## Task 5 - 230935

In a triangle $A B C$, a line parallel to $A B$, whose position is otherwise not specified, intersects the side $A C$ at a point $A_{1}$ between $A$ and $C$, and it intersects the side $B C$ at $B_{1}$. Furthermore, let $P$ be a point on $A B$ between $A$ and $B$, whose position is also not otherwise specified.

The area of triangle $A B C$ is $F_{0}$, and the area of triangle $A_{1} B_{1} C$ is $F_{1}$.

Determine the area $F$ of the quadrilateral $A_{1} P B_{1} C$ in terms of $F_{0}$ and $F_{1}$!
The triangles $\triangle A_{1} B_{1} C$ and $\triangle A B C$ are similar due to the parallelism of $A_{1} B_{1}$ and $A B$, as their corresponding interior angles at $A_{1}$ and $A$ as well as $B_{1}$ and $B$ are alternate interior angles, and thus equal. Therefore, there exists a real number $k>0$ such that $\triangle A_{1} B_{1} C$ is obtained from $\triangle A B C$ by a dilation with factor $k$ and center $C$.

Let $h$ be the length of the altitude from $C$ to $A B$ and $h_{1}$ the length of the altitude from $C$ to $A_{1} B_{1}$. It follows that $h_{1}=k \cdot h$ and $\left|A_{1} B_{1}\right|=k \cdot|A B|$, thus $F_{1}=k^{2} \cdot F_{0}$, i.e., $k=\sqrt{\frac{F_{1}}{F_{0}}}$.

The diagonal $A_{1} B_{1}$ divides the quadrilateral $A_{1} P B_{1} C$ into the two triangles $\triangle A_{1} B_{1} C$ and $\triangle A_{1} B_{1} P$, so its area is the sum of the areas of these two sub-triangles.

Let $h_{2}$ be the length of the altitude from $P$ to $A_{1} B_{1}$, and $F$ the foot of the perpendicular from $C$ to $A B$, and $F_{1}$ the foot of the perpendicular from $C$ to $A_{1} B_{1}$. Since $A_{1} B_{1} \| A B$, $F_{1}$ lies on $C F$ and we have

$$
h=|C F|=\left|C F_{1}\right|+\left|F_{1} F\right|=h_{1}+\left|F_{1} F\right|
$$

Again, because $A_{1} B_{1} \| A B$, $h_{2}=\left|F_{1} F\right|$, since $F_{1}$ is also the foot of the perpendicular from $F$ to $A_{1} B_{1}$. Therefore, $h_{2}=h-h_{1}=(1-k) \cdot h$.

For the triangle $\triangle A_{1} B_{1} P$, the area is thus

$$
F_{P}:=\frac{1}{2}\left|A_{1} B_{1}\right| \cdot h_{2}=\frac{1}{2} \cdot k \cdot|A B| \cdot(1-k) \cdot h=F_{0} \cdot\left(k-k^{2}\right)
$$

and thus for the area $F$ of the quadrilateral $A_{1} P B_{1} C$

$$
F=F_{P}+F_{1}=F_{0} \cdot\left(k-k^{2}+k^{2}\right)=F_{0} \cdot k=F_{0} \cdot \sqrt{\frac{F_{1}}{F_{0}}}=\sqrt{F_{1} \cdot F_{0}}
$$

#### Abstract