Lauburu Geometry

For the problem, we use the following formula

• Circle matrix : If the equation is $(x - h)^2 + (y - k)^2 = r^2$ , then the matrix is $\begin{bmatrix} 1 & 0 & -h\\ 0 & 1 & -k\\ -h & -k & h^2 + k^2 - r^2 \end{bmatrix}$
• Transformation 1: $L^TC^{-1}L = 0$
• Transformation 2: If $ax + by + c = 0$ is the formula of a straight line, then $\theta$ -rotation is $\begin{array}{rl} x' &= a\cos(\theta) + b\sin(\theta)\\ y' &= -a\sin(\theta) + b\cos(\theta) \end{array}$

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Let

• The respecting point be at the origin $(0,0)$
• Unit circle equation denote $(x + 3)^2 + y^2 = 1$ , so the matrix is $C_{1} = \begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & 0 \\ 3 & 0 & 8 \end{bmatrix}$
• Large circle equation denote $x^2 + (y - 2)^2 = 4$ , so the matrix is $C_{2} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & -2 & 0 \end{bmatrix}$
• Arbitrary point to be of the matrix form $P = \begin{pmatrix}x & y & 1 \end{pmatrix}^T$
• The equation of the tangent line be of the matrix form $L = \begin{pmatrix}a & b & c \end{pmatrix}^T$ , where $P^TL = 0$ gives $ax + by + c = 0$ . With the angle $\theta$ (respect to the horizontal) and horizontal distance to the origin $d$ , we can write $L = \begin{pmatrix}-\sin\theta & \cos\theta & -d \end{pmatrix}^T$

From $L^TC_1^{-1}L = 0$ and $L^TC_2^{-1}L = 0$ , the tangent line satisfies the following system of equations $\begin{array}{rrl} C_1:& d^2 - 6d\sin(\theta) + 8\sin^2(\theta) - \cos^2(\theta) &= 0\\ C_2:& d^4 - 4d\cos(\theta) - 4\sin^2(\theta) &= 0 \end{array}$ where the possible solutions are

• $d = \dfrac{2}{13}$ and $\theta = \arctan(5)$
• $d = 2$ and $\theta = \dfrac{\pi}{2}$
• $d = -\dfrac{2}{13}\left(6\sqrt{3} - 11\right)$ and $\theta = -2\arctan(3 + 2\sqrt{3})$
• $d = \dfrac{2}{13}\left(6\sqrt{3} + 11\right)$ and $\theta = -2\arctan(3 - 2\sqrt{3})$

Since the fourth solution resembles the equation of outer tangent line, the equation of $L$ is $\dfrac{2\sqrt{3}-3}{6\sqrt{3}-11}x - \dfrac{2(3\sqrt{3} - 5)}{6\sqrt{3} - 11}y - \dfrac{2}{13}\left(6\sqrt{3} + 11\right) = 0$ Then, since the formed figure is symmetric about the origin, apply Transformation 2 (see above solution) for $L$ , rotating about $-90^{\circ}$ and $90^{\circ}$ to get $\begin{array}{rl} \dfrac{2\sqrt{3}-3}{6\sqrt{3}-11}y + \dfrac{2(3\sqrt{3} - 5)}{6\sqrt{3} - 11}x - \dfrac{2}{13}\left(6\sqrt{3} + 11\right) &= 0\\ -\dfrac{2\sqrt{3}-3}{6\sqrt{3}-11}y - \dfrac{2(3\sqrt{3} - 5)}{6\sqrt{3} - 11}x - \dfrac{2}{13}\left(6\sqrt{3} + 11\right) = 0 \end{array}$ where the intersection points are $\left(-\dfrac{38 + 16\sqrt{3}}{169}, \dfrac{382 + 232\sqrt{3}}{169}\right)$ and $\left(-\dfrac{382 + 232\sqrt{3}}{169}, -\dfrac{38 + 16\sqrt{3}}{169}\right)$ . So the distance between these two intersection points is $\dfrac{4}{13}\left(11 + 6\sqrt{3}\right)$ . Thus, since each of the four shaded regions has the area of $2\pi$ , $\text{Area} = \left(\dfrac{4}{13}\left(11 + 6\sqrt{3}\right)\right)^2 - 8\pi \approx \boxed{18.19}$ .