# Odd Man Out

This week’s Classic asks about the probability of a triangle constructed with vertices randomly located on each side of an equilateral triangle containing the center point (centroid). Below is an example of one good and one bad triangle:

Looking at the figure we see that one line of the red triangle creates a boundary excluding the center point, placing it in one of the three regions of the equilateral triangle not inside the red triangle. We can see that the probablity of a random triangle including the center point will be 1 minus the probability of any of the 3 sides of a random triangle excluding the center point. Since only one side of a particular random triangle can create a region excluding the center point, the probabilities of each side doing so are independent and also by symmetry equal. Therefore:

$P(C \in T) = 1- 3*P(C \in R)$

Where C is the center point, T is any random triangle, and R is the region outside the random triangle created by one of of its sides. We set up the problem for calculating the probability of exclusion for one side of the random triangle created by points on the left and bottom sides of the equilateral triangle:

Given the above set up, we note that the center point can only be excluded if x > 0 . Possible ranges for the point on the opposite side are in the range of ‘d’. Valid extrema for the sides for this x point that exclude the center point are shown in green. Noting that the probability distributions for the location of the vertices on all the equilateral sides are uniform, the probability of the center point being included becomes:

$P(C \in T) = 1-3* \int_{0}^{1/2} d \cdot \, dx$

Using similar triangles in the above diagram and the geometry of the equilateral triangle we see that:

$i+j = 1/ \sqrt{3}, \hspace{.5cm} h/j = 2\sqrt{3}x,\hspace{.5cm} i = \sqrt{3}h , \hspace{.5cm} d = 2h$

And therefore:

$d = \frac{4x}{6x+1}$

$P(C \in T) = 1-3* \int_{0}^{1/2} \frac{4x}{6x+1} \, dx$
$P(C \in T) = 1-\frac{1}{3}\Bigl|_{0}^{1/2} [6x-ln(6x+1)+1]$
$P(C \in T) = \frac {1}{3}ln(4)$