The Starry Rebound

An (infinitely small) ball starting out in the middle of a 5 pointed star table (outer 5 points 10m radius, inner 5 points 5m radius) has a starting angle of a random value from 0 to 360 degrees. The ball is now set loose and travels around the table. On average, how many sides will have been hit once the ball has traveled 1000 m ?

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This is kind of a cool question. – Alexander Gruber Apr 16 '13 at 5:45
The question has turned up on various websites, going back at least to 2005, but I haven't seen any answers. Is it really homework? It should have an additional tag. I'll edit in "probability", but if someone has a better tag idea, go for it. – Gerry Myerson Apr 16 '13 at 5:51
Thanks Gerry Myerson – Emmanuel Apr 17 '13 at 5:35

To kick start people's thinking.

Each point of the pentacle is an isosceles triangle with base angles $64^\circ$, top angle $52^\circ$. The lengths of the sides are 6.6407, base 5.8779 and height 5.9549.

Without loss of generality, scale the pentacle by $\frac{1}{5.9549}$ and consider the ball travelling a distance of $\frac{1000}{5.9549}$.

If we imagine the point lying sideways with the midpoint of its base at $(0,0)$ and the top at $(1,0)$, the region is bounded by the lines

$$\begin{cases} x=0\\ y=\tan(154^\circ)x-\tan(154^\circ)\\ y=\tan(26^\circ)x-\tan(26^\circ)\\ \end{cases}$$

The ball can enter this region with an angle to the x-axis of $-90^\circ$ to $90^\circ$ and with a y-value of $-\tan(26^\circ)$ to $\tan(26^\circ)$. Due to symmetry we only need to consider positive angles i.e. the ball is heading up.

Let the angle be $\theta_0$ and the position of the ball be $y_0$. The ball travels along the line

$$y=\tan(\theta_0)x+y_0$$

For brevity, let $\tau=\tan(26^\circ)=-\tan(154^\circ)$ and $\gamma_n=\tan(\theta_n)$. Giving $$\begin{cases} x=0\\ y=-\tau x+\tau\\ y=\tau x-\tau\\ \end{cases}$$

$$y=\gamma_0x+y_0$$

The ball strikes the side at the point

$$\left(x_1,y_1\right)=\left(\frac{\tau-y_0}{\gamma_0+\tau},\gamma_0\frac{\tau-y_0}{\gamma_0+\tau}+y_0\right)$$

Having traveled a distance of

\begin{align} d_0&=\sqrt{\left(\frac{\tau-y_0}{\gamma_0+\tau}\right)^2+\left(\gamma_0\frac{\tau-y_0}{\gamma_0+\tau}\right)^2}\\ &=\sqrt{(1+\gamma_0^2)\left(\frac{\tau-y_0}{\gamma_0+\tau}\right)^2} \end{align}

It rebounds along the line

$$y-y_1=\gamma_1(x-x_1)$$

where

$$\gamma_1=\tan(\theta_1)=\tan(154^\circ-\theta_0)$$

Now if $\theta_0\ge64^\circ$ then $\theta_1\le90^\circ$ and the ball is headed back out after 1 bounce, travelling a further distance of

\begin{align} d_1&=\sqrt{x_1^2+\left(y_1-(\gamma_1(0-x_1)+y_1)\right)^2}\\ &=\sqrt{(1+\gamma_1^2)x_1^2}\\ &=\sqrt{(1+\gamma_1^2)\left(\frac{\tau-y_0}{\gamma_0+\tau}\right)^2} \end{align}

if $\theta_0\lt64^\circ$, the ball is still headed in and will strike to other side of the point.

• Where will it hit?
• At what angle will it bounce?
• What are the critical angles for continuing in or coming out?
• Once it is heading out how many more times will it bounce?

Knowing how it behaves in the point, how it behaves in the pentagram is easier and then it hits another point.

Rinse and repeat.

Note also that the initial conditions and the symmetry of the problem $0\le\theta_0\le36^\circ$ and $y_0=\frac{\tan(26^\circ)}{tan(36^\circ)}\tan(\theta_0)$

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