Triangle problem with 3 unknowns.

Question

I've just been sent this question by somebody. It's been a while, but I'm not sure this can be done with the information available.

The questions says to find a value for $k$:

So far all I've been able to prove is that

$$OB=6b-kb+kb=6b$$

From here all equations of a triangle (sine rule, cosine rule etc.) would still leave two unknowns so I'm not sure how to find a value for $k$.

Answer 1

Apply Menelaus's Theorem to get $$\frac{3a}{3a} \times \frac{2x}{x} \times \frac{(6-k)b}{kb}=1$$

This gives us that $$\frac{(6-k)}{k}=\frac{1}{2}$$ So we have $k=4$.

Another way to see this, of course, is to notice that $N$ is the centroid of $\triangle{OAC}$, which gives us that $\overline{ON} : \overline{NB}=2:1$.