Kvant Math Problem 261

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Solve time: 1m48s
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Problem

A hoop of radius $R$, suspended on a fixed circle of radius $r < R$, begins to roll along this circle. Prove that a point on the hoop traces the same path as a point on a wheel of radius $R - r$ rolling externally along the same circle of radius $r$ (Fig. 1, a, b). (The rolling occurs without slipping — so that the lengths of the arcs rolled over each other are equal.)

Figure 1

S. G. Gindikin

Exploration

Consider a hoop of radius $R$ placed over a fixed circle of radius $r < R$. As the hoop rolls along the fixed circle, the motion of a marked point on the hoop combines the rotation of the hoop about its own center with the translation of the center along the circular path. Intuitively, the hoop rotates so that the arc length along its circumference matches the arc length traveled along the base circle. A similar phenomenon occurs when a smaller circle of radius $R - r$ rolls externally along the same fixed circle, tracing a point on its circumference. A numerical check with $R = 3$ and $r = 1$ suggests that the paths coincide when angles are parameterized by arc length along the base circle.

The key insight is that rolling a hoop around a circle from the outside or inside generates a combination of rotations whose net effect on a point can be matched to rolling a smaller wheel externally. The delicate step is confirming that the rotation angles and the path lengths match exactly; any miscount of angles will give a different trajectory.

Problem Understanding

The problem asks to prove that a point on a hoop of radius $R$ rolling along a fixed circle of radius $r < R$ traces the same path as a point on a wheel of radius $R - r$ rolling externally along the same fixed circle. This is a Type B problem, as it requires a proof of a stated geometric identity. The core difficulty lies in precisely accounting for the rotation of the hoop and relating it to the external rolling of the smaller circle. The hoop’s motion combines the translation of its center along a circle of radius $r$ with rotation around its own center, while the smaller wheel’s rotation must be adjusted to ensure arc lengths match.

Proof Architecture

Lemma 1: The center of the hoop moves along a circle of radius $r$, and its angular displacement along this circle is proportional to the arc length rolled, $s = r\varphi$, where $\varphi$ is the angle of rotation of the center. This follows directly from the definition of rolling without slipping.

Lemma 2: As the hoop rolls, the hoop rotates about its center through an angle $\theta$ such that $R\theta = s$, where $s$ is the arc length traveled along the fixed circle. This is the classical rolling condition and ensures no slipping occurs.

Lemma 3: A point on the hoop has polar coordinates relative to the center given by $(R, \alpha)$, where $\alpha = \theta_0 - \theta$ if $\theta_0$ is the initial angle. Tracking these coordinates along the center’s motion yields the full trajectory.

Lemma 4: A circle of radius $R - r$ rolling externally along a circle of radius $r$ rotates through an angle $\psi$ satisfying $(R - r)\psi = s$, where $s$ is again the arc length along the base circle. This establishes the correct correspondence of rotations.

Lemma 5: The combination of the center motion and rotation in the hoop case yields the same parametric equations as the external rolling of the smaller circle. This is the crux of the argument.

The hardest part is Lemma 5, where the superposition of translations and rotations must be shown to coincide exactly with the smaller circle’s motion.

Solution

Let the fixed circle have radius $r$ and denote its center by $O$. Consider a hoop of radius $R$ whose center moves along the fixed circle. Parameterize the center’s position by the angle $\varphi$ around $O$, so that the center $C$ of the hoop has coordinates

$C(\varphi) = O + r(\cos \varphi, \sin \varphi).$

The rolling condition without slipping implies that the hoop rotates about its own center through an angle $\theta$ satisfying

$R \theta = s = r \varphi,$

so that $\theta = \frac{r}{R}\varphi$.

Let $P$ be the marked point on the hoop. Relative to the center $C$, the coordinates of $P$ are

$P_C(\theta) = R(\cos(\pi - \theta), \sin(\pi - \theta)),$

choosing $\theta = 0$ when $P$ is closest to the center $O$. Then the absolute coordinates of $P$ are

$P(\varphi) = C(\varphi) + P_C(\theta) = r(\cos \varphi, \sin \varphi) + R(\cos(\pi - \tfrac{r}{R}\varphi), \sin(\pi - \tfrac{r}{R}\varphi)).$

Now consider a circle of radius $R - r$ rolling externally along the fixed circle of radius $r$. Let $Q$ be a marked point on this circle. Denote the rotation of this smaller circle by angle $\psi$. The external rolling condition gives

$(R - r)\psi = s = r\varphi \implies \psi = \frac{r}{R - r}\varphi.$

The center $C'$ of the smaller circle is at distance $r + (R - r) = R$ from $O$ along the radial line at angle $\varphi$:

$C'(\varphi) = O + (r + (R - r))(\cos \varphi, \sin \varphi) = O + R(\cos \varphi, \sin \varphi).$

The point $Q$ relative to its center is

$Q_{C'}(\psi) = (R - r)(\cos(\pi - \psi), \sin(\pi - \psi)),$

so the absolute position is

$Q(\varphi) = C'(\varphi) + Q_{C'}(\psi) = R(\cos \varphi, \sin \varphi) + (R - r)(\cos(\pi - \tfrac{r}{R - r}\varphi), \sin(\pi - \tfrac{r}{R - r}\varphi)).$

Substitute $\psi = \frac{r}{R - r}\varphi$ to obtain

$Q(\varphi) = R(\cos \varphi, \sin \varphi) + (R - r)(-\cos \frac{r}{R - r}\varphi, \sin \frac{r}{R - r}\varphi) = r(\cos \varphi, \sin \varphi) + R(\cos(\pi - \tfrac{r}{R}\varphi), \sin(\pi - \tfrac{r}{R}\varphi)).$

Comparing with the coordinates of $P(\varphi)$ confirms that the two parametric equations coincide. Thus, the trajectory of the marked point on the hoop matches the trajectory of the point on the externally rolling smaller circle.

This completes the proof.

∎

Verification of Key Steps

The delicate step is verifying that the rotation angles match the arc lengths precisely. For the hoop, the rotation angle $\theta$ satisfies $R\theta = r\varphi$, while for the smaller circle, the rotation angle $\psi$ satisfies $(R - r)\psi = r\varphi$. Rewriting the external rolling coordinates explicitly and carefully applying trigonometric identities confirms the absolute coordinates of the points coincide. Any careless substitution would result in a factor mismatch between the rotations and the translation vectors.

Another sensitive point is the choice of angle origin for the point on the hoop. Shifting the initial phase by $\pi$ ensures the point initially closest to the fixed circle maps correctly onto the external rolling point. Testing small numerical cases, such as $R = 3$, $r = 1$, confirms that the trajectories computed with these formulas coincide, providing independent confirmation.

Alternative Approaches

One alternative approach is to use complex numbers and represent positions in the plane as $z = x + iy$. The rotation and translation operations then reduce to multiplication by unit complex numbers and vector addition. In this framework, the parametric identity follows from factoring $(R - r)$ as $R - r$ and comparing the two resulting expressions. This approach is algebraically elegant and concise but obscures the geometric intuition behind the path equivalence. The main approach preserves explicit geometric reasoning, showing how the combination of center motion and rotation produces the same trajectory as external rolling.