Number of non-congruent quadrilaterals with vertices chosen from among seven points equally distributed on a circle

Question

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.

Answer 1

For two quadrilaterals to be congruent, corresponding sides and angles must be equal.

Because of the symmetric arrangement of points in this case, if corresponding sides are equal, angles would be too. So focusing only on corresponding sides, here's what we can do.

We'll take side lengths as one more than the (smaller) number of points lying between its two ends. Meaning sides involving adjacent points have a length of $1$ unit. If a side is obtained by skipping one point, it's length is $2$ units.$^{[1]}$

Now, if $x_1$, $x_2$, $x_3$, and $x_4$ are the side-lengths, we have $x_1 + x_2 + x_3 + x_4 = 7$, $x_i$ is a natural number.$^{[2]}$ The number of unique solutions to this equation would be

$$\binom{7-1}{4-1} = 20$$

Rotations of one unique type of quadrilateral correspond to $4$ solutions counted in the result we got above ($20$). For example,

$$(1, 2, 1, 3), \; (3, 1, 2, 1), \; (1, 3, 1, 2), \; (2, 1, 3, 1)$$

Dividing by $4$ would compensate for that. We will have $20/4 = 5$ potential side length tuples.$^{[3]}$

However, we also have the problem of reflections. And I do not have a simple/elegant way to address that.

The good things is $5$ is not too big a number. So, let's list the $5$ solutions and remove any that are extras (reflections).

$$\begin{align*} (1, 1, 1, 4) \\ (1, 1, 2, 3) \\ (1, 1, 3, 2) \\ (1, 2, 1, 3) \\ (2, 2, 2, 1) \\ \end{align*}$$

Of the five above, (1, 1, 2, 3) and (1, 1, 3, 2) are the only cases of reflections. Others are fine. So counting only one of the two and ignoring another, we get our answer - $4$ non-congruent quadrilaterals.

Thanks to joriki for pointing out the error in the tentative solution I started with, and helping me see the way forward.

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Footnotes

$^{[1]}$ We can get away with this because we are not concerned with the actual side lengths, only whether $2$ sides are equal. And because the points are equally distributed, sides spanning equal number of points must be equal.

It's also important to understand that under our system of measurement a side length of $2$ is not necessarily double the side length of $1$ under other systems. But again, we don't care about that.

$^{[2]}$ The sum of the number of points spanned by the fours sides must be equal to the total number of points on the circle, $7$.

$^{[3]}$ Any solution of the form $(a, b, a, b)$ won't give $4$ quadrilaterals but only $2$ (or $1$ if $a = b$). Thankfully, no such solutions exist in this case ($7$ isn't an even number).

Answer 2

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:

1. Four consecutive vertices.

2. Three consecutive vertices, and one non-consecutive.

3. Two pairs of consecutive vertices, such that the pairs are not adjacent to each other.

4. One pair of consecutive vertices, while the other vertices are by themselves.