QUESTION: Point $A$ is chosen randomly from the circumference of the unit circle, while point $B$ is chosen randomly in the interior. A rectangle is then constructed using $A, B$ as opposite vertices, with sides parallel or perpendicular to the coordinate axes. What is the probability that the rectangle lies entirely inside the circle?
MY ANSWER: Here's what I have done..
Without loss of generality, assume that the circle is centered at the origin ($O$). Now, consider the point $A$ in the first quadrant and we will find out in which region point $B$ has to lie in order to satisfy the given condition.. Clearly, the condition is satisfied if $B$ lies anywhere inside the area of the rectangle formed by the points $A$, it's reflection on the $x$ and $y$ axes and reflection about the origin.
Let $\vec{OA}$ make an angle of $\theta$ with the positive direction of the $x$ axis. By symmetry, one side of the above proposed rectangle is $2\cos \theta$ (one which is parallel to the $x$ axis) and the other is $2\sin \theta$ (that which is parallel to $y$ axis).
Therefore the area of the rectangle when $\vec{OA}$ makes an angle of $\theta$ with the positive direction of the $x$ axis is $$A_{\theta}=2\sin \theta . 2 \cos \theta$$ $$A_{\theta}=2 \sin {2 \theta}$$
This is the feasible region for point $B$. Obviously, the total area to choose from is nothing but the total area of the circle and that is $\pi \times 1^2 = \pi$
Hence, the probability in this case becomes $$P_{\theta}=\frac{2 \sin 2 \theta}\pi$$
Now, as we have considered the first quadrant, $\theta$ can vary from $0^{\circ}$ to $90^{\circ}$. We simply need to integrate $P_{\theta}$ from $0$ to $\frac{\pi}2$. That's what we do -
$$\int_{0}^{\frac{{\pi}}2} P_{\theta} \text{ }d \theta$$ $$= \int_{0}^{\frac{\pi}2} \frac{2 \sin 2 \theta}\pi d \theta $$ $$=\frac{2}\pi \int_{0}^{\frac{\pi}2} \sin 2 \theta \text{ } d \theta $$ $$=\frac{2}\pi$$
Now this is the required probability when $A$ lies in the first quadrant. Clearly, according to the symmetry of the problem the probability will be same for all the four quadrants.. And since $A$ can lie in any one of the four quadrants, the answer must simply be
$$P_{\text{total}}=\frac{2}\pi \times 4$$ $$\therefore P_{\text{total}} = \frac{8}\pi$$
$\big($ or one may even think in this way
$A$ can lie either in the $1^{\text{st}}$ quadrant OR $2^{\text{nd}}$ OR $3^{\text{rd}}$ OR $4^{\text{th}}$
$\therefore P_{\text{total}} = \frac{2}\pi +\frac{2}\pi +\frac{2}\pi +\frac{2}\pi = \frac{8}\pi$ $\big)$
Wait........ What!!??
Probability greater than one!!... Reasons?
Either we have just (unknowingly though) made a grand discovery!
Or, we have just made a mistake :P..
And I know that the later case is more believable (hehe :') )... But I cannot find any mistake in this..
Can anyone help me out ?
Thank you so much for your kind support..
Your $P_\theta$ is actually a conditional probability. More precisely if we let $C$ be the event that the rectangle is contained in the circle, then $$P_\theta = \mathbb{P}(C \: | \: A=\theta),$$ where we just represent $A$ by its angle rather than its coordinates. To find $\mathbb{P}(C)$ we use the law of total probability $$\mathbb{P}(C) = \int_0^{2\pi} \mathbb{P}(C \: | \: A=\theta) f_A(\theta) \: d\theta,$$ where $f_A$ is the density function of the angle, which is $\frac{1}{2\pi}$ assuming a uniform distribution. So using the same symmetry arguments that you did, we find that \begin{align*} \int_0^{2\pi} \mathbb{P}(C \: | \: A=\theta) f_A(\theta) \: d\theta &= \frac{1}{2\pi}\cdot4\cdot \int_0^{\pi/2} \frac{2 \sin(2\theta)}{\pi} \: d\theta \\ &= \frac{4}{\pi^2} \end{align*} which is a valid probability.