Math Competition Question Hong Kong I was (as again) doing practise papers, when i saw this question:
Triangle ABC is divided by segments BD, DF and FE into four triangles. E and D lie on CA and F lies on BC. These four small triangles have equal areas. If BF=2DE, find the ratio of AC:BC.
I have completely no idea on how to solve it. Anyone help me?
 A: 
Since $\triangle CEF$ and $\triangle EDF$ have equal area and equal height, you get $CE=ED$. Without loss of generality, we can assume this length to be equal to $1$. I will also assume the height of these two triangles to be equal to $2$ so that all four areas should be equal to $1$, but read below for comments on this.
The whole triangle $\triangle ABC$ has a base of length $AC=b$ and a height $h$. The resulting area should be $\frac12bh=4$ while the area of $\triangle ABD$ should be $\frac12(b-2)h=1$. You can combine these two to get $b=4(b-2)$, so $b=\frac83$ and threfore $h=3$.
So now you can assume some coordinates:
$$
C=\begin{pmatrix}0\\0\end{pmatrix}\quad
E=\begin{pmatrix}1\\0\end{pmatrix}\quad
D=\begin{pmatrix}2\\0\end{pmatrix}\\
A=\begin{pmatrix}\tfrac83\\0\end{pmatrix}\quad
F=\begin{pmatrix}2x\\2\end{pmatrix}\quad
B=\begin{pmatrix}3x\\3\end{pmatrix}
$$
From this you get
\begin{align*}
\lVert B-F\rVert &= 2\lVert D-E\rVert \\
\lVert B-F\rVert^2 &= 4\lVert D-E\rVert^2 \\
x^2+1 &= 4 \\
x^2 &= 3 \\
x_{1,2} &= \pm\sqrt3
\end{align*}
So there are two solutions, but both lead to the same ratio
$$\frac{\lVert A-C\rVert}{\lVert B-C\rVert}=\frac49$$
You could have choosen a different height for point $F$ up front, which would have led to different solutions for $x$. But the final ratio would nevertheless have been the same. If you wanted to, you could formulate the generic solution like this:
$$
C=\begin{pmatrix}0\\0\end{pmatrix}\quad
E=\begin{pmatrix}1\\0\end{pmatrix}\quad
D=\begin{pmatrix}2\\0\end{pmatrix}\\
A=\begin{pmatrix}\tfrac83\\0\end{pmatrix}\quad
F=\begin{pmatrix}2x\\2y\end{pmatrix}\quad
B=\begin{pmatrix}3x\\3y\end{pmatrix}
$$
\begin{align*}
\lVert B-F\rVert &= 2\lVert D-E\rVert \\
\lVert B-F\rVert^2 &= 4\lVert D-E\rVert^2 \\
x^2+y^2 &= 4
\end{align*}
$$\frac{\lVert A-C\rVert}{\lVert B-C\rVert}=
\frac{\tfrac83}{3\sqrt{x^2+y^2}} =
\frac{8}{3\cdot 3\cdot \sqrt4} =
\frac{4}{9}
$$
A: The area of the triangle $DBC$ is thrice the area of the triangle $ABD$, and they have the same height, so $DC=3AD$, and $AC=4AD$. The triangles $EFC$ and $DFE$ have the same area and the same height, so $CE=ED$ and $BF=DC$. The area of the triangle $CDB$ is thrice the area of the triangle $FDB$, and they have the same height, so $BC=3BF$. But $BF=DC=AC-AD=3AD$, hence $BC=9AD$. Thus $\frac{AC}{BC}=\frac 4 9$.
A: From the ratio $3:1$ of the areas of the triangles $BDC$ and $BDF$, which are of the same height, we have $BC=3BF$. Similarly, by considering the same-height triangles $DFE$ and $EFC$, we get $DE=EC$. And likewise, looking at triangles $ABD$ and $DBC$ gives $DC=3AD$. Now, noting that $AC=AD+DE+EC$ and $DC=DE+EC$, while $BF=2DE$, some simple arithmetic yields $AC:BC=4:9$.
