A right-angled triangle has acute angles of $60^{\circ}$ and $30^{\circ}$. Two equal-radius circles are inscribed in the triangle, touching the hypotenuse, each other, and one of the legs. How many times is the smaller leg the radius of the circles?
In the right-angled triangle $\mathrm{ABC}$, let $\angle BAC = 60^\circ$ and choose the hypotenuse $AB$ to be 2 units long. It is easy to notice that the triangle is exactly half of an equilateral triangle, so $AC = 1$ and $BC = \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5b00d1f125b2820d967cg-1.jpg?height=466&width=339&top_left_y=274&top_left_x=882)

The projections of the centers $O_{1}$ and $O_{2}$ of the two equal circles with radius $r$ onto the hypotenuse are $E$ and $D$, respectively. Let $AE = x$ and $BD = y$, then

$$
x + y + 2r = 2
$$

From the triangle $O_{1}AE$: $x = \frac{r}{\tan 30^\circ} = \frac{3r}{\sqrt{3}}$. From the triangle $O_{2}BD$: $y = \frac{r}{\tan 15^\circ}$.

$$
\tan 15^\circ = \frac{\sin 30^\circ}{1 + \cos 30^\circ} = \frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}} = 2 - \sqrt{3}, \quad \text{so} \quad y = \frac{r}{2 - \sqrt{3}}
$$

Substituting the obtained values into (1):

$$
\frac{3r}{\sqrt{3}} + \frac{r}{2 - \sqrt{3}} + 2r = 2, \quad \text{thus} \quad r\left(\frac{4\sqrt{3} + 6}{\sqrt{3}}\right) = 2, \text{ hence } \quad r = 2 - \sqrt{3}.
$$

The desired ratio:

$$
\frac{CA}{r} = \frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}
$$