Probability of forming a trapezium using $4$ of the $12$ equally spaced points on the circumference

This is a question from the $$2022$$ APMOPS competition online paper $$1$$ last question :

There are $$12$$ equally spaced point on the circumference. If $$4$$ of the $$12$$ points are to be chosen at random, what is the probability that a quadrilateral having the $$4$$ points chosen as vertices will be trapezium ( Trapezium is a quadrilateral with one pair of parallel sides)?

My attempts:

We choose $$2$$ points as a vertices. We have $$\binom{12}{2}=66$$ ways to choose the vertice. Let us label the vertices as vertices $$1$$.

In the remaining $$10$$ points, we need to choose $$2$$ of the points to form a vertex parallel to vertices $$1$$.

For every vertices $$1$$, we have $$5$$ vertices formed by the remaining $$10$$ points that parallel to the vertices $$1$$.

So the total number of ways to form a trapezium is $$66\times5\div2=165$$ as we double counted the two vertices we choose.

But we also double counted the total number of ways having $$4$$ points chosen as vertices will be a rectangle.

We can still choose a vertices from $$12$$ equally spaced points on the circumference. We can form $$\binom{12}{2}=66$$ vertices by the points.

In this $$66$$ vertices, $$6$$ of the vertices(the $$6$$ of the vertices are all the axis of symmetry of the regular dodecagon)can’t form a rectangle.

For every of the remaining $$60$$ vertices, we have another $$1$$ vertices formed by the remaining $$10$$ points that parallel and equal to the previous vertices we choose.

We counted four times the total number of ways ( two times for the two vertices we chose, and the remaining two times for the remaining two vertices was formed) having $$4$$ points chosen as vertices will be a rectangle. So we need to divide $$4$$ and we get the answer $$15$$ ($$60\div4=15$$).

$$\frac{\text{P(four points form a trapezium)}}{\text{P(four points randomly chosen from the 12 points)}}=\frac{165-15}{\binom{12}{4}}=\frac{150}{495}=\frac{10}{33}$$

Is my approach correct?

• You're saying "vertex 1" means a pair of vertices? Jun 28, 2023 at 12:12
• "parallel to vertex 1" doesn't make sense. Lines or line segments are parallel, not points. Jun 28, 2023 at 12:34

1. You can't say "a vertex parallel to vertices 1". A vertex can't be parallel to another vertex.

2. Are you sure about "For every vertices 1, we have 5 vertices formed by the remaining 10 points that parallel to the vertices 1"? (Assuming that you meant sides)

There are only $$4$$ in this case.

Hints to a Solution

Label the vertices of the $$12$$-gon $$A_1, A_2, A_3, \dots A_{12}$$.

Consider a diagonal $$A_i A_j$$ where $$i$$ and $$j$$ have the same parity. There are $$30$$ of these diagonals and there are $$4$$ other diagonals that are parallel to each of these diagonals, giving a total of $$120$$ trapeziums.

Then, consider a diagonal where $$A_i A_j$$ where $$i$$ and $$j$$ have different parity. There are $$36$$ of these diagonals and there are $$5$$ other diagonals that are parallel to each of these diagonals, giving a total of $$180$$ trapeziums.

However, any trapezium $$A_i A_j A_k A_m$$ where $$A_i A_j \parallel A_k A_m$$ is counted twice. Once when considering $$A_i A_j$$ and once when considering $$A_k A_m$$. Similarly, any parallelogram $$A_i A_j A_k A_m$$ is counted four times. I'd leave the rest to you.

• So, how do we approach this question? Jun 28, 2023 at 13:36
• Then that’s difficult to consider the two parallel lines Jun 28, 2023 at 13:37
• Wait, I am editing my post to include the solution. Jun 28, 2023 at 13:37
• Ok, thanks for your hints. I will try this question again. Jun 28, 2023 at 14:39
• If you can't figure out the solution, tell me and I'll include the full solution. Jun 29, 2023 at 2:08