Ratio in triangle $\Delta ABC$. Find ratio $|CF|:|FP|$.

Question

Let $\Delta ABC$ be triangle and $P$ and $Q$ points on $AB$ and $AC$ such that $|AP|:|PB|=1:5$ and $|AQ|:|QC|=2:1$. Lines $BQ$ and $CP$ intersect in point $F$. Find ratio $|CF|:|FP|$.

I want to solve this problem without using triangle similarities or some theorems, but only vectors. I tried to write vector $\vec{CF}$ and $\vec{CP}$ using for example $\vec{AB}$ and $\vec{AC}$ that would solve my problem, but can't do it.

Answer 1

Use a reference system centered at $O$ and let for simplicity $\overrightarrow{OA}=:\vec{a}, \overrightarrow{OB}=:\vec b, \overrightarrow{OC}=:\vec c,\overrightarrow{OP}=:\vec p$, etc. Notice that, using the given ratios, we have $$\vec p=\frac56\cdot \vec a+\frac16\cdot \vec b, \qquad\vec q=\frac13\cdot \vec a+\frac23\cdot \vec c$$ Since $F$ is collinear with $B,Q$ and $P,C$, consider the real numbers $\lambda, \mu$ satisfying $$\begin{cases}\overrightarrow{CF}=\lambda \cdot \overrightarrow{CP}=\lambda\cdot \left(\frac56\cdot \vec a+\frac16\cdot \vec b-\vec c\right)\\\overrightarrow{BF}=\mu\cdot \overrightarrow{BQ}=\mu\cdot \left (\frac13\cdot \vec a+\frac23\cdot \vec c-\vec b\right)\end{cases}$$ We deduce that $$\begin{align*}\overrightarrow{CF}+\overrightarrow{FB}&=\overrightarrow{CB}\\\iff\lambda\cdot \left(\frac56\cdot \vec a+\frac16\cdot \vec b-\vec c\right)+ \mu\cdot \left (-\frac13\cdot \vec a-\frac23\cdot \vec c+\vec b\right)&=\vec b-\vec c \end{align*}$$ This yields $$\begin{cases}\frac56\cdot \lambda-\frac13\cdot \mu=0\\\frac16\cdot \lambda+\mu=1\\-\lambda -\frac23\mu=-1\end{cases}$$ Solve the system to obtain $\lambda=\frac38$ and $\mu=\frac{15}{16}$. Hence $$\overrightarrow{CF}=\frac38 \cdot \overrightarrow{CP}\implies \lvert CF\rvert:\lvert FP\rvert=3:5$$

Answer 2

There is a one way... let's hang 1 kg item on A after that you can easily say that u must hang 2 kg item on C to balance AC line where Q is center of mass. ( mass(A) * AQ = QC * mass(c) ) using the same way find mass of P mass(A) * AP = PB mass(B) $ \rightarrow $ mass(B) = 1\5 so the center of mass$ (P) = (1 + 1/5) $

Now you know the mass of P and C now write mass(P)*PF = mass(C) * CF $ \rightarrow $ $ \frac{CF}{FP} = \frac{mass(P)}{(mass(C)} $

Answer 3

$\vec{AC} = \vec{b}, \vec{AB} = \vec{c}$

Say, $|\vec{PF}|:|\vec{FC}| = (1-x):x$, $|\vec{BF}|:|\vec{FQ}| = (1-y):y$

$\vec{FC} = x \,\vec{PC} = x(\vec{PA} + \vec{AC}) = x(\vec{b} - \frac{\vec{c}}{6})$

Similarly,

$\vec{FC} = (\vec{FQ} + \vec{QC}) = y \vec{BQ} + \frac{\vec{b}}{3} = y(\vec{BA} + \vec{AQ}) + \frac{\vec{b}}{3} = y(\frac{2\vec{b}}{3} -\vec{c}) + \frac{\vec{b}}{3} \,$

Equating the coefficients of $\vec{b}$ and $\vec{c}$,

$x = \frac{2y+1}{3}, \frac{x}{6} = y$

Solving both, $x = \frac{3}{8}$, so $\, 1-x = \frac{5}{8}$

$CF:FP = 3:5$