Minimization over two lines This is a minimization question where the minimizing points can be chosen freely on two lines:
$$\mbox{minimize}\, \prod_{i=1}^K {y_i}\quad \mbox{such that}\quad \prod_{i=1}^K {y_i}=1-\prod_{i=1}^K(1-x_i)$$
where every $(x_i,y_i)$ is either a point on the line $l_1:y=\theta(1-x)$ or a point on the line $l_2:y=\theta^{-1}(\theta-x)$. Note that $l_2$ is the inverse function of $l_1$.
The lines $l_1$ and $l_2$ are restricted to the window $[0,1]^2$. $\theta,x_i$, and $y_i$ can take values only in $(0,1)$
If we can select all $(x_i,y_i)$ only on $l_1$ then $y_i=\theta(1-x_i)$. Writing this in the condition of minimization above
$$\prod_{i=1}^K {y_i}=1-\prod_{i=1}^Ky_i/\theta$$ Then simply it gives $$\prod_{i=1}^K {y_i}={\theta^K\over 1+\theta^K}$$
I wonder if we are free to choose every $(x_i,y_i)$ not only on $l_1$ but also on $l_2$, can we get a lower $$\prod_{i=1}^K {y_i}$$ in terms of $\theta$ and $K$.

From  Yulia V's answer:
It seems that the problem further reduces to 
$$\mbox{minimize}\, \frac{\prod_{i=1}^M ((1-\theta)/y_i + \theta)}{\theta^{K-M}}\quad \mbox{such that}\quad \theta^{K-M}\prod_{i=1}^M {y_i}+\prod_{i=1}^M (1- \theta+\theta y_i)>1$$
 A: This is the beginning of your problem study. I believe you can finish it using this approach, but there are many cases to study... Please check my algebra.
Let the answer include $M$ points such that $1-y_i=\frac{x_i}{\theta}$, $K-M$ points such that to $1-x_i=\frac{y_i}{\theta}$.
We will compute the answer for each $M\in\{0,1,\dots,K\}$, then pick $M$ that yields overall minimum product of $y$.
$M<K$ case
Let us do the optimization in $2$ steps. Firstly, for a given set of $y_1,\dots,y_M$ let us find the optimal $y_{M+1},\dots,y_K$. Then, let us optimise the result over $y_1,\dots,y_M$.
The first step consists of minimizing $\prod_{k=1}^K y_k$ given $y_1,\dots,y_M$, $x_1,\dots,x_M$,  and $$\prod_{k=1}^K y_k = 1 - \prod_{k=1}^M (1-x_k) 
\frac{\prod_{k=M+1}^K y_k}{\theta^{K-M} }$$
on $y_{M+1}\in[0,\theta], \dots, y_{K}\in[0,\theta]$. This is equivalent to the following constaint
$$\prod_{k=M+1}^K y_k = \frac{1}{\prod_{k=1}^M y_k +  
\frac{\prod_{k=1}^M (1-\theta + \theta y_k)}{\theta^{K-M} }} < \theta^{M}$$
Thus, we need to minimize 
$\prod_{k=1}^K y_k = \frac{1}{1 +  
\frac{\prod_{k=1}^M (1-\theta + \theta y_k)/y_k}{\theta^{K-M} }}$, or maximize $$\frac{\prod_{k=1}^M (1-\theta)/y_k + \theta }{\theta^{K-M} }$$ conditional on $$\prod_{k=1}^M y_k +  
\frac{\prod_{k=1}^M (1-\theta + \theta y_k)}{\theta^{K-M} } > \theta^{-M}$$
Maximum value can be achieved if $y_i=0$, and this is only possible if $ (1-\theta)^{M}  > \theta^{K-2M}$; the minimum of $\prod_{k=1}^K y_k$ will be $0$.
