Acute Triangle Probability

For the first angle $\theta$ , a random number must be chosen between $0^\circ$ and $180^\circ$ . The probability that it is less than $90^\circ$ is $\frac 1 2$ , assuming that the probability of getting a right angle is exactly zero. The second angle $\phi$ is a random number between $0^\circ$ and $180^\circ - \theta$ , but must be between $90^\circ - \theta$ and $90^\circ$ in order for the triangle to remain acute. Therefore, the probability that $\phi$ is acute is $\frac{90^\circ-(90^\circ-\theta)}{180^\circ-\theta}=\frac{\theta}{180^\circ-\theta}$ . Since there are an infinite number of possibilities, integrate from $\theta = 0^\circ$ to $90^\circ$ to find the average value: $\frac {1}{90} \times \int_{0}^{90} \frac {\theta}{180-\theta} \ d \theta=0.386$ . Multiply by the $\frac 1 2$ probability that angle $\theta$ was acute in the first place and you get $19.3 \%$ .

For triangle to be acute, all its three angles must be less than 90, which provides us following three contraints: $\theta < 90$ $\phi < 90$ $180 - \theta - \phi < 90$ or $\phi > 90 - \theta$

Now making use of Law of Total Expectation , the probability that triangle is acute can be expressed and evaluated as follows:

$P(\theta < 90, 90 - \theta < \phi < 90) = \int_{0}^{90}f_\theta \int_{90-\theta}^{90}f_{\phi|\theta} d\phi d\theta$ where $f_\theta$ is Probability Mass Function of $\theta$ , and $f_{\phi|\theta}$ is Probability Mass Function of $\phi$ conditioned on $\theta$ .

$f_{\theta} = \frac{1}{180},$ $f_{\phi|\theta} = \frac{1}{180-\theta}$ .

The above integral is easy to evaluate and gives the following answer $19.3\%$

Python 2.7:

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from random import random
from time import time
def acute():
    angle_1 = random() * 180
    angle_2 = random() * (180 - angle_1)
    angle_3 = 180 - (angle_1 + angle_2)
    return angle_1 < 90 and\
        angle_2 < 90 and\
        angle_3 < 90
acute_count = 0.0
trials = 0.0
end = time() + 30
while time() < end:
    if acute():
        acute_count += 1
    trials += 1
print "Answer:", 100*acute_count/trials

For those who are confused why the answer is not $25\%$ (as I admit I first was), I'd like to propose a slightly different algebra- and calculus-based solution from how people usually pose this problem (which has been making the Internet rounds for a few years now). We let $\mathrm{d}P$ be the probability differential associated with the portion of the plane bounded by $(\theta, \phi)$ and $(\theta+\mathrm{d}\theta, \phi+\mathrm{d}\phi)$ . As we know that the $\theta$ are equally weighted, we must have that $\mathrm{d}P = f(\theta, \phi)\,\mathrm{d}\theta\,\mathrm{d}\phi$ satisfies $\int_0^{180-\theta} f(\theta, \phi)\,\mathrm{d}\theta\,\mathrm{d}\phi = c\,\mathrm{d}\theta$ for any $\theta$ , where $c$ is a constant. This is satisfied by $f(\theta, \phi) = \frac{c}{180-\theta}$ . Then, as the sum of probabilities over all $(\theta, \phi)$ must be $1$ , we have $\iint \mathrm{d}P = \int_0^{180} \frac{c}{180-\theta}\,\mathrm{d}\theta = 1$ , which gives us $c = \frac{1}{180}$ . Thus, $\mathrm{d}P = \frac{\mathrm{d}\theta\,\mathrm{d}\phi}{180(180-\theta)}$ . Our bounds for the problem are $0 < \theta < 90$ , $0 < \phi < 90$ , and $0 < 180-\theta-\phi < 90$ . The third inequality gives us $180-\theta < 90+\phi$ , or $90-\theta < \phi$ . Thus, we reduce our bounds to $0 < \theta < 90$ , $90-\theta < \phi < 90$ . We then take the integral $P = \int_0^{90}\int_{90-\theta}^{90} \frac{\mathrm{d}\phi\,\mathrm{d}\theta}{180(180-\theta)}$ . We perform the inside integral to get $P = \int_0^{90} \frac{\theta\,\mathrm{d}\theta}{180(180-\theta)}$ . Rewrite this using partial fractions as $P = \int_0^{90} \frac{1}{180-\theta}-\frac{1}{180}\,\mathrm{d}\theta$ . Finally, the last integral gives us $P = \left[-\frac{\theta}{180}-\log(180-\theta)\right]_0^{90} = \log(2)-\frac{1}{2} \approx 0.193$ , or $19.3\%$ .

Note that the equal weighting of the $\theta$ values is not preserved by the integral $\frac{\int_0^{90}\int_{90-\theta}^{90} \mathrm{d}\phi\,\mathrm{d}\theta}{\int_0^{180}\int_{0}^{180-\theta} \mathrm{d}\phi\,\mathrm{d}\theta} = \frac{1}{4}$ . This is what @Calvin Lin means when he says that the "rod" method mentioned below does not use the right probability distribution.