Here’s a math puzzle.

Walking Ant Probability Problem

Chris Lomont, Nov 2017

An ant is dropped onto a 3-4-5 right triangle, then walks in a straight line until he leaves the triangle. What is the probability of exiting on the length 5 side?

We assume all probabilities are uniform: the area of the triangle and the direction as an angle in $[0,2\pi]$.

Let the triangle be $\Delta ABC$ with point $D=(x,y)$ chosen uniformly by area on the triangle, with $A=(0,3), B=(0,0), C=(4,0)$. Then the ant walks off the 5 length edge with probability $p_x=\frac{\angle ADC}{2\pi}$.

Let $\alpha=\angle DAC,\beta=\angle DCA,\gamma=\angle ADC$. Then $\alpha + \beta + \gamma = \pi$, $\tan(\angle A-\alpha)=\frac{x}{3-y}$, and $\tan(\angle B-\beta)=\frac{y}{4-x}$. Using $\angle A + \angle B + \pi/2 = \pi$, and combining, gives

$$\gamma=\frac{\pi}{2}+\tan^{-1}\frac{x}{3-y}+\tan^{-1}\frac{y}{4-x}$$

Integrating over the triangle, and dividing by the triangle area and $2\pi$, gives the desired probability

$$p=\frac{1}{2\pi}\times\frac{1}{6}\int_\Delta\frac{\pi}{2}+\tan^{-1}\frac{x}{3-y}+\tan^{-1}\frac{y}{4-x} dA$$.

This gives

$$p=\frac{1}{4}+\frac{1}{12\pi}\int_0^3\int_0^{\frac{4}{3}(3-y)} \tan^{-1}\frac{x}{3-y} dx dy +\frac{1}{12\pi}\int_0^4\int_0^{\frac{3}{4}(4-x)} \tan^{-1}\frac{y}{4-x} dy dx $$

The last 2 terms have the same form: $\int_0^a\int_0^{m(a-t)} \tan^{-1}\frac{s}{a-t} ds dt $. Solving,

$$ \begin{aligned} \int_0^a\int_0^{m(a-t)} \tan^{-1}\frac{s}{a-t} ds dt &= \int s\tan^{-1}\frac{s}{a-t}-\frac{a-t}{2}\ln((a-t)^2+s^2)|^{m(a-t)}_0 dt \ &= [m\tan^{-1}m - \frac{1}{2}\ln(m^2+1)]\int_0^a (a-t) dt \ &= [m\tan^{-1}m - \frac{1}{2}\ln(m^2+1)] a^2/2 \end{aligned} $$

Plugging in and simplifying gives

$$p=\frac{12\pi + 16 \ln(4/5)+9\ln(3/5)}{24\pi}\approx 0.39$$