Seven points are equally distributed on a circle. How many non-congruent quadrilaterals can be drawn with vertices chosen from among the seven points?
My approach:
If there are $7$ points and we have to choose 4 points then there are $\binom{7}{4}$ ways to construct a quadrilateral.
Hence, there are a total of $35$ ways in a circle to construct a quadrilateral using $7$ points. But these $35$ quadrilaterals include both congruent and non-congruent quadrilaterals. From this how can we find no of non-congruent quadrilaterals?
Also, I am unclear about the concept of non-congruent and congruent quadrilaterals. So please help me understand this with your clear explanation.
Best Answer
This can be answered with Burnside's lemma, because counting quadrilaterals up to congruence is the same as counting the number of orbits of quadrilaterals under the symmetry group of the regular heptagon.
For each symmetry, we must add up the total number of fixed points for that symmetry, and then divide by the number of symmetries. There are 14 symmetries, consisting of 7 rotations and 7 reflections.
The trivial rotation has $\binom74=35$ fixed points. All 6 other rotations have no fixed points.
Each reflection has $\binom 32=3$ fixed points. A quadrilateral with reflection symmetry is uniquely specified by choosing two out of three points which are on one side of the line of symmetry, and then choosing those corresponding points on the other side.
Putting this altogether, the number of quadrilaterals up to congruence is $$ \frac1{14}\left(35 +\color{gray}{6\cdot0}+7\cdot 3 \right)=4. $$ Here are what the four quadrilaterals look like:
Four consecutive vertices.
Three consecutive vertices, and one non-consecutive.
Two pairs of consecutive vertices, such that the pairs are not adjacent to each other.
One pair of consecutive vertices, while the other vertices are by themselves.