, 2022

Let $l = VA = VB = VC = VD$, $\angle AVC = 2\varphi$, and denote by $E$ the intersection point of the altitude $VO$ with $\lambda$. Since $\lambda \cap (ACV) = MP$, $\lambda \cap (BDV) = NQ$, $(ACV) \cap (BDV) = VO$, $E$ will be the intersection point of the diagonals of $MNPQ$ ($E = MP \cap NQ$). Then $VM = \frac{2}{3}l$, $VN = \frac{1}{2}l$, $VP = \frac{1}{3}l$, $VQ = \frac{p}{p+q}l$, and $VE$ is an angle bisector for $\triangle MPV$ and $\triangle NQV$.

Applying the formula $\left(l_c = \frac{2ab \cos \frac{\gamma}{2}}{a+b}\right)$ we derive
$$VE = \frac{2 \cos \varphi \cdot \frac{2}{3} l \cdot \frac{1}{3} l}{\frac{2}{3} l + \frac{1}{3} l} = \frac{2 \cos \varphi \cdot \frac{1}{2} l \cdot \frac{p}{p+q} l}{\frac{1}{2} l + \frac{p}{p+q} l},$$
thus
$$\frac{2}{3} \cdot \frac{1}{3} = \frac{\frac{1}{2} \cdot \frac{p}{p+q}}{\frac{1}{2} + \frac{p}{p+q}},$$
and conclude that $\frac{p}{p+q} = \frac{2}{5}$, i.e., $\frac{p}{q} = \frac{2}{3}$.