3. A trapezoid $A B C D$ is given with bases $A B, C D$, such that $2|A B|=3|C D|$.

a) Find a point $P$ inside the trapezoid such that the areas of triangles $A B P$ and $C D P$ are in the ratio $3: 1$ and the areas of triangles $B C P$ and $D A P$ are in the ratio $3: 1$.

b) For the found point $P$, determine the sequential ratio of the areas of triangles $A B P, B C P$, $C D P$ and $D A P$.
3. Let's assume that point $P$ has the required properties. A line parallel to the bases of the trapezoid and passing through point $P$ intersects sides $A D$ and $B C$ at points $M$ and $N$ respectively (Fig. 1). Let $v$ be the height of the given trapezoid, $v_{1}$ the height of triangle $C D P$, and $v_{2}$ the height of triangle $A B P$.

![](https://cdn.mathpix.com/cropped/2024_04_17_699b1054d41f58a5f164g-3.jpg?height=314&width=1016&top_left_y=511&top_left_x=523)

Fig. 1

a) Since the areas of triangles $A B P$ and $C D P$ are in the ratio $3: 1$, we have

$$
\frac{|A B| v_{2}}{2}: \frac{|C D| v_{1}}{2}=3: 1 \quad \text { or } \quad \frac{v_{1}}{v_{2}}=\frac{1}{3} \cdot \frac{|A B|}{|C D|}=\frac{1}{3} \cdot \frac{3}{2}=\frac{1}{2}
$$

From the marked pairs of similar right triangles, it follows that the heights $v_{2}$ and $v_{1}$ are divided in the ratio 2: 1 by point $M$ on side $A D$ and by point $N$ on side $B C$ (in the case of a right angle at one of the vertices $A$ or $B$, this is evident). This determines the construction of points $M$ and $N$, and thus the segment $M N$. Now we will determine in what ratio the point $P$ divides this segment.

Since the areas of triangles $B C P$ and $D A P$ are in the ratio $3: 1$, we have

$$
\begin{gathered}
\left(\frac{|N P| v_{1}}{2}+\frac{|N P| v_{2}}{2}\right):\left(\frac{|M P| v_{1}}{2}+\frac{|M P| v_{2}}{2}\right)=3: 1, \\
\frac{|N P|\left(v_{1}+v_{2}\right)}{2}: \frac{|M P|\left(v_{1}+v_{2}\right)}{2}=3: 1, \quad|N P|:|M P|=3: 1 .
\end{gathered}
$$

This fully describes the construction of the (unique) point $P$.

b) Complete the triangle $D A C$ to form a parallelogram $D A X C$. Its side $C X$ will divide the segment $M N$ into two parts, and since $v_{1}=\frac{1}{3} v$, the length of the segment $M N$ can be expressed as $|M N|=|M Y|+|Y N|=|A X|+\frac{1}{3}|X B|=|C D|+\frac{1}{3}(|A B|-|C D|)=\frac{1}{3}|A B|+\frac{2}{3}|C D|=$ $=\frac{7}{6}|C D|$, because according to the problem, $|A B|=\frac{3}{2}|C D|$. Therefore,

$$
|M P|=\frac{1}{4}|M N|=\frac{1}{4} \cdot \frac{7}{6} \cdot|C D|=\frac{7}{24}|C D|,
$$

so the ratio of the areas of triangles $C D P$ and $D A P$ is

$$
\frac{|C D| v_{1}}{2}: \frac{|M P|\left(v_{1}+v_{2}\right)}{2}=\left(|C D| v_{1}\right):\left(\frac{7}{24} \cdot|C D| \cdot 3 v_{1}\right)=1: \frac{7}{8}=8: 7
$$

The ratio of the areas of triangles $B C P$ and $C D P$ is therefore $21: 8$ and the ratio of the areas of triangles $A B P$ and $B C P$ is $24: 21$. The consecutive ratio of the areas of triangles $A B P, B C P, C D P$, and $D A P$ is thus $24: 21: 8: 7$.

For a complete solution, award 6 points. In part a), award 1 point for determining the ratio $v_{1}: v_{2}=1: 2$, 1 point for determining the segment $M N$, and 1 point for the final determination of point $P$. In part b), award 3 points, deducting 1 point if the consecutive ratio $24: 21: 8: 7$ is not explicitly stated.