Problem 1. Consider three collinear points $A, B$, and $C$ such that $\frac{A B}{B C}=\frac{3}{5}$ and $A C=40 \text{ cm}$. Let $M$ and $P$ be the midpoints of segments $[A B]$, respectively $[B C]$.

a) Determine the lengths of segments $[A B]$ and $[B C]$.

b) Determine the value of the ratio $\frac{A P}{M C}$.
## Solution and Scoring

The lengths of the segments $[A B],[B C]$, and the value of the ratio $\frac{A P}{M C}$ will be calculated based on the position of point $C$ relative to points $A$ and $B$, using the relationship $A B=\frac{3}{5} \cdot B C$

Thus:

a) If $B \in[A C]$, then $A C=A B+B C=\frac{3}{5} \cdot B C+B C=\frac{8}{5} \cdot B C$; we obtain the equation $\frac{8}{5} \cdot B C=40$, from which $B C=25 \mathrm{~cm}$, and $A B=15 \mathrm{~cm}$.

If $C \in(B A)$, since $B A<B C$ it follows that $A \in[B C]$ and then $A C=B C-A B=B C-\frac{3}{5} \cdot B C=\frac{2}{5} \cdot B C$.

We obtain the equation $\frac{2}{5} \cdot B C=40$, from which $B C=100 \mathrm{~cm}$, and $A B=60 \mathrm{~cm}$. 3 points

b) If $B \in[A C]$, then $\frac{A P}{M C}=\frac{A B+\frac{B C}{2}}{B C+\frac{A B}{2}}=\frac{11}{13}$. 2 points

If $A \in[B C]$, we get $C A<C P<C M<C B$, meaning the points are in the order $C-A-P-M-B$, and $\frac{A P}{M C}=\frac{1}{7}$

1 point