## SUBJECT IV

Consider the triangle $\mathrm{ABC}$ where $\mathrm{D}$ is the midpoint of segment $\mathrm{BC}$. If E is the midpoint of segment $\mathrm{AD}$, F is the midpoint of segment $\mathrm{BE}$, and $\mathrm{G}$ is the intersection point of lines $\mathrm{AB}$ and $\mathrm{CF}$, calculate $\frac{A G}{A B}$.

Mathematical Gazette

## Note.

All problems are mandatory.

Each problem is worth 7 points.

Working time 3 hours.

Proposers: Prof. Stoica Angela, Curta Doina, Danciu Alin, Gîņa Vasile

## MURES COUNTY SCHOOL INSPECTORATE

S.S.M.R - MURES BRANCH

Mathematics Olympiad

Local phase 14.02.2014

8th grade

Grading rubric


## SUBJECT IV

Consider the triangle $\mathrm{ABC}$ where $\mathrm{D}$ is the midpoint of segment $\mathrm{BC}$. If E is the midpoint of segment $\mathrm{AD}$, $\mathrm{F}$ is the midpoint of segment $\mathrm{BE}$, and $\mathrm{G}$ is the intersection point of lines $\mathrm{AB}$ and $\mathrm{CF}$, calculate $\frac{A G}{A B}$.

## Solution

Construct $F M \| A B$ with $M \in A D$ and $E N \| A B$ with $N \in B C$. (1p)

We will have $F M$ as the midline in $\triangle E A B \Rightarrow F M=\frac{A B}{2}$ and $E N$ as the midline in $\triangle D A B \Rightarrow N E=\frac{A B}{2}$

Thus, $F M=N E=\frac{A B}{2}(1 \mathrm{p})$

$\triangle F G B \equiv \triangle F P E(U L . U) \Rightarrow[E P] \equiv[B G](1 \mathrm{p})$

NOTE: Any other correct solution is evaluated with maximum points.

From (t.f.a) $\Rightarrow \triangle C P N \approx \triangle C G B \Rightarrow \frac{N P}{B G}=\frac{C N}{B C}=\frac{3}{4} \Rightarrow N P=\frac{3 \cdot B G}{4}$ (2p)

But, $N E=E P+N P \Rightarrow \frac{A B}{2}=B G+\frac{3 \cdot B G}{4} \Rightarrow A B=\frac{7 \cdot B G}{2} \Rightarrow A G=\frac{5 \cdot B G}{2}$ (1p)

Finally, $\frac{A G}{A B}=\frac{5}{7} \cdot(1 \mathrm{p})$