Distribution of distance of random point on side of equilateral triangle to opposite vertex

Question

I'm trying to find the distribution of the distance from a random (uniformly chosen) point on a side of an equilateral triangle (of side length = 1) to the opposite vertex.

I did the following:

From a triangle with vertexes on points $(0,0),(1,0),(1/2,\sqrt{3}/2)$. Assuming this distribution will be the same for any side on any equilateral triangle of side length = 1, I got a point on the side $(0,0),(1,0)$. So, the distance from any point on this side to the opposite vertex will be:

$$ d^2 = (x-1/2)^2+(y-\sqrt{3}/2)^2$$ $$ d^2 = (x^2 - x + 1)$$

As $y = 0$ on this side.

And $X \sim U(0,1)$

And I need now the distribution for the random variable $d = \sqrt{x^2-x+1}$. However, I can't find a way to do this transformation, as there is no obvious way to isolate x and apply formulas like this: $f_y(y) = f_x[g^{-1}(y)]|\frac{d}{dy}g^{-1}(y)|$

Is there any other way to approach this problem or am I just missing how to transform this random variable?

Answer 1

I propose a solution based:

• on the cdf (cumulative distribution function), the pdf being obtained by differentiation of the cdf.

• on the use of a slightly simpler equilateral triangle with vertices $A(-1,0), B(1,0),C(0,\sqrt{3}).$

Let $M=(X,0)$ with $X$ R.V. uniformly distributed on $(-1,1)$ (with pdf $\frac{1}{2}\mathbb{1_{(-1,1)}}(x)$).

Let $D$ be the random variable distance $CM:=D=\sqrt{X^2+3}$ (Pythagoras).

By definition, the cdf of D is:

$$F_D(d)=P(D<d)=P(D^2<d^2)=P(X^2+3<d^2)=P(X^2<d^2-3)$$

$$F_D(d)=P(-\sqrt{d^2-3}<X<\sqrt{d^2-3})=\frac{1}{2}2\sqrt{d^2-3}=\sqrt{d^2-3}$$

(because $X$ is uniformly distributed on $(-1,1)$). By differentiation:

$f_D(d)=\dfrac{d}{\sqrt{d^2-3}}$ or more precisely $\boxed{f_D(d)=\dfrac{d}{\sqrt{d^2-3}}\mathbb{1_{(\sqrt{3},2)}}(d)}$ by integrating the range of values of $D$ into the expression of the pdf.

Remark: For an equilateral triangle with a different size, one has to replace $f_D(d)$ by $\frac{1}{a}f_D(\frac{d}{a})$ where $a$ is a scaling factor.

Answer 2

It should be intuitively clear that the transformation of $D$ from $X$ is not monotone along a side length of the triangle, since for any such point that is not a midpoint, there is another point whose distance to the opposite vertex is equal.

Consequently, you can exploit additional symmetries of the triangle and consider only the half-edge. That is to say, suppose your $X$ is uniformly chosen on $(0,1/2)$ rather than $(0,1)$ and the resulting transformation $$D = g(X) = \sqrt{X^2 - X + 1}$$ is now monotone on this interval, giving us $$f_D(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right|,$$ and since $f_X = 2$ is uniform on $(0,1/2)$ and $g^{-1}(y) = \frac{1}{2}\sqrt{4y^2-3}$, we easily obtain $$f_D(y) = \frac{4y}{\sqrt{4y^2-3}}, \quad y \in \left[\frac{\sqrt{3}}{2}, 1\right).$$