3. In the rectangle $A P C D$, let $c=|C D|=|A P|$ and $v=|A D|=|C P|$ (Fig. 4, where we have also marked additional lengths that we will derive during the solution). ${ }^{1}$

![](https://cdn.mathpix.com/cropped/2024_04_17_b5b9f153edf17d9fdfd1g-4.jpg?height=359&width=711&top_left_y=877&top_left_x=678)

Fig. 4

From the assumption $S(A P C D)=\frac{1}{2} S(A B C D)$, it follows for the second half of the area of $A B C D$ that $\frac{1}{2} S(A B C D)=S(P B C)$, hence $S(A P C D)=S(P B C)$ or $c v = \frac{1}{2}|P B| v$. Given that $v \neq 0$, it follows that $|P B|=2 c$, and consequently $|A B|=3 c$.

Triangles $C D K$ and $P B K$ have right angles at vertices $C, P$ and congruent (vertical) angles at the common vertex $K$, so they are similar by the AA similarity criterion, with a similarity ratio of $|P B|:|C D|=2 c: c=2$. Therefore, $|P K|:|C K|=2: 1$, from which $|K P|=\frac{2}{3} v$ and $|C K|=\frac{1}{3} v$.

The areas of triangles $A B C$ and $B C K$ are thus given by

$$
S(A B C)=\frac{|A B| \cdot|C P|}{2}=\frac{3 c v}{2} \quad \text { and } \quad S(B C K)=\frac{|C K| \cdot|B P|}{2}=\frac{\frac{1}{3} v \cdot 2 c}{2}=\frac{c v}{3},
$$

so their ratio is

$$
\frac{S(B C K)}{S(A B C)}=\frac{\frac{1}{3} c v}{\frac{3}{2} c v}=\frac{2}{9}
$$

Answer. Triangle $B C K$ occupies $\frac{2}{9}$ of the area of triangle $A B C$.

For a complete solution, award 6 points, of which 1 point for deriving $|B P|=2|C D|$, 2 points for identifying the similarity of triangles $C D K$ and $P B K$ with a ratio of 2, 1 point for determining $|C K|=\frac{1}{3}|A D|$, and the remaining 2 points for formulating and calculating the desired ratio. Tolerate the absence of the argument from the footnote in the students' protocols. [^0]
3. In a right trapezoid $A B C D$ with a right angle at vertex $A$, the base $A B$ contains a point $K$ which is the intersection of the height $C P$ of the trapezoid with its diagonal $B D$. The area of quadrilateral $A P C D$ is half the area of trapezoid $A B C D$. Determine what fraction of the area of triangle $A B C$ is occupied by triangle $B C K$.

