## Task B-2.5.

Point $E$ is the midpoint of side $\overline{A B}$ of square $A B C D$. A point $F$ is chosen on diagonal $\overline{A C}$ such that triangle $E F D$ is a right triangle with a right angle at $F$. In what ratio does point $F$ divide diagonal $\overline{A C}$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_1020492bf6e1d2c949fdg-08.jpg?height=625&width=623&top_left_y=1835&top_left_x=708)

$|D E|=\sqrt{a^{2}+\left(\frac{a}{2}\right)^{2}}=\frac{a}{2} \sqrt{5}$.
$\overline{D E}$ is the diameter of the circle on which points $A$ and $F$ lie (Thales).

Angles $D A F$ and $D E F$ are inscribed angles over the chord $D F$ and are equal to $45^{\circ}$, so triangle $E F D$ is isosceles.

$|D F|=\frac{|D E|}{\sqrt{2}}=\frac{a \sqrt{10}}{4}$.

From triangle $D G F$ we get

$(a-x)^{2}+x^{2}=|D F|^{2}$, i.e., the quadratic equation $16 x^{2}-16 a x+3 a^{2}=0$.

The solutions are $x_{1}=\frac{3 a}{4}$ and $x_{2}=\frac{a}{4}$.

Thus, $|G C|=\frac{a}{4}$.

Triangle $F C G$ is an isosceles right triangle, so $|F C|=x \sqrt{2}=\frac{a \sqrt{2}}{4}=\frac{|A C|}{4}$.

Therefore, $|A F|=\frac{3|A C|}{4}$, or point $F$ divides the diagonal in the ratio $3: 1$.

## NATIONAL MATHEMATICS COMPETITION

3rd grade - high school - B variant

Trogir, April 9, 2015.