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Trapezium in polygon (Posted on 2021-02-18) Difficulty: 2 of 5
Find the number of trapeziums that it can be formed with the vertices of a regular polygon of n sides.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.0000 (1 votes)

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Solution computer exploration and solution (spoiler) | Comment 1 of 3
I'll be using the American "trapezoids".

This looked quite horrendous when in the queue, much more than D2. Differing angles incorporating 1, 2, ... sides' endpoints, and then having to account for trapezoids that occur in more than one angle (slope), make it more difficult.  I'm sure some people would prefer to leave out those trapezoids that are also parallelograms, while others would include them. Those parallelograms also need weeding out of duplicates, so they need to be recognized as parallelograms.

So when it went "live" I wrote a program to actually examine each set of four vertices for a given n-gon, assuring that no trapezoid is either missed or counted twice. Each trapezoid is counted once, and if it's also a parallelogram its also counted in a count of those.

The MATLAB program:

clc
for sides=4:50
   tot=0; para=0;
   deltaAngle=360/sides;
   for vertex=1:sides
      angle=(vertex-1)*deltaAngle+3;   % tilted 3° to avoid div by zero later
      x=cosd(angle); y=sind(angle);
      coord(vertex,1)=x; coord(vertex,2)=y;
   end
   quads=nchoosek([1:sides],4);
   for i=1:size(quads)
      quad=quads(i,:);
      parallelCt=0;
      slope1=(coord(quad(1),2)-coord(quad(2),2))/(coord(quad(1),1)-coord(quad(2),1));
      slope2=(coord(quad(3),2)-coord(quad(4),2))/(coord(quad(3),1)-coord(quad(4),1));
      if abs(slope1-slope2) < .0000001
         parallelCt=parallelCt+1; 
      end
      slope1=(coord(quad(1),2)-coord(quad(4),2))/(coord(quad(1),1)-coord(quad(4),1));
      slope2=(coord(quad(2),2)-coord(quad(3),2))/(coord(quad(2),1)-coord(quad(3),1));
      if abs(slope1-slope2) < .0000001
         parallelCt=parallelCt+1; 
      end
      if parallelCt>0
         tot=tot+1;
         if parallelCt>1
            para=para+1; 
         end
      end
   end
  % disp([sides tot para])
   fprintf('%3d %6d %4d ',sides,tot,para)
end

The figure for trapezoids includes the count of parallelograms from column 3.

  n   trap  para
  4      1    1
  5      5    0
  6      9    3
  7     21    0
  8     30    6
  9     54    0
 10     70   10
 11    110    0
 12    135   15
 13    195    0
 14    231   21
 15    315    0
 16    364   28
 17    476    0
 18    540   36
 19    684    0
 20    765   45
 21    945    0
 22   1045   55
 23   1265    0
 24   1386   66
 25   1650    0
 26   1794   78
 27   2106    0
 28   2275   91
 29   2639    0
 30   2835  105
 31   3255    0
 32   3480  120
 33   3960    0
 34   4216  136
 35   4760    0
 36   5049  153
 37   5661    0
 38   5985  171
 39   6669    0
 40   7030  190
 41   7790    0
 42   8190  210
 43   9030    0
 44   9471  231
 45  10395    0
 46  10879  253
 47  11891    0
 48  12420  276
 49  13524    0
 50  14100  300

I tried looking up 1,5,9,21,30,54 in the OEIS, put came up with nothing.

However, I tried subtracting the parallelograms from the total in each case, and looked up 5,6,21,24 and found A060423,   Number of obtuse triangles made from vertices of a regular n-gon. The triangles are apparently those using the base of the trapezoid and one of the other two vertices. At first I thought that my count of trapezoids should be half of their value, but then realized there are odd values in their list, so the symmetry of the regular figures apparently would find only duplicates if the remaining vertex of the trapezoid were counted separately. (I haven't verified this, but the match in numbers with mine indicate this is so.)

The OEIS gives formulae for the triangles and therefore the non-parallelogram trapezoids:

a(n) = n*(n-1)*(n-3)/8 when n odd; n*(n-2)*(n-4)/8 when n even.

Then to this must be added the parallelograms: (n/2 - 1)*n/4, but only for even n. That adds up to:

n*(n-1)*(n-3)/8 when n is odd

n*(n-2)*(n-4)/8 + (n/2-1)*n/4 when n is even.

Wolfram Alpha simplifies the formula for even to

n * (n - 2) * (n - 3) / 8 when n is even

which works even for n=4.

Edited on February 18, 2021, 12:46 pm
  Posted by Charlie on 2021-02-18 10:54:02

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