Exercise 4. Four circles $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}, \mathrm{C}_{4}$ of identical radius $\mathrm{r}$ are internally tangent to a circle of radius $R$. We set $C_{5}=C_{1}$. We assume that for all $i=1,2,3,4$, the circles $C_{i}$ and $C_{i+1}$ are tangent. Determine the value of the ratio $\frac{r}{R}$.
## Solution to Exercise 4

![](https://cdn.mathpix.com/cropped/2024_06_04_0d2b7ff8eaebf8b8e86fg-3.jpg?height=501&width=553&top_left_y=492&top_left_x=778)

Let A, B, C, D be the centers of the small circles and O be the center of the large circle. Since the small circles are internally tangent to the large circle, the points A, B, C, D are at a distance R - r from O.

Furthermore, since \( C_1 \) and \( C_2 \) are tangent, we have \( AB = 2r \), and similarly \( BC = CD = DA = 2r \), so the angles \( \widehat{AOB}, \ldots, \widehat{DOA} \) are equal. Since their sum is \( 360^\circ \), these are right angles.

By the Pythagorean theorem, we have \( AB^2 = OA^2 + OB^2 = 2OA^2 = 2(R - r)^2 \). Since \( AB = 2r \), we get \( 2r^2 = (R - r)^2 \). Therefore, \( r\sqrt{2} = R - r \), and thus \( \frac{r}{R} = \frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1 \).