Answer
Result \[ \frac{\theta}{3} \]
Work Step by Step
Step 1
An angle $\theta$ is bisected using a straightedge and compass to produce ray $\vec{R}_1$, then the angle between $\vec{R}_1$ and the initial side is bisected to produce ray $\vec{R}_2$. Thereafter, rays $\vec{R}_3, \vec{R}_4, \vec{R}_5, \ldots$ are constructed in succession by bisecting the angle between the preceding two rays. Continuing in this way, the angle between $\vec{R}_k$ and the initial side is: \[ \frac{\theta}{2} - \frac{\theta}{2^2} + \frac{\theta}{2^3} - \frac{\theta}{2^4} + \frac{\theta}{2^5} - \ldots + (-1)^{k+1} \frac{\theta}{2^k} \]
Step 2
As $k$ tends to infinity, the angle between $\vec{R}_k$ and the initial side is: \[ \begin{align*} &\frac{\theta}{2} - \frac{\theta}{2^2} + \frac{\theta}{2^3} - \frac{\theta}{2^4} + \frac{\theta}{2^5} - \ldots + (-1)^{k+1} \frac{\theta}{2^k} + \ldots \\ &= \sum_{k=1}^\infty (-1)^{k+1} \frac{\theta}{2^k} \end{align*} \] It is a geometric series with $a = \frac{\theta}{2}$ and $r = -\frac{1}{2}$. Its value is: \[ \begin{align*} &\sum_{k=1}^\infty (-1)^{k+1} \frac{\theta}{2^k} \\ &= \frac{a}{1-r} = \frac{\frac{\theta}{2}}{1 + \frac{1}{2}} = \frac{\theta}{3} \end{align*} \] Thus, we proved that the sequence of angles that rays $\vec{R}_1, \vec{R}_2, \vec{R}_3, \ldots$ make with the initial side has a limit of $\frac{\theta}{3}$. Result \[ \frac{\theta}{3} \]
