9.3. Given a triangle $A B C$. A line parallel to $A C$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively, and the median $A M$ at point $Q$. It is known that $P Q=3$, and $Q T=5$. Find the length of $A C$.
Answer: $AC=11$.

Solution. First method. Draw a line through point $Q$ parallel to $BC$ (where $N$ and $L$ are the points of intersection of this line with sides $AB$ and $AC$ respectively, see Fig. 9.3a). Since $AM$ is the median of triangle $ABC$, then $LQ=NQ$. Moreover, $PT \| AC$, so $PQ$ is the midline of triangle $ANL$. Therefore, $AL=2PQ=6$. Additionally, $QL \| TC$ and $QT \| LC$, hence $LQTC$ is a parallelogram, from which $LC=QT=5$. Thus, $AC=AL+LC=6+5=11$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f1014a27e7f65c25d39cg-1.jpg?height=380&width=505&top_left_y=2306&top_left_x=133)

Fig. 9.3a

![](https://cdn.mathpix.com/cropped/2024_05_06_f1014a27e7f65c25d39cg-1.jpg?height=372&width=506&top_left_y=2304&top_left_x=775)

Fig. 9.3b

![](https://cdn.mathpix.com/cropped/2024_05_06_f1014a27e7f65c25d39cg-1.jpg?height=385&width=505&top_left_y=2292&top_left_x=1432)

Fig. 9.3c

Second method. Draw the midline $XM$ in triangle $ABC$ (see Fig. 9.3b). Then $XM \| PT \| AC$, hence $\triangle APQ \sim \triangle AXM$ and $\triangle QMT \sim \triangle AMC$, from which it follows that $\frac{PQ}{XM}=\frac{AQ}{AM}$ and $\frac{QT}{AC}=\frac{QM}{AM}$, that is, $\frac{PQ}{XM}+\frac{QT}{AC}=1$. Substituting the values from the problem and considering that $AC=2MX$, we get $\frac{6}{AC}+\frac{5}{AC}=1$, from which $AC=11$.

Third method. Draw the midline $MN$ in triangle $ABC$ (see Fig. 9.3c). Since $QT \| AC$, then $QT$ is bisected by segment $MN$. From the similarity of triangles $APQ$ and $MXQ$, we get $\frac{AQ}{QM}=\frac{3}{2.5}=\frac{6}{5}$. Then $\frac{MQ}{AM}=\frac{5}{11}$, from which $\frac{QT}{AC}=\frac{5}{11}$, that is, $AC=11$.

Fourth method. Write the Menelaus theorem for triangle $BPT$ and line $AM: \frac{PQ}{QT} \cdot \frac{TM}{MB} \cdot \frac{BA}{AP}=1$. Let $BP=x, AP=y$. Then $\frac{BT}{TC}=\frac{x}{y}$ and $\frac{TM}{MB}=\left(\frac{x+y}{2}-y\right): \frac{x+y}{2}=\frac{x-y}{x+y}$. Thus, $\frac{3}{5} \cdot \frac{x-y}{x+y} \cdot \frac{x+y}{y}=1$, from which $\frac{x-y}{y}=\frac{5}{3}$, that is, $\frac{x}{y}=\frac{8}{3}$. Since $\frac{PT}{AC}=\frac{x}{x+y}$, then $AC=11$.

Grading criteria:

+ fully justified solution

± a correct, overall, solution with an arithmetic error

干 the problem is not solved, but there is a correct idea for an additional construction

- only the correct answer is provided
- the problem is not solved or solved incorrectly
- a special case is considered (e.g., the problem is solved for an isosceles triangle)