The astronomers of Planet Xiddler are back!
This time, they have identified three planets that circularly orbit a neighboring star. Planet A is three astronomical units away from its star and completes its orbit in three years. Planet B is four astronomical units away from the star and completes its orbit in four years. Finally, Planet C is five astronomical units away from the star and completes its orbit in five years.
They report their findings to Xiddler’s Grand Minister, along with the auspicious news that all three planets are currently lined up (i.e., they are collinear) with their star. However, the Grand Minister is far more interested in the three planets than the star and wants to know how long it will be until the planets are next aligned.
How many years will it be until the three planets are again collinear (not necessarily including the star)?
Assume the sun is always centered at $(0, 0)$, and Planets A, B, and C are currently at $(3, 0)$, $(4, 0)$, and $(5, 0)$ respectively, measured in astronomical units.
Since the radius and period were give for each planet, their coordinates along the system ecliptic can be determined.
$$A_x = 3\cos\left(\frac{2\pi t}{3}\right)$$$$A_y = 3\sin\left(\frac{2\pi t}{3}\right)$$
$$B_x = 4\cos\left(\frac{2\pi t}{4}\right)$$
$$B_y = 4\sin\left(\frac{2\pi t}{4}\right)$$
$$C_x = 5\cos\left(\frac{2\pi t}{5}\right)$$
$$C_y = 5\sin\left(\frac{2\pi t}{5}\right)$$
There are various equations one can use, but the slope of Planet A and Planet B should be equal to the slope of Planet A and Planet C to have collinearity on their ecliptic. Thus,
$$\dfrac{A_y - B_y}{A_x - B_x} = \dfrac{A_y - C_y}{A_x - C_x}$$
$$(A_y - B_y)(A_x - C_x) = (A_y - C_y)(A_x - B_x)$$
$$A_y \cdot A_x - A_y \cdot C_x - B_y \cdot A_x + B_y \cdot C_x = A_y \cdot A_x - A_y \cdot B_x - C_y \cdot A_x + C_y \cdot B_x$$
$$0 = (B_{y}A_x - B_{x}A_y) + (A_{y}C_x - A_{x}C_y) + (C_{y}B_x - C_{x}B_y)$$
$$0 = 12\big(\sin(\theta_B)\cos(\theta_A)-\cos(\theta_B)\sin(\theta_A)\big) + 15\big(\sin(\theta_A)\cos(\theta_C)-\cos(\theta_A)\sin(\theta_C)\big) + 20\big(\sin(\theta_C)\cos(\theta_B)-\cos(\theta_C)\sin(\theta_B)\big)$$
Using the difference identity $\sin(m)\cos(n) - \cos(m)\sin(n) = \sin(m - n)$,
$$0 = 12\sin(\theta_B-\theta_A) + 15\sin(\theta_A-\theta_C) + 20\sin(\theta_C-\theta_B)$$
$$0 = 12\sin\left(-\frac{2\pi t}{12}\right) + 15\sin\left(\frac{4\pi t}{15}\right) + 20\sin\left(-\frac{2\pi t}{20}\right)$$
$$0 = -12\sin\left(\frac{2\pi t}{12}\right) + 15\sin\left(\frac{4\pi t}{15}\right) - 20\sin\left(\frac{2\pi t}{20}\right)$$