B5. A square $A B C D$ with side 8 is folded in such a way that vertex $A$ coincides with the midpoint of $C D$ (see figure). What is the area of the gray triangle?

![](https://cdn.mathpix.com/cropped/2024_04_17_e8fb87dfe5880fdb9bf7g-1.jpg?height=391&width=505&top_left_y=2237&top_left_x=1392)

## C-Problems

For the C-Problems, not only the answer is important; your reasoning and the method of solving must also be clearly written. Solve each C-Problem on a separate sheet of paper. Each correctly worked-out C-Problem is worth 10 points. Partial solutions can also earn points. Therefore, write everything clearly and submit your scratch paper (for each problem!) as well.
B5.

We introduce the points $E, F$, and $G$ as indicated in the figure. Suppose $|D E|=x$. Then $|A E|=8-x$, because the side of the square has length 8. Using the Pythagorean theorem, we get $(8-x)^{2}=|A E|^{2}=|D E|^{2}+|A D|^{2}=x^{2}+16$. Solving this gives $x=3$. Note that $\angle C A F=180^{\circ}-\angle D A E-\angle E A B=90^{\circ}-\angle D A E=\angle D E A$. Furthermore, $\angle A D E=90^{\circ}=\angle F C A$. Therefore, triangles $D E A$ and $C A F$ are similar (AA). This implies $\frac{1}{4}|A F|=\frac{|A F|}{|A C|}=\frac{|E A|}{|D E|}=\frac{5}{3}$ and thus $|A F|=\frac{20}{3}$. We also get $\frac{1}{4}|C F|=\frac{|C F|}{|A C|}=\frac{|A D|}{|D E|}=\frac{4}{3}$ and

![](https://cdn.mathpix.com/cropped/2024_04_17_b513c92c69bbbad6ea97g-2.jpg?height=437&width=502&top_left_y=204&top_left_x=1391)
thus $|C F|=\frac{16}{3}$. Now it follows that $|B F|=8-|A F|=\frac{4}{3}$. Angles $\angle C F A$ and $\angle B F G$ are equal (vertical angles). Moreover, angles $\angle A C F$ and $\angle G B F$ are both $90^{\circ}$ and thus equal. Therefore, triangles $C F A$ and $B F G$ are similar (AA). This implies $\frac{3}{4}|B G|=\frac{|B G|}{|B F|}=\frac{|A C|}{|C F|}=\frac{4}{\frac{16}{3}}=\frac{3}{4}$ and thus $|B G|=1$. The area of the gray triangle is now $\frac{1}{2} \cdot|B G| \cdot|B F|=\frac{1}{2} \cdot 1 \cdot \frac{4}{3}=\frac{2}{3}$.

## C-problems