Official Solution. Since the length of the trajectory and the rotation angle are known we can find the sum of arc of each radius. Hence reformlate the problem. A point $A$ and a ray $\ell$ with origin $A$ are given on the plane. Also two numbers $r_{1}>r_{2}$ and two angles $\alpha_{1}, \alpha_{2}, \alpha_{1}+\alpha_{2}<\pi$ are given. Find the locus of the endpoints $B$ of the following trajectories $\Gamma$ : (1) $\Gamma$ has the begin point $A$ and touches $\ell$ at $A$; (2) $\Gamma$ is the union of arcs with radii $r_{1}$ and $r_{2}$, and the sums of angle measures of these arcs are equal to $2 \alpha_{1}$ and $2 \alpha_{2}$ respectively; (3) two adjacent arcs have the common tangent at their common endpoint and lie on the same side of this tangent. Answer. Let $O_{1}, O_{2}$ be the endpoints of arcs satisfying to the first condition with radii $r_{1}, r_{2}$ respectively and angle measures $2\left(\alpha_{1}+\alpha_{2}\right)$. Consider two discs $W_{1}, W_{2}$ centered at $O_{1}, O_{2}$ with radii $2\left(r_{1}-r_{2}\right) \sin \alpha_{2}, 2\left(r_{1}-r_{2}\right) \sin \alpha_{1}$ respectively. The required locus is the intersection of these discs. Proof. Let $P Q$ be an arbitrary arc of $\Gamma$ with radius $r_{1}$, and $R$ be a point of segment $P Q$ such that $P R: P Q=r_{2}: r_{1}$. An arc $P R$ touching $\Gamma$ at $P$ has radius $r_{2}$ and the tangent to it at $R$ is parallel to the tangent to $\Gamma$ at $Q$. Replace the arc $P Q$ to $P R$ and translate the part $Q B$ of $\Gamma$ to $\overrightarrow{Q R}$. Repeating this operation for each arc with radius $r_{1}$ we translate $B$ to $O_{2}$, and $\overrightarrow{O_{2} B}=\frac{r_{1}-r_{2}}{r_{1}} \sum \overrightarrow{P_{i}Q_{i}} .$ Similarly $O_{1} B=\frac{r_{2}-r_{1}}{r_{2}} \sum \overrightarrow{Q_{i}P_{i+1}}$ (we suppose that $\left.Q_{0}=A, P_{n+1}=B\right)$. Put off all arcs of $\Gamma$ on the unit circle starting from point $X$. The endpoint of the last arc is a point $Y$ such that $\smile X Y=2\left(\alpha_{1}+\alpha_{2}\right)$ and the homothety $H$ with coefficient $1 /\left(r_{1}-r_{2}\right)$ maps vector $O_{1} O_{2}$ to $X Y$. Color all arcs obtained from the arcs with radius $r_{1}$ red, and color blue the remaining arcs. Compare to each arc the vector from its begin point to the endpoint. The homothety $H$ maps vector $O_{2} B$ to the sum of red vectors and maps $B O_{1}$ to the sum of blue vectors. Let $Z$ be a point on arc $X Y$ such that $\smile X Z=2 \alpha_{1}$. The homothety $H$ maps the boundary circles of $W_{2}$ and $W_{1}$ to the circles centered at $X, Y$ respectively and passing through $Z$. Prove that the length of $O_{2} B$ is maximal if the red vector is unique. Let the tangent $m$ to the unit circle parallel to $O_{2} B$ toche the circle at $E$. Consider an arc $T$ with length $2 \alpha_{1}$ and midpoint $E$. If the endpoints of $T$ lie on red arcs divide these arcs into two parts. The projection to $m$ of a red vector with length $\varphi$ lying outside $T$ is less than $\varphi \cos \alpha_{1}$, therefore the sum of projections to $m$ of all red vectors is less than the projection of $T$. Similarly the length of $O_{1} B$ is maximal if the blue vector is unique. Thus $B$ lies inside the intersection of two discs. Clearly the common points of boundary circles of $W_{1}$ and $W_{2}$ correspond to the trajectories containing exactly one arc of each radius, and the points of these circles correspond to the trajectories containing one arc of some radius and two arc of the remaining one. Prove that each point $B$ inside both discs correspond to the unique trajectory containing four arc and starting from the arc with radius $r_{1}$. Let $D$ be the image of $B$ in homothety $H$. Find on arc $X Y$ such points $E, F, G$ that $\overrightarrow{X E}+\overrightarrow{F G}=\overrightarrow{X D}$. Let $E, G$ be the common points of the perpendicular bisector to $D Z$ with arcs $X Z, Y Z$, and $F$ be the second common point of arc $X Y$ with the line passing through $Z$ and parallel to $E G$. Then $D E F G$ is a parallelogram which is equivalent to the required equality. It is easy to see that $E, F, G$ are uniquely defined by $D$.

parmenides51 wrote: On the attraction "Merry parking", the auto has only two position* of a steering wheel: "right", and "strongly right". So the auto can move along an arc with radius $r_1$ or $r_1$. The auto started from a point $A$ to the Nord, it covered the distance $\ell$ and rotated to the angle $a < 2\pi$. Find the locus of its possible endpoints. and the problem has a typo: So the auto can move along an arc with radius $r_1$ or $r_2$.