[ Ratio of areas of triangles with a common angle ]

On the lateral sides $A D$ and $B C$ of trapezoid $A B C D$, points $P$ and $Q$ are taken respectively, such that $A P: P D=3: 2$. Segment $P Q$ divides the trapezoid into parts, one of which is twice the area of the other. Find the ratio $C Q: Q B$, if $A B: C D=3: 2$.
Let's continue the lateral sides of the trapezoid to their intersection at point $K$. Triangle $KDC$ is similar to triangle $KAB$ with a similarity ratio of $\frac{CD}{AB} = \frac{2}{3}$, hence $\frac{KD}{KA} = \frac{2}{3}$. Let $KD = 2x, AK = 3x$. Then $AD = x$, and since $AP: PD = 3: 2$, we have

$$
PD = \frac{2}{5} AD = \frac{2}{5} x, \quad KP = KD + PD = 2x + \frac{2}{5} x = \frac{12}{5} x
$$

Suppose the area of quadrilateral $PDCQ$ is twice the area of quadrilateral $APQB$. Let $S_{\triangle KDC} = S$. Then

$$
S_{\triangle KAB} = \left(\frac{2}{3}\right)^2 S_{\triangle KDC} = \frac{9}{4} S, \quad S_{ABCD} = \frac{9}{4} S - S = \frac{5}{4} S
$$

$$
S_{PDCQ} = \frac{2}{3} S_{ABCD} = \frac{2}{3} \cdot \frac{5}{4} S = \frac{5}{6} S, \quad S_{\triangle KPD} = S + \frac{5}{6} S = \frac{11}{6} S, \quad \frac{S_{\triangle KDC}}{S_{\triangle KPQ}} = \frac{S}{\frac{11}{6} S} = \frac{6}{11},
$$

and since

$$
\frac{S_{\triangle KDC}}{S_{\triangle KPQ}} = \frac{KD}{KP} \cdot \frac{KC}{KQ} = \frac{2}{\frac{12}{5}} \cdot \frac{KC}{KQ} = \frac{5}{6} \cdot \frac{KC}{KQ} = \frac{6}{11}
$$

we have $\frac{KC}{KQ} = \frac{36}{55}$. Therefore, point $Q$ lies on the extension of side $BC$ beyond point $B$, since $\frac{KC}{KB} = \frac{2}{3} > \frac{36}{55}$.

Now suppose the area of quadrilateral $PDCQ$ is half the area of quadrilateral $APQB$. Then

$$
S_{PDCQ} = \frac{1}{3} S_{ABCD} = \frac{1}{3} \cdot \frac{5}{4} S = \frac{5}{12} S, \quad S_{\triangle KPD} = S + \frac{5}{12} S = \frac{17}{12} S, \quad \frac{S_{\triangle KDC}}{S_{\triangle KPQ}} = \frac{S}{\frac{17}{12} S} = \frac{12}{17}
$$

and since

$$
\frac{S_{\triangle KDC}}{S_{\triangle KPQ}} = \frac{KD}{KP} \cdot \frac{KC}{KQ} = \frac{2}{\frac{12}{5}} \cdot \frac{KC}{KQ} = \frac{5}{6} \cdot \frac{KC}{KQ} = \frac{12}{17}
$$

we have $\frac{KC}{KQ} = \frac{72}{95}$. Therefore, point $Q$ lies on side $BC$, since $\frac{KC}{KB} = \frac{2}{3} < \frac{72}{95}$, and

## Answer

$23: 13$.