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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)as the picture linkedcan’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 $15$ ($60\div4=15$).

So the final answer is

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

Is my approach correct?

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

1 Answer 1

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  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) enter image description here

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.

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

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