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What is the number of rectangles that can be obtained by joining four of the 12 vertices of a 12-sided regular polygon?

The thing that I've tried is to find out at what intervals should I connect to vertices of the polygon so that a rectangle can be formed. But that didn't seem to help at all .

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  • $\begingroup$ Triangles ? Or rectangles ? $\endgroup$
    – Jean Marie
    Mar 3, 2016 at 15:38
  • $\begingroup$ Rectangles ...sorry everyone . $\endgroup$
    – Tejus
    Mar 3, 2016 at 15:49

4 Answers 4

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$3$ squares and $6+6$ non-squares, making $15$ rectangles

enter image description here

Each non-square can be rotated $30^\circ$ to produce a different rectangle, but $180^\circ$ produces the original version. So six purple possibilities and six red possibilities.

Each square can be rotated $30^\circ$ to produce a different square, but $90^\circ$ produces the original version. So three green possibilities.

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Ok, I think you tried to form rectangles. And, I am assuming all the vertices are distinct. So when you are forming a rectangle, it is tantamount to finding the number of ways you can draw its diagonals.

Now, choose the first vertex, that can be done in $\color{red}{12}$ ways(assuming all the vertices are distinct).

So, in order to find diagonally opposite vertex, you can have only one way(because each vertex has unique diametrically opposite vertex in an even sided regular polygon), so, in $\color{red}{1}$ way.

Now, you have to find another vertex. As, only $\color{red}{10}$ points are left, you can do this in $\color{red}{10}$ ways. And its diametrically opposite vertex in $\color{red}{1}$ way.

So, the number of way is $$\color{blue}{12\times 1\times 10\times 1=120}$$

And, here I forgot to remove the repeatings. In this form of counting, each rectangle is counted $8$ times. So divide the above number by $8$. So, the ultimate result is $$\color{blue}{\frac {120}8=15}.$$

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  • $\begingroup$ Answer is 15 and I'm going to explain it int the comments .. $\endgroup$
    – Tejus
    Mar 3, 2016 at 15:58
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    $\begingroup$ You have counted each rectangle $8$ times. $4$ of them come from picking the first vertex and $2$ come from which is the third vertex (which can also be the fourth to give the same rectangle.) $\endgroup$
    – Ross Millikan
    Mar 3, 2016 at 16:01
  • $\begingroup$ @RossMillikan, ok fixed this. Thank you. $\endgroup$
    – user249332
    Mar 3, 2016 at 16:03
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Ok I got it ... Divide the dodecagon into $2$ halves each with $6$ vertices . Take any half and select $2$ points from those $6$ . That would be $\binom 62 = 15$ . The next step is to choose the points from the other half and these points have to be exactly opposite to the ones chosen in the first half ,hence there would be only $1$ way of selecting them and answer would be $15\times 1 =15$

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  • The two ends of each diagonal of the rectangle must be diametrically opposite.

  • Give identical #s to diametrically opposite vertices, viz. $1-2-3-4-5-6-1-2-3-4-5-6$

  • Choose a pair of two different #s, thus $\binom62$ = 15 rectangles.

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