5. (8 points) On the sides $A B, B C$ and $C A$ of triangle $A B C$, points $C_{1}, A_{1}$ and $B_{1}$ are marked respectively. Segments $A A_{1}, B B_{1}$ and $C C_{1}$ intersect at point $P$. Find the ratio $A P: P A_{1}$, if $A B: A C_{1}=\alpha>1$ and $A C: A B_{1}=\beta>1$.
# Solution (Ceva's and Menelaus' Theorems):

![](https://cdn.mathpix.com/cropped/2024_05_06_e38b8a325740b6549434g-05.jpg?height=557&width=1020&top_left_y=401&top_left_x=541)

Since \( A B: A C_{1}=\alpha \) and \( A C: A B_{1}=\beta \), then \(\frac{A C_{1}}{C_{1} B}=\frac{1}{\alpha-1}\) and \(\frac{A B_{1}}{B_{1} C}=\frac{1}{\beta-1}\).

We write Ceva's theorem for triangle \( A B C \) and cevians \( A A_{1}, B B_{1} \) and \( C C_{1} \):

\[
\frac{A C_{1}}{C_{1} B} \cdot \frac{B A_{1}}{A_{1} C} \cdot \frac{C B_{1}}{B_{1} A}=1
\]

Thus,

\[
\frac{B A_{1}}{A_{1} C}=\frac{A B_{1}}{B_{1} C} \cdot \frac{B C_{1}}{C_{1} A}=\frac{\alpha-1}{\beta-1}
\]

Now, we use Menelaus' theorem for \( A A_{1} C \) and line \( B B_{1} \):

\[
\frac{B A_{1}}{B C} \cdot \frac{C B_{1}}{B_{1} A} \cdot \frac{A P}{P A_{1}}=1
\]

or

\[
\frac{A P}{P A_{1}}=\frac{B C}{B A_{1}} \cdot \frac{A B_{1}}{B_{1} C}=\left(1+\frac{\beta-1}{\alpha-1}\right) \cdot \frac{1}{\beta-1}=\frac{1}{\alpha-1}+\frac{1}{\beta-1}
\]

Solution (Van Aubel's Theorem): Apply Van Aubel's theorem:

\[
\frac{A P}{P A_{1}}=\frac{A C_{1}}{C_{1} B}+\frac{A B_{1}}{B_{1} C}=\frac{1}{\alpha-1}+\frac{1}{\beta-1}
\]

Answers:

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