JMO geometry Problem.

Question

$ABCD$ is a parallelogram. $E$ intersects $AD$ as $AE:ED =1:3$ and $F$ intersects $AB$ as $AF:FB=7:1$. $CE$ and $DF$ meets at point $P$. $CP:PE$ =?

Answer 1

Let $DF\cap BC=\{K\}$, $AE=a$ and $BF=b$.

Thus, $AF=7b$, $ED=3a$ and $BC=4a$.

Now, since $\Delta KBF\sim \Delta DAB$, we obtain: $$\frac{KB}{4a}=\frac{b}{7b},$$ which gives $$KB=\frac{4a}{7}$$ and since $\Delta KPC\sim\Delta DPE$, we obtain: $$\frac{PC}{PE}=\frac{4a+\frac{4a}{7}}{3a},$$ which gives $$\frac{PC}{PE}=\frac{32}{21}.$$

Answer 2

Without loss of generality, I assumed that parallelogram is a rectangle (to be honest, I couldn't find how to draw a parallelogram in Paint so I made this assumption. But it doesn't affect the answer). Then by similarity of $\Delta FAD$ and $\Delta FBG$, we have $$\frac{|FB|}{|AF|} = \frac{1}{7} = \frac{|GB|}{|AD|}$$So, $|GB| = \frac{4a}{7}$. Then by similarity of $\Delta PED$ and $\Delta PCG$, we have $$\frac{|CG|}{|AD|} = \frac{|CP|}{|PE|} = \frac{\frac{32a}{7}}{3a} = \frac{32}{21}$$

Answer 3

Using vectors . . .

Regarding all points as vectors, place the origin at $A$.

Since $|AE|\,{:}\,|ED|=1:3$, it follows that $E=\frac{1}{4}D$.

Since $|AF|\,{:}\,|FB|=7:1$, it follows that $F=\frac{7}{8}B$.

Since $ABCD$ is a parallelogram, it follows that $C = B+D$.

Since $P$ is on the segment $DF$, it follows that $$P = D+s(F-D)$$ for some real number $s$, with $0 < s < 1$, where $s$ is the ratio $|DP|\,{:}\,|DF|$. \begin{align*} \text{Then}\;\;P &= D+s(F-D) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \\[4pt] &=D+s({\small{\frac{7}{8}}}B - D)\\[4pt] &=\left({\small{\frac{7}{8}}}s\right)B + (1-s)D\\[4pt] \end{align*}

Since $P$ is on the segment $CE$, it follows that $$P = C+t(E-C)$$ for some real number $t$, with $0 < t < 1$, where $t$ is the ratio $|CP|\,{:}\,|CE|$. \begin{align*} \text{Then}\;\;P &= C+t(E-C)\\[4pt] &=(B+D) + t({\small{\frac{1}{4}}}D-(B+D)\bigr)\\[4pt] &=(1-t)B+(1-{\small{\frac{3}{4}}}t)D\\[4pt] \end{align*} Equating the two expressions for $P$, we get $$\left({\small{\frac{7}{8}}}s\right)B + (1-s)D = (1-t)B+(1-{\small{\frac{3}{4}}}t)D$$ Since $B,D$ are linearly independent, we get the two equations \begin{align*} {\small{\frac{7}{8}}}s &= 1-t\\[4pt] 1-s &= 1-{\small{\frac{3}{4}}}t \end{align*} Solving the above system yields $s = {\large{\frac{24}{53}}},\;t = {\large{\frac{32}{53}}}$.

Since $|CP|\,{:}\,|CE|=t={\large{\frac{32}{53}}}$, it follows that $|CP|\,{:}\,|PE|=32\,{:}\,21$.