Halving a vector and halving an angle 
An object is subjected to two equal forces along two different directions. If the magnitude of one of them is halved, the angle which the new resultant makes with the other component force is also halved. What is the angle between the forces?


Well I'm not sure whether the vector notation in the second figure is right or not – if wrong please correct it (I'm learning vectors).

Let's set
$$
\left\{
\begin{array}{ll}
\vec{p} &= p \cos{\alpha} \hat{i} + p \sin{\alpha}\hat{j} \\
\vec{p^*} &=p \cos{\alpha} \hat{i} - p \sin{\alpha}\hat{j} \\ 
\end{array}
\right.
$$
$\Rightarrow \vec{p} + \vec{p^*} = 2p \cos{\alpha}\hat{i}$. But
$\frac{1}{2}\vec{p} + \vec{p^*} = \frac{3}{2}p \cos{\alpha} \hat{i} - \frac{1}{2} p \sin{\alpha} \hat{j} = \vec{p}_N$. 
Now, it was defined that the angle between $\vec{p}_N$ and $\vec{p^*}$ is $\frac{\alpha}{2}$:
$$
\cos{\frac{\alpha}{2}} = \frac{\vec{p}_N \cdot \vec{p^*}}{\left|\vec{p}_N \right| \left|\vec{p^*} \right|} = \frac{\frac{3}{2}p^2 \cos^2{\alpha} + \frac{1}{2}p^2 \sin^2{\alpha}}{p^2 \sqrt{  \frac{9}{4}\cos^2{\alpha} + \frac{1}{4} \sin^2{\alpha}  }} = \frac{ \frac{1}{2} + \cos^2{\alpha} }{\sqrt{ \frac{1}{4} + 2 \cos^2{\alpha}}} = \frac{ 1 + 2\cos^2{\alpha} }{\sqrt{ 1 + 8 \cos^2{\alpha}}}
$$
We also note that $\cos{2x} = 2\cos^2{x} - 1 \Rightarrow \cos{\frac{\alpha}{2}} = \sqrt{ \frac{\cos{\alpha} + 1}{2} }$. Inserting this result into the previous equation, and taking the notation $\cos{\alpha} = \eta$, we obtain the equation
$$
\frac{\eta+1}{2} = \left( \frac{1+ 2\eta^2}{\sqrt{1+8\eta^2}} \right)^2
=\frac{(1+ 2\eta^2)^2}{1+8\eta^2} = \frac{4\eta^4 + 4\eta^2 +1}{8\eta + 1}
$$
$$
\Rightarrow 8\eta^3 + 3\eta^2 + \eta +1 = 8\eta^4 + 8 \eta^2 +2
$$
$$
\Rightarrow   8\eta^4 - 8\eta^3 - \eta  + 1= 0
$$
The solutions to this equation are $\eta = 1$ and $\eta= \frac{1}{2}$, giving $\alpha = 0$ (trivial) or $\alpha = 60^{\circ}$. The angle between the forces is therefore $2\alpha = 120^{\circ}$.
