Probability of Rectangle in Unit Circle - Explain Solution Please

Question

I have a bit of an odd request regarding this problem:

Let $C$ be the unit circle. Point $a$ is chosen randomly on the boundary of $C$ and another point $b$ is chosen randomly from the interior of $C$ (points are chosen independently and uniformly over their domains). Let $R$ be the rectangle with the sides parallel to the $x$- and $y$- axes with diagonal $ab$. What is the probability that no point of $R$ lies outside of $C$.

Already answered here - This makes sense...

However, when trying to understand a slightly more rigorous solution set out by the course, I couldn't quite wrap my head around the proof:

Let $a = (cos( \theta ),sin(θ))$ and $b = (b_x, b_y)$. We will show that no point of $R$ lies outside $C$ if and only if:

$|b| ≤ |sin(θ)|$ and $|a| ≤ | cos(θ)|$ (1 for reference)

The other two vertices of $R$ are $(cos(θ), b_y)$ and $(b_x,sin(θ))$. If $|b_x| ≤ | cos(θ)|$ and $|b_y| ≤ |sin(\theta)|$, then each vertex $(x, y)$ of $R$ satisfies $x^2 + y^2 ≤ cos^2(θ) + sin^2(θ) = 1$ and no points of $R$ can lie outside of $C$. Conversely if no points of $R$ lie outside $C$, then applying this to the two vertices other than a and b, we find

$cos^2(θ) + b^2 ≤ 1$ and $a^2 + sin^2(θ) ≤ 1$. (2 for reference)

Then, 2 is equivalent to 1 (labeled above)

Would anyone be able to break down why:

1. Part 2 is true: $cos^2(θ) + b^2 ≤ 1$ and $a^2 + sin^2(θ) ≤ 1$
2. Optionally, but appreciated (I can probably work this out given the above is answered). How inequalities in 2 are equivalent to inequalities in 1.

I've been going in circles trying to make this work... pun absolutely intended.

Answer 1

Starting with your equations 1 you can square both sides leaving you with $b^2 \le \sin^2(\theta)$, and $a^2 \le \cos^2(\theta)$.

Recall that $\cos^2(\theta) + \sin^2(\theta) = 1$

So we can add $\cos^2(\theta)$ to the first equation, giving us $\cos^2(\theta) + b^2 \le \cos^2(\theta) + \sin^2(\theta)$, which from the identity above, simplifies to $\cos^2(\theta) + b^2 \le 1$. The same can be done for the other equation by adding $\sin^2(\theta)$, which will leave you with $a^2 + \sin^2(\theta) \le 1$.

Answer 2

Let $a = (\cos(\theta),\sin(\theta))$ and $b = (b_x, b_y)$. We will show that no point of $R$ lies outside $C$ if and only if:

$$\begin{gather}|b_y| \leq |\sin(\theta)| \\ \quad\text{and}\quad \\ |b_x| \leq |\cos(\theta)|. \end{gather}\tag1$$

I have changed the last part a little to use the names $b_x$ and $b_y$ in the inequalities instead of $a$ and $b.$

The proof assumes that as long as all four corners of the rectangle are inside the circle, the rest of the rectangle is too. This is a fact that can be proved separately.

Two corners are inside or on the circle due to the way they were chosen. The coordinates of the other two vertices, as stated in the solution set, are $(\cos(\theta), b_y)$ and $(b_x,\sin(\theta))$.

Any point with coordinates $(x,y)$ is in or on (not outside) the unit circle if and only if $x^2+y^2\leq 1.$ Plug in the coordinates of each of the two questionable corners for $x$ and $y$ and you get the conditions that these two corners are in or on the circle. These conditions are precisely $$ \begin{align} \cos^2(\theta) + b_y^2 &\leq 1,\\ b_x^2 + \sin^2(\theta) &\leq 1. \end{align} \tag2 $$

The rest of the proof (which you already understand, thanks to another answer) is to show that the conditions $(1)$ are also exactly equivalent to the conditions $(2)$ and therefore $(1)$ tells you exactly when the rectangle is completely contained in and on the circle.