3. Consider the arbitrary triangle $\triangle A B C, N$ the midpoint of $[A C], E$ the symmetric point of $B$ with respect to $N$ and $P$ the symmetric point of $C$ with respect to $B$. If the line $P E$ intersects the sides $[A B]$ and $[A C]$ at points $M$, respectively $F$, calculate the ratio $\frac{F M}{M P}$.

Prof. Ispăsoiu Dorel
Solution: Since $N$ is the midpoint of $\quad[A C] \Rightarrow[A N] \equiv[N C](1), \quad$ and $\quad E=\operatorname{sim}_{N} B \Rightarrow B-N-E$ are collinear and $[B N] \equiv[N E](2)$. From (1) and (2) we have $A B C E$ is a parallelogram $\Rightarrow A E \| B C,[A E] \equiv[B C](3)$. Since $P=\operatorname{sim}_{B} C \Rightarrow P-B-C$ are collinear and $[P B] \equiv[B C](4)$. From (3) and (4) $\Rightarrow[A E] \equiv[P B]$ and since $A E \| P B$ we have $A E B P$ is a parallelogram $\Rightarrow[A M] \equiv[M B],[P M] \equiv[M E](5)$. Therefore, $E M$ and $A N$ are medians in $\triangle A B E$, $A N \cap E M=\{F\} \Rightarrow F$ is the centroid in $\triangle A B E \Rightarrow \frac{F M}{M E}=\frac{1}{3}$. From (5) $\Rightarrow \frac{F M}{M P}=\frac{1}{3}$.

## Scoring:

Figure

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$N$ is the midpoint of $[A C] \Rightarrow[A N] \equiv[N C](1) ; E=\operatorname{sim}_{N} B \Rightarrow B-N-E$ are collinear and $[B N] \equiv[N E](2)$

From (1) and (2) we have $A B C E$ is a parallelogram.

| $A B C E$ is a parallelogram $\Rightarrow A E \| B C,[A E] \equiv[B C](3) ; \quad P=\operatorname{sim}_{B} C \Rightarrow P-B-C \quad$ are collinear and | $\mathbf{1} \mathbf{p}$ |
| :--- | :--- |
| $[P B] \equiv[B C](4)$. From (3) and (4) $\Rightarrow[A E] \equiv[P B]$ and since $A E \| P B$ we have $A E B P$ is a parallelogram. | $1 \mathbf{1} \mathbf{p}$ |
| $A E B P$ is a parallelogram $\Rightarrow[A M] \equiv[M B],[P M] \equiv[M E](5)$ | $\mathbf{1} \mathbf{p}$ |
| $E M$ and $A N$ are medians in $\triangle A B E, A N \cap E M=\{F\} \Rightarrow F$ is the centroid in $\triangle A B E$ | $\mathbf{1} \mathbf{p}$ |
| $F$ is the centroid in $\triangle A B E \Rightarrow \frac{F M}{M E}=\frac{1}{3}$ | $\mathbf{1} \mathbf{p}$ |
| From $(5) \Rightarrow \frac{F M}{M P}=\frac{1}{3}$. |  |