# Possible shorter solution to this problem?

How many pairs of diagonals of a regular decagon are parallel?

The answer is $45$, and is computed via: $$5\binom{4}{2} + 5 \binom{3}{2}$$

which comes as a result of fixing one vertex and choosing $2$ of the $4$ possible parallel diagonals with that vertex fixed, and then fixing a side and choose $2$ of the $3$ possible parallell diagonals with that side fixed.

However, the answer $45$ is also just $\binom{10}{2}$. Is this just a coincidence or is there another, shorter, way to approach this problem?

• Law of small numbers? – Hagen von Eitzen Apr 13 '14 at 18:11
• Well, for each pair of parallel sides there is exactly one pair of parallel diagonal lines. Thus the number of ways to choose 2 sides of the 10 is the same as the number of parallel diagonal lines is what I'd reason is happening. – Ukhrir Apr 13 '14 at 18:13
• @Ukhrir not necessarily. Check the edit to the OP post. – 1110101001 Apr 13 '14 at 18:20
• Why not try it in hexagon? – evil999man Apr 13 '14 at 18:42
• Doing it in octagon and 12-gon will lead you to believe that it is just another mathematical co-incidence – evil999man Apr 13 '14 at 18:45

In a decagon, there are $10\choose2$ ways to pick pairs of vertices that are 1 side apart. Connecting these pairs gives pairs of parallel diagonals. But for each pair, only 5 of the other 9 pairs don't share a share a vertex or result in a diagonal that is just a side of the decagon, so there are $\frac59{10\choose2}$ ways here.

Now we check pairs of vertices where the vertices are 2 sides apart. There are still $10\choose2$ ways to do this where only $\frac39$ don't share vertices, don't give us sides of the decagon, and don't intersect with each other. By intersect, I mean that if you draw a segment that conjoins one pair's points and another for the other pair, then the two don't intersect. Allowing for this would make us double count some pairs of diagonals. So there are $\frac39{10\choose2}$ ways here.

We do the same for pairs of vertices that are 3 sides apart and find that there are $\frac19{10\choose2}$ possibilities

There are none for pairs of vertices that are further apart.

$$\left(\frac59+\frac39+\frac19\right){10\choose2}={10\choose2}$$
$$\left(\frac1{n-1}\sum_{i=2}^{n/2-1}n-2i-1\right)\times{n\choose2}$$