Problem 1582

1. Define the problem and setup the equations:
We have a fixed circle with radius $a$ and a rolling circle with radius $b = \frac{a}{2}$. The rolling circle rolls without slipping on the inner surface of the fixed circle. We need to prove that a point marked on the circumference of the rolling circle traces a straight line as the circle rolls.

2. Equation of the fixed circle:
The fixed circle can be described by the equation:
$$x^2 + y^2 = a^2$$

3. Parameterize the position of the rolling circle:
Let $\phi$ be the angle parameterizing the position of the rolling circle's center as it rolls inside the fixed circle. The center of the rolling circle will be at:
$$\left( (a - b) \cos \phi, (a - b) \sin \phi \right)$$
Since $b = \frac{a}{2}$, this becomes:
$$\left( \frac{a}{2} \cos \phi, \frac{a}{2} \sin \phi \right)$$

4. Parameterize the position of the marked point:
Let $\theta$ be the angle parameterizing the position of the marked point on the rolling circle relative to its center. Since the rolling circle rolls without slipping, $\theta$ is related to $\phi$ by:
$$\theta = \frac{a - b}{b} \phi = \phi$$
because $\frac{a - b}{b} = 1$.

5. Position of the marked point:
The coordinates of the marked point on the rolling circle are given by:
$$\left( \frac{a}{2} \cos \phi + \frac{a}{2} \cos \theta, \frac{a}{2} \sin \phi - \frac{a}{2} \sin \theta \right)$$
Substituting $\theta = \phi$, we get:
$$\left( \frac{a}{2} \cos \phi + \frac{a}{2} \cos \phi, \frac{a}{2} \sin \phi - \frac{a}{2} \sin \phi \right)$$
Simplifying, we obtain:
$$\left( a \cos \phi, 0 \right)$$

6. Conclusion:
The coordinates of the marked point are $(a \cos \phi, 0)$, which lies on the x-axis. Therefore, as the rolling circle makes a turn, the marked point moves along the x-axis, which is a straight line.

$\blacksquare$