4. Consider the segment $[A B]$. We denote $M$ as the midpoint of $[A B]$, $M_{1}$ as the midpoint of $[B M]$, $M_{2}$ as the midpoint of $\left[M_{1} A\right]$, $M_{3}$ as the midpoint of $\left[M_{2} B\right]$, and $M_{4}$ as the midpoint of $\left[A M_{3}\right]$. It is known that $\mathrm{M}_{1} \mathrm{M}_{3}=\mathrm{M}_{2} \mathrm{M}_{4}+2 \mathrm{~cm}$.

a) Find the length of the segment $[A B]$.

b) Denote by $M_{a}$ and $M_{b}$ the midpoints of the segments $\left[M_{2} M_{4}\right]$ and $\left[M_{1} M_{3}\right]$. Show that $M_{a}$ is the midpoint of $\left[A M_{b}\right]$.

Note: Working time 2 hours.

Each sub-question is worth 7 points.

## GRADING-SOLUTIONS-PROBLEMS-MATHEMATICS OLYMPIAD-

## LOCAL PHASE - FEBRUARY 14, 2014
4. Notam $A B=x, A M=B M=x / 2, M_{1}=M_{1} B=x / 4, A M_{1}=x / 2+x / 4=3 x / 4, A_{2}=M_{2} M_{1}=3 x / 8$, $A M_{3}=3 x / 8+x / 8+3 x / 16=11 x / 16, A M_{4}=11 x / 32=M_{4} M_{3}, M_{2} M=3 x / 8-x / 4=x / 8$, $M_{2} B=x / 8+x / 2=5 x / 8, M_{2} M_{3}=M_{3} B=5 x / 16, M_{3} M_{1}=5 x / 16-x / 4=x / 16$. $(5 x / 16>x / 8$, $5 x / 16<3 x / 8), M_{3}=x / 4-x / 16=3 x / 16$.

3 points.

a)Avem atunci $x / 16=x / 32+2$, deducem $x=64, A B=64 \mathrm{~cm}$.

1 point

b) $M_{2} M_{a}=M_{a} M_{4}=x / 64, M_{1} M_{b}=M_{b} M_{3}=x / 32 . A M_{a}=11 x / 32+x / 64=23 x / 64$ and $M_{a} M_{b}=x / 64+x / 8+3 x / 16+x / 32=(x+8 x+12 x+2 x) / 64=23 x / 64$, so $M_{a}$ is the midpoint of $\left[\mathrm{AM}_{\mathrm{b}}\right]$

3 points.