# Equilateral triangle in a Brownian path

brownian motionreal-analysis

I am curious about the following simple problem but I couldn't do any progress on it. I would like to know whether it is possible to prove (with probabilistic proof) that a brownian trajectory contains the vertices of a equilateral triangle.

## Mathematical Formulation

Let $$(W_t)_{t\in[0,1]}$$ a standard Wiener process on the interval $$[0,1]$$. Does it exist $$\ell>0$$ and $$(t_1,t_2,t_3)\in [0,1]^3$$ such that:
$$\begin{cases} ||(t_1,W_{t_1})-(t_2,W_{t_2})|| = \ell\\ ||(t_3,W_{t_3})-(t_2,W_{t_2})|| = \ell\\ ||(t_1,W_{t_1})-(t_3,W_{t_3})|| = \ell\\ \end{cases} ~~~~\mathrm{?}$$

## Illustration

The problem I mention is then to find a equilateral triangle of length $$\ell$$ as in the drawing.

The answer is yes, here is a sketch of proof. Take a decreasing sequence $$\{t_n\}_{n \ge 0}$$ such that $$W_{t_n} = 0$$ and such that $$\lim_{n \to \infty}t_n = 0$$. Consider the triangles formed by $$(0,0)$$, $$(t_n,0)$$, $$(t_n/2, \sqrt 3 t_n/2)$$. Then, for large values of $$n$$, the probability of lying below the graph of $$W$$ approaches $$1/2$$ and these events are approximately independent for values such that the ratio of the corresponding times is large enough. Pick now two values $$n$$ and $$m$$ such that the third vertex lies above the graph of $$W$$ at $$t_n$$ and below at $$t_m$$. The claim then follows from the intermediate value theorem by having the second vertex tracing out the graph of $$W$$ for $$t \in [t_n,t_m]$$.