GCSE Vector Question Being Pulled For Being 'Too Hard'

Question

So apparently this was lined up as the last question on a GCSE paper but is being pulled because it was deemed 'too hard'. I personally have studied it for over an hour and my feeling is not that it's just too hard, rather that it's unsolvable without more information. Or am I missing something obvious?

Here is the full question:

All I managed to glean are the following expressions:

$\vec{CX} = \frac{m}{m+1}(a+b)$

$\vec{XE} = \frac{1}{m+1}(a+b)$

$\vec{FX} = \frac{n}{n+1}(2a-\frac{1}{2}b)$

$\vec{XM} = \frac{1}{n+1}(2a-\frac{1}{2}b)$

I've no idea how one would find values for $m$ (the scalar ratio of $\vec{CX}$ to $\vec{XE}$) or $n$ without being given the vectors $a$ and $b$.

Answer 1

Without loss of generality, translate the entire diagram so that $F$ is at the origin.

$X$ lies on two lines. One through $C$ and $E$, i.e. on the line $(a-b)+s(a+b)$, where $s$ ranges through the real numbers, and one through $F$ and $M$, i.e. on the line $t(a-b+a+b/2) = t(2a-b/2)$, where $t$ ranges through the real numbers. Equating these and collecting coefficients of $a$ and $b$, we have $$ (1+s)a + (-1+s)b = (2t)a + (-t/2)b $$

Since $a$ and $b$ are independent, we have $1+s = 2t$ and $-1+s = -t/2$. Solving this system, we obtain $s = 3/5$, $t = 4/5$. Note that $X$ is $t$ multiples of $|2a-b/2|$ from $F$ and $1-t$ multiples of $|2a-b/2|$ from $M$, so $n = t/(1-t) = 4$.

Answer 2

By the Ratio Theorem $\left(\text{or using }\,\vec{EX}=\vec{EM}+\frac1{n+1}\vec{MF}\right),$ $$\vec{EX}=\frac{-1}{n+1}\left(2\mathbf{a}+\frac{n}2 \mathbf{b}\right).$$

Since $CXE$ is a straight line, for some real $\lambda,$ $$\begin{aligned}\vec{EX}&=\lambda\vec{EC}\\&=-\lambda\left(\mathbf{a}+\mathbf{b}\right).\end{aligned}$$

Equating the above expressions for $\vec{EX}:$ $$\frac{2}{n+1}\mathbf{a}+\frac{n}{2(n+1)} \mathbf{b}=\lambda\mathbf{a}+\lambda\mathbf{b}.$$

Since $\mathbf{a}$ and $\mathbf{b}$ are not collinear, $$\frac{2}{n+1}=\lambda=\frac{n}{2(n+1)}\\n=4.$$