Random Triangles – II

Last week’s Riddler from Allen Gu was about determining the probability of a random triangle T inscribed in an equilateral triangle containing the centroid point C of the main triangle. The result as determined by multiple solvers was:

P(C \in T) = \frac{1}{3} ln(4)

You can see our solution to the equilateral problem in the post previous to this.

As a follow up we looked at how this probability changes as the shape of the main triangle was allowed to change. The surprising (?) result is that it does not change, and is independent of the main triangle shape.

To set this up, we fix the base of the main triangle on the x-axis with one vertex on the origin, and allow the other two sides to be determined by an arbitrary point P with coordindates (Px, Py). The centroid of the traingle is C= ((L1+Px)/3,Py3). The sides of the triangle now have lengths L1, L2, L3 consistent with forming the triangle established by the base length L1 and arbitrary point P.

We see that the probability set up is identical to the equilateral case with a few modifications to normalize the probability distributions for sides that have lengths not necessarily equal to 1.0 :

P = 1-\frac{1}{L1\cdot\,L2} \int_{L1/2}^{L1} d2(x1)  dx1 -\frac{1}{L2\cdot\,L3} \int_{L2/2}^{L2} d3(x2) dx2 -  \frac{1}{L1\cdot\,L3} \int_{L3/2}^{L3} d1(x3)  dx3

We will evaluate the first integral in the above expression. Using equations for the intersection of the dotted line L4 in the picture and L2 we find:

Qx = \frac{x1}{3x1-L1} Px

We then calculate the distance d2(x1) from point P to point Q and find:

d2(x1) = [(Px-\frac{Px\cdot\,x1}{3x1-L1})^2 - (Py-\frac{Py\cdot\,x1}{3x1-L1})^2 ]^\frac{1}{2}

d2(x1) = \frac {2x1-L1}{3x1-L1} L2

And evaluating the first integral we see a familiar result from the original problem:

\frac{1}{L1\cdot\,L2} \int_{L1/2}^{L1} \frac {2x1-1}{3x1-1} L2 \cdot \,dx1 = \frac{3-ln(4)}{9}

We can rotate the triangle and set up the other integrals similarly and will obtain the same result. Therefore the total probability is identical to the original problem:

P(C \in T) = \frac{1}{3} ln(4)

It is somewhat surprising that the result does not depend on the triangle shape at all and is therefore identical to the result from the original problem. Additionally surprising is that the contribution to the probability is the same for each integral term. Some intuitions behind this are that the integration limits along sides are always over half of each side due the centroid constraint and the probability mapping of x to d is independent of the side lengths due to the ratiometric, and eventually logarithmic, nature of the integrals.

Random Triangles – II

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