TITLE: Prove that the shortest side of one triangle is the longest side of another, given 3 pairs of points.
QUESTION [6 upvotes]: I have received this question and am having great difficulties. I don't even know how to try to solve it.
It goes like this:
6 points are given in a room. These points are pairwise differently distant from each over. However no 3 points may lie on a straight line. Observe all triangles whose corners are the aforementioned points. 
Now prove that there is always one triangle whose shortest side is the longest side of another triangle.

REPLY [2 votes]: COMMENT.- Let $P_1,P_2,\cdots P_6$ be the six points. Because of there are no three collinear points, each triple $(P_i,P_j,P_k)$ determines a triangle so there are $\dbinom 63=20$ distinct triangles. Besides, by hypothesis, there are $\dbinom62=15$ distinct sides and numbering points conveniently we can put for these fifteen sides
$$\overline {P_1P_2}\lt\overline {P_1P_3}\lt\overline {P_1P_4}\lt\overline {P_1P_5}\lt\overline {P_1P_6}\lt\overline {P_2P_3}\lt\overline {P_2P_4}\lt\overline {P_2P_5}$$ $$\overline {P_2P_5}\lt\overline {P_2P_6}\lt\overline {P_3P_4}\lt\overline {P_3P_5}\lt\overline {P_3P_6}\lt\overline {P_4P_5}\lt\overline {P_4P_6}\lt\overline {P_5P_6}$$
Each side
 $\overline {P_iP_j}$ gives rise to four triangles $\triangle{P_iP_jP_k}$ for $k$ distinct of $i$ and $j$.
So there are fifteen sides of different length  with which twenty triangles are formed.
What else? I believe the end is almost given so I invited young beginners to win the offered bounty. Go ahead!