What is the probability that the angle is a right angle?

Question

Let's say you have a square, say $ABCD$, with side length 1. A point $P$ is randomly chosen from inside the square, with any point being equally probable of being chosen. What is the probability that the angle $\angle APB$ is a right angle?

This looks simple, and I do this problem for the case of $\angle APB$ being acute or obtuse, and that the set of all points satisfying the condition above is a semicircle inside the square centered at the midpoint of $AB$. How do you do this, or is it impossible?

Answer 1

This question doesn't really make any sense, but the way I would interpret 'probability' of being chosen in this case is in terms of area. Since the area of the region which would fulfil the required criterion is zero (a line), the associated probability is zero.

As for the angle being obtuse, the associated area is a semi-circle centred at the midpoint of the line $AB$ with radius $0.5$ hence its total area is $$\frac12\pi(0.5)^2=\frac18\pi$$ as the total area of the shape is $1\times1=1$ we have $$\text{probability }\angle APB\text{ obtuse}=\frac18\pi$$ $$\text{probability }\angle APB\text{ acute}=1-\frac18\pi$$

Answer 2

The probability is zero. You've correctly identified the set of all points satisfying the desired condition -- the locus is a semicircle. In the acute or obtuse case, you were able to find the locus, and it turned out to have positive two-dimensional area. You were able to use this to calculate a probability. For the right angle case, the locus isn't "two dimensional" -- it's one dimensional. Thus its two-dimensional area is $0$, so the probability is $0$. Intuitively, it is impossible to randomly pick a point exactly on that locus.

To make these ideas more rigorous, you must take a course on measure theory.