Find scalar quantity linking two lengths The question is a GCSE maths question per the picture below. A friend asked me and I can't figure it out.. and I have a maths degree.. embarrassing! What would be a good approach to this question?

 A: 
$|AP|=k\,|AB|$, $\angle NOB=\alpha$.
Let $[OBN]$ denote the area of $\triangle OBN$.
\begin{align} 
[AOM] &=[AMB]= \tfrac12\,ab\sin\alpha
,\\
[OBN]&=\tfrac12\cdot3a\cdot2b\,\sin\alpha=3ab\sin\alpha
,\\
[ABN]&=[OBN]-2[AOM]=2\,ab\sin\alpha=4[AOM] 
,\\
[OMN]&=\tfrac12[OBN]=\tfrac32 ab\sin\alpha
,\\
[AMP]&=k[AMB]
,\\
[APN]&=k[ABN]
,\\
[AMP]+[APN]&=
[OMN]-[AOM]=ab\sin\alpha
,\\
[AMP]+[APN]&=
k([AMB]+[ABN])=\tfrac52\,k\,ab\sin\alpha
,\\
k&=\tfrac25
.
\end{align}  
A: Just using vectors:
$$
\vec{MN}=\vec{ON}-\vec{OM}=3\vec a-{1\over2}\vec b;
$$
$$
\vec{MP}=\vec{OA}+\vec{AP}-\vec{OM}=\vec a+k(\vec b-\vec a)-{1\over2}\vec b
=(1-k)\vec a+\left(k-{1\over2}\right)\vec b.
$$
But $\vec{MN}$ and $\vec{MP}$ have the same direction, hence their coefficients must be proportional:
$$
3:\left(-{1\over2}\right)=(1-k):\left(k-{1\over2}\right),
\quad\hbox{whence}\quad
k={2\over5}.
$$
A: ![enter image description -
By putting X just in middle, I require this area in exact two equal area.
Means where Y fall ?
Means shaded area is equal to unshaded area.
](https://i.stack.imgur.com/c6fgP.jpg)
