}{(1-\sqrt{2})^{2}}=1
$$

the equation of the ellipse is given.

![](https://cdn.mathpix.com/cropped/2024_05_02_5c0facb06f15585edbbbg-1.jpg?height=523&width=1068&top_left_y=2183&top_left_x=515)

Point $A$ must traverse the unit circle. This means that the angle $\alpha$ must vary over a $360^{\circ}$ interval starting from some value. We previously saw that during this interval, point $C$ travels along one of the two curves. However, the path of point $C$ will be a continuous line, so it can only switch from one curve to the other if for some $\alpha$, equations (1) and (2) give the same point:

$$
\sin \alpha=(\sqrt{2}+1) \cos \alpha, \quad \text{and} \quad -\cos \alpha=(1-\sqrt{2}) \sin \alpha
$$

Both equalities hold if

$$
\tan \alpha=\sqrt{2}+1, \quad \alpha=67.5^{\circ}+k \cdot 180^{\circ}
$$

The point corresponding to $\alpha=67.5^{\circ}$ is denoted by $C_1$, and the point corresponding to $\alpha=67.5^{\circ}+180^{\circ}$ is denoted by $C_2$. If $\alpha$ varies from $67.5^{\circ}$ to $67.5^{\circ}+180^{\circ}$, then point $C$ travels from $C_1$ to $C_2$ in the positive direction along the (1) curve; on the (2) curve, point $C$ travels from $C_1$ to $C_2$ in the negative direction. Similarly for the other interval.
![](https://cdn.mathpix.com/cropped/2024_05_02_5c0facb06f15585edbbbg-2.jpg?height=578&width=1180&top_left_y=827&top_left_x=450)

3. Figure

In summary, if we assume that point $A$ travels in the positive direction around the unit circle, then the path of point $C$ can be one of the following (Figure 3) (depending on the starting point); there are a total of 20 possibilities.