Equilateral Triangle Billiards

In my approach, I turn the triangle so that $C$ is the origin of a coordinate system and $AB$ lies to the right, parallel to the $y$ -axis. The sides of the triangle have length 1, so that the coordinates are $A\ (\tfrac12\sqrt 3, \tfrac12)$ , $B\ (\tfrac12\sqrt 3, -\tfrac12)$ , $C\ (0,0)$ . The ball is launched from point $O\ (\tfrac13\sqrt 3,0)$ .

The bounces may be modeled by reflecting the billiard table in all directions, as shown below. The side $AB$ and its reflections form a regular hexagon. The ball can be thought of as traveling in a straight line through this hexagon until it hits one of its sides.

The points of interest are $P = A\ (\tfrac12\sqrt 3, \tfrac 12)$ ; $Q\ (0,1)$ ; $R\ (-\tfrac12\sqrt 3, \tfrac 12)$ .

Let the random variable $\Theta$ be the direction angle in which the ball is launched (counterclockwise, relative to the $+x$ -axis). We limit ourselves to the upper half of the plane, as the situation is symmetric. Thus $0 \leq \Theta < \pi$ . Let $B = b(\Theta)$ be the number of bounces that takes place: $\begin{array}{ll} 0 \leq \Theta < \theta_P & B = 0\ \text{dark green region} \\ \theta_P \leq \Theta < \theta_Q & B = 1\ \text{medium green region} \\ \theta_Q \leq \Theta < \theta_R & B = 2\ \text{light green region} \\ \theta_R \leq \Theta \leq \pi & B = 3\ \text{white region} \end{array}$ We must determine $\mathbb E B = \sum_{b = 0}^3 b\cdot \mathbb P(B = b) = 3 - \left(\mathbb P(B \leq 2) + \mathbb P(B \leq 1) + \mathbb P(B = 0)\right)$ and since $\mathbb P(\Theta < \theta) = \theta/\pi$ this becomes $\mathbb E B = 3 - \frac 1 \pi (\theta_R + \theta_Q + \theta_P).$

Now $\tan \theta_P = \frac{\tfrac12 - 0}{\tfrac12\sqrt 3 - \tfrac13\sqrt 3} = \frac{\tfrac12}{\tfrac16\sqrt 3} = \sqrt 3; \\ \tan \theta_Q = \frac{1 - 0}{0 - \tfrac13\sqrt 3} = -\frac{1}{\tfrac13\sqrt 3} = -\sqrt 3; \\ \tan \theta_R = \frac{\tfrac12 - 0}{-\tfrac12\sqrt 3 - \tfrac13\sqrt 3} = -\frac{\tfrac12}{\tfrac56\sqrt 3} = \tfrac15\sqrt 3.$

Choosing all $\theta$ s between $0$ and $\pi$ , we obtain $\theta_P = \text{inv tan}\ \sqrt 3;\ \ \ \theta_Q = \pi - \text{inv tan}\ \sqrt 3;\ \ \ \theta_R = \pi - \text{inv tan}\ \tfrac15\sqrt 3.$ Thus we obtain $\mathbb E B = 3 - \frac 1 \pi (\text{inv tan}\ \sqrt 3 + \pi - \text{inv tan}\ \sqrt 3 + \pi - \text{inv tan}\ \tfrac15 \sqrt 3) \\ = 1 + \frac 1 \pi\text{inv tan}\ \tfrac15 \sqrt 3 \approx \boxed{1.106}.$

Let $AB$ be on the left with $BC$ lying horizontally at the bottom. If $M$ denotes the midpoint of side $BC$ and $O$ denotes the centre of the triangle, then it is evident that the only way to obtain $3$ rebounds is when the ball is hit between $\theta$ and $\frac{\pi }{3}$ measured anticlockwise from $OM$ about $O$ . Note that this can also be achieved on side $AC$ , but due to symmetry of the two sides, this can be accounted for in the final calculation.

By forming an equation (using coordinate geometry) for the path after the second rebound, we can solve for the value of $\theta$ in

$\left(\sqrt{3}+\tan\left(\theta -\dfrac{\pi }{6}\right)\right)\left(\dfrac{\tan\left(\frac{\pi }{4}-\theta \right)\left(\frac{\sqrt{3}}{6}\tan(\theta )+\frac{1}{2}\right)+\sqrt{3}}{\tan\left(\frac{\pi }{4}-\theta \right)+\sqrt{3}}\right)-\sqrt{3}=0, \;\; 0\leq \theta \leq \dfrac{\pi }{3}$

and then calculate

$\dfrac{1}{3}(0)+2\times \dfrac{1}{6}(1)+2\times \dfrac{\theta }{2\pi }(2)+2\times \dfrac{\frac{\pi }{3}-\theta }{2\pi }(3)\approx \boxed{1.106}.$

The number of bounces for each section of the table is shown. The region for zero and one hits is straightforward. The boundary between one hit and two occurs when you hit it at a midpoint. It will rebound back on itself and hit the opposite corner.

The boundary between two and three is a little more difficult. I did it algebraically, but below I outline how the geometry works out for those interested. I am assuming side lengths of one. The boundary occurs at $\frac{1}{4}$ the length of the right and bottom sides, halfway between the midpoint and corner.

The angular spread for zero hits is $\ang{120}$ , the one hit regions each span $\ang{60}$ (totaling $\ang{120}$ ), the two hits regions each span $\ang{40.893}$ (totaling $\ang{81.786}$ ), and the three hits region spans $\ang{38.213}$ . The expected value for the hits therefore works out to

$\frac{1}{360} \left(0 \cdot 120 + 1 \cdot 120 + 2 \cdot 81.786 + 3 \cdot 38.213 \right) \approx \boxed{1.106}$

$\large \textbf{Where the three hits boundary occurs:}$

First, I have shown the ray that just intersects the corner when the boundary is reached.

$O$ is where the ball orginates, $V_1$ and $V_2$ are vertices, $M_1$ and $M_2$ are the midpoints of the sides, and lastly $R_1$ and $R_2$ are the points of reflection. Examining the $O M_1 R_1$ triangle, which was posited to have a length of $\frac{1}{4}$ along $M_1 R_1$ and a given length of $\frac{1}{\sqrt{12}}$ along $O M_1$ . It is a right triangle so the other side length can be calculated by Pythagoras, angles given by the law of cosines. The angle of incidence $\theta_1$ :

$\theta_1 = \arccos\left( \frac{\frac{1}{16 }+ \frac{7}{48} - \frac{1}{12}}{2 \cdot \frac{1}{4} \cdot \sqrt{\frac{7}{48}}} \right) \approx \ang{49.107}$

This informs the angle of reflection in triangle $R_1 V_1 R_2$ . The side length $R_1 V_1$ is again assumed to be $\frac{1}{4}$ . The angle at the vertex is known to be $\ang{60}$ . This permits the calculation of the new incidence angle $180 - 60 - 49.107 = 70.893$ . The law of cosines once again comes into play to calculate the other side lengths, though these are not important to the problem at hand.

In the last triangle of interest, $R_2 M_2 V_2$ is a right triangle with one leg being the altitude of the whole table $\left(\frac{\sqrt{3}}{2} \right)$ and other leg length $\frac{3}{10}$ . This gives the appropriate reflection angle of $\ang{70.893}$ . A ball aimed at halfway between the midpoint of a side and the corner will hit the corner after two bounces.

If hit more toward the corner, the ball will bounce three times before reaching the left leg. As it goes further into the corner, the triangle $R_1 V_1 R_2$ diminishes until the ball bounce off the corner and to the midpoint on the left leg. This I also verified algebraically, but the work is very long and tedious.

Hint: Tile the plane with equilateral triangles. Then the problem becomes looking at where a line intersects a triangular lattice. Note that when you reflect the triangle ABC, all reflections of AB form a hexagon.

The answer is

$\frac{\frac{\pi }{3}+2 \left(-\frac{2 \pi }{3}+\pi -\tan ^{-1}\left(\frac{\sqrt{3}}{5}\right)\right)+3 \left(\pi -\left(\pi -\tan ^{-1}\left(\frac{\sqrt{3}}{5}\right)\right)\right)}{\pi }=\frac{\pi +\tan ^{-1}\left(\frac{\sqrt{3}}{5}\right)}{\pi }$

Bonus :

This problem has many extensions:

• You can look at other shapes that tile the plane (triangle, parallelograms, hexagons). The hexagons case seems difficult.

• Look at a circular, elliptical billiard, etc.

• You can ask the expected time (or distance) traveled by the ball before hitting side AB. The integral is not hard but a little annoying to set up.

• You can consider different regions on the side of the triangle for "pockets" instead of AB.

• Look at different starting positions of the ball.

• Look at how the expected number of sides hit or the expected distance traveled varies with the starting position.

  For example, if the ball starts near C, it is easy to see that the distance traveled by the ball will be very close to the sidelength of the triangle, regardless of which direction the ball is shot in.
  

Edit:

Below is a graph of the expected number of sides hit (on the z-axis) versus the starting position of the ball.

A is located at $(1,0)$ , B is located at $(1/2,\sqrt{3}/2)$ , and C is located at $(0,0)$ .