On a sheet of paper, Emanuelle draws two circles of radius 1 that are tangent at one point. Then, she draws a third circle of radius $1-\sqrt{2}$ that tangentially touches the previous two externally, as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-07.jpg?height=452&width=727&top_left_y=1112&top_left_x=736)

Emanuelle calculates the area of the region bounded and exterior to the three circles, which is shown in gray in the figure above. What value did she find?

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Solution

a) Let $A, B$ and $C$ be the centers of the circles $C_{1}, C_{2}$ and $C_{3}$ respectively. Consider the triangle whose vertices are given by these points as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-28.jpg?height=465&width=727&top_left_y=510&top_left_x=546)

Let $P$ be the point where $C_{1}$ touches $C_{2}$, $Q$ the point where $C_{2}$ touches $C_{3}$, and $R$ the point where $C_{3}$ touches $C_{1}$. Note that the area of the region in question can be found by subtracting the sum of the areas of the circular sectors $A P R, B P Q$, and $C Q R$ from the area of the triangle $A B C$, as shown in the figure below:
![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-28.jpg?height=466&width=1316&top_left_y=1308&top_left_x=248)

Since the radii of the circles $C_{1}$ and $C_{2}$ are both equal to 1, the lengths of the segments $A P$ and $P B$ are given by $|A P|=1$ and $|P B|=1$. Therefore, the segment $A B$ has a length of $|A B|=|A P|+|P B|=2$. Since the circle $C_{3}$ has a radius of $\sqrt{2}-1$, we have $|C Q|=\sqrt{2}-1$. Therefore, the length of the segment $C B$ is given by $|C B|=|C Q|+|Q B|=\sqrt{2}$. Similarly, we have $|C A|=\sqrt{2}$. It follows that the triangle $A B C$ is isosceles with two sides of length $\sqrt{2}$ and one side of length 2. Since $2^{2}=\sqrt{2}^{2}+\sqrt{2}^{2}$, we have $|A B|^{2}=|B C|^{2}+|C A|^{2}$, which means the sides of the triangle $A B C$ satisfy the Pythagorean theorem. Thus, this triangle is a right triangle with the hypotenuse being the side $A B$. Therefore, the angle $c$ shown in the figure above measures $90^{\circ}$. Since it is also isosceles, the other angles measure $45^{\circ}$. In particular, its area is equal to half the product of the legs, i.e., $(\sqrt{2} \times \sqrt{2}) / 2=1$. We can now calculate the areas of the circular sectors, as shown in the figure below:
![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-29.jpg?height=384&width=614&top_left_y=231&top_left_x=800)

Since $a=45^{\circ}$, the area of the circular sector $A P R$ is equal to $1 / 8$ of the area of the disk bounded by $C_{1}$, i.e.,

$$
\text { Area of sector } A P R=\frac{1}{8} \pi 1^{2}=\frac{\pi}{8}
$$

Similarly, we have

$$
\text { Area of sector } B P Q=\frac{1}{8} \pi 1^{2}=\frac{\pi}{8}
$$

Since $c=45^{\circ}$, the area of the circular sector $C Q R$ is equal to $1 / 4$ of the area of the disk bounded by the circle $C_{3}$, so:

$$
\text { Area of sector } C Q R=\frac{1}{4} \pi(\sqrt{2}-1)^{2}=\frac{(3-2 \sqrt{2}) \pi}{4}
$$

Therefore, the area of the region found by Emanuelle is equal to

$$
1-\pi\left(\frac{3-2 \sqrt{2}}{4}+\frac{1}{8}+\frac{1}{8}\right)=1-\pi \frac{(2-\sqrt{2})}{2}
$$