[Math] Probability of the Center of a Square Being Contained in A Triangle With Vertices on its Boundary

calculusgeometric-probabilityprobability

Background : I happen to love solving tough problems. Problem is, I simply cannot answer some! It happened again today, as I attempted to solve the questions in this site: http://www.skytopia.com/project/imath/imath.html

This site seems to have really, really difficult questions, and in fact, I'm struggling to answer the second question! That struggle is not without reason, as you shall understand after reading the below question…

The Question is: 3 points are chosen at random along the square's outline. They then combine to form a triangle. What is the probability that the centre of the square will be 'engulfed' by this triangle?

My effort: I divided the square as below (I tried my best people,- assume the gaps between the line segments are points; I've been only ably to name the vertices of the square), –

                                 a1 a2 a3 a4
                               A __ __ __ __ B
                            d1  |           |  b1
                            d2  |           |  b2
                            d3  |           |  b3
                            d4  |__ __ __ __|  b4
                               D             C
                                 c1 c2 c3 c4

After having made these divisions and naming them accordingly…

  • Considering point to be random, evaluated the probability of a point lying on any one of the sixteen segments to be 1/16.

  • Realised centre of the square is engulfed in some cases of 3 points lying on (some examples):- 

    _(a1,b3,c3);(a2,b4,c4);(a3,c1,d1);(a4,c2,d2)_
_(b1,c3,d3);(b2,c4,d4);(b3,d1,a1);(b4,d2,a2)_
_(c1,d3,a3);(c2,d4,a4);(c3,a1,b1);(c4,a2,b2)_
_(d1,a3,b3);(d2,a4,b4);(d3,b1,c1);(d4,b2,c2)_

  • Realised that I'm not accounting for so many more favourable cases (Not able to aptly apply concept of permutation and combination in this question).

  • Realised that I'm not able to calculate the sample space.

  • Realised that the answer is much tougher to compute because I'm not accounting for special cases of collinearity, as well as cases where the vertices and points of separation in my diagram (Not able to assign those points any segment; even if one for each, square's vertices cause problems when evaluating cases, preventing successful generalisation)

A convincing, ingenious, enlightening and accurate solution to question bolded above will be immensely appreciated.

Best Answer

The answer is $\frac{1}{4}$. Given any two points $P$ and $Q$ on the square of side $s$ (regardless of whether they are on the same, different, or adjacent sides), consider the shortest path (of length $d$) along the square between $P$ and $Q$. (This is green in the picture below.) For a random pair of points, $d$ is on average length $s$. For a third point $R$ on the square, $\Delta PQR$ will enclose the center if and only if $R$ is on the reflection (purple) of the path of length $d$ through the center. The purple path has average length $s$, so the probability that $\Delta PQR$ encloses the center is $\frac{s}{4s}=\frac{1}{4}$.

enter image description here

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