3. Rectangle $A B C D$ has side $A B$ of length $2 a$, and side $A D$ of length $a$. Let the midpoint of side $A B$ be denoted by $E$ and choose an arbitrary point $F$ on side $A D$. The area of triangle $E C F$ depends on the choice of point $F$. What is the smallest and the largest area of triangle $E C F$ as the position of point $F$ changes?
I/3. 1. way The area of triangle $E C F$ is equal to $\frac{|E C| \cdot v}{2}=\frac{a v \sqrt{2}}{2}$, where we denote the length of the height to side $E C$ with $v$. As the height increases when point $F$ travels from $A$ to $D$, the area correspondingly increases. When $F=A$, $v_{\text {min }}=\frac{a \sqrt{2}}{2}$, and the area of triangle $E C F$ is the smallest. It is equal to $\frac{a^{2}}{2}$, which is a quarter of the area of rectangle $A B C D$. When $F=D$, $v_{\max }=a \sqrt{2}$, and the area of triangle $E C F$ is the largest. Then the area of triangle $E C F$ is equal to $a^{2}$, which is half of the area of

![](https://cdn.mathpix.com/cropped/2024_06_07_af92d6d33b11d7748e87g-06.jpg?height=429&width=597&top_left_y=2104&top_left_x=1232)
rectangle $A B C D$.

2. way If we denote $|F A|=x$, the area of triangle $E C F$ is equal to $2 a^{2}-\frac{1}{2} a x-\frac{1}{2} a^{2}-$ $\frac{1}{2}(a-x) 2 a=\frac{1}{2} a^{2}+\frac{1}{2} a x$. Since $a>0$ and $0 \leq x \leq a$, the smallest area of triangle $E C F$ is $\frac{1}{2} a^{2}$ and is achieved at $x=0$, the largest area is $a^{2}$ and is achieved at $x=a$.

Correctly determined minimum area: 2 points. Correctly determined maximum area: 2 points. Geometric justification that these are indeed the extreme possibilities, or an explicit formula for the area involving 1 independent variable: 3 points.