Project Euler Problem 894
Consider a unit circlecircle with radius 1 C0 on the plane that does not enclose the origin.
Solution
Answer: 0.7718678168
Let the scaling factor be $q\in(0,1)$, rotation angle $\theta$, and let the distance from the origin to the centre of $C_0$ be $a$. Then circle $C_n$ has:
- radius $r_n = q^n$,
- centre at distance $a q^n$ from the origin,
- angle $n\theta$.
Because $C_0$ is externally tangent to $C_1,C_7,C_8$,
$$|O_0O_n| = 1+q^n \qquad (n=1,7,8)$$
and geometrically,
$$|O_0O_n|^2 = a^2\left(1+q^{2n}-2q^n\cos(n\theta)\right).$$
So we solve the system
$$a^2(1+q^2-2q\cos\theta)=(1+q)^2,$$
$$a^2(1+q^{14}-2q^7\cos(7\theta))=(1+q^7)^2,$$
$$a^2(1+q^{16}-2q^8\cos(8\theta))=(1+q^8)^2.$$
Numerically,
$$a \approx 2.4739899466740867, \quad q \approx 0.9063314061485950, \quad \theta \approx 0.8267295394140587.$$
The tangency pattern repeats self-similarly:
- one family of circular triangles formed by $(C_n,C_{n+1},C_{n+8})$,
- another by $(C_n,C_{n+7},C_{n+8})$.
For $n=0$, their areas are computed as:
$$A_1 \approx 0.08231076980835923, \qquad A_2 \approx 0.05551655820073314.$$
Each subsequent triangle is scaled by factor $q^{2n}$ in area, so the total green area is the geometric sum
$$\sum_{n\ge0}(A_1+A_2)q^{2n} = \frac{A_1+A_2}{1-q^2}.$$
Evaluating:
$$\text{Total area} \approx 0.7718678168436464.$$
Rounded to 10 decimal places:
Answer: 0.7718678168