Connect $n$ white and $n$ black points $n$ black and $n$ white points are drawn on plane, so that no three of them lay on one line. How to prove that we can connect each white point to some black point by  straight segment so that no two segments intersect? Each black point should be connected to exactly one white point. 
 A: Consider the vertices of the convex hull of the set. If it contains both black and white points, it also contains two consecutive black and white points that we can connect and then proceed by induction on the rest of the set.
So suppose that all the vertices are black. By rotation, we may assume that the $x$-coordinates of the points are all different. Now the left-most point is black and so is the right-most point. If we pick points in order from left to right, and let $w_k$ be the number of white points among the first $k$ points and $b_k$ be the number of black points among the first $k$ points, then $w_1 = 0$, $b_1 = 1$, $b_{n-1} = n-1$ and $w_{n-1} = n$. Thus the difference $b_k - w_{k}$ goes from $1$ to $-1$, and it must be $0$ in between. Thus there exists a separating line, and we can proceed by induction.
A: Use the extremal principle.
Consider all $n!$ the ways that we can connect the white dots to the black dots. Because this is finite, there exists a way which has minimal total length. I claim that this set satisfies the condition. (Note: Assuming that the problem statement is true, This set is a natural candidate to satisfy the conditions.)
Proof: Suppose not, then there is some pair of line segments which intersect at a point $P$. Let the corresponding dots be $W_N, W_M$ and $B_N, B_M$, where $W_NB_N$ are connected and $W_MB_M$ are connected. By the triangle inequality, 
$$ W_NB_N + W_MB_M \\ = (W_NP + PB_N) + (W_M P + P B_M) \\ = W_NP + PB_M + W_M P + PB_N \\> W_N B_M + W_M B_N$$
The inequality is strict since no three points lie on a line. This contradicts the assumption that the line segments had minimal total length.
