# Probability that rectangle is inside a circle

We randomly uniformly pick point $x$ on a unit circumference $\{x^2 + y^2 = 1\}$. Then we randomly uniformly pick point $y$ on a unit circle $\{x^2 + y^2 \leq 1\}$. Let $R$ be a rectangle with diagonal $xy$, which sides are parallel to coordinate axes. What is the probability that $R$ lies completely inside the unit circle?

We see, that after $x$ is chosen we have a rectangle $A$, within which our point $y$ should fall to satisfy the condition of the task.

The area of the rectangle $A$ equals $S_A(x) = 4 \sin{(2x)}$ and the probability $P(\{y \in A\}) = \frac{S_A}{S_{circle}} = \frac{S_A}{\pi}.$ How can I find this probability?

The problem for me is that $S_A(x)$ is the transformation of random variable $x \sim unif(0, \frac{\pi}{2})$. For some reason I think that the answer is $\frac{\mathbb{E}[S_A(x)]}{\pi} = \frac{1}{\pi} \cdot \frac{1}{\pi/2 - 0} \cdot \int_0^{\pi/2}S_A(x)dx$, but I do not know why I think so.

• Where did this problem come from? Commented May 30, 2017 at 13:37
• @Frpzzd It's from an old entrance exam to MS CS program in some russian university Commented May 30, 2017 at 13:44
• Do you have a way to access the answer? I just posted my answer, but I'm not sure if it is correct. Commented May 30, 2017 at 13:49

For a given position of $X$, say at coordinates $(\cos\theta,\sin\theta)$ in the first quadrant, the area of the rectangle is indeed

$$A=4\sin\theta\cos\theta.$$

As $Y$ is drawn uniformly in the disk, the probability to fall in $A$ is $A/\pi$.

Then as $X$ is drawn uniformly on the circumference and independently of $Y$, you take the expectation on the first quadrant,

$$P=\frac4\pi\frac 2\pi\int_{\theta=0}^{\pi/2}\sin\theta\cos\theta\,d\theta=\frac4{\pi^2}.$$

• Our answers to this question seem to conflict. Can you find a fault in my answer? You're probably right (I'm not that experienced with probability using integrals) but I would like to know what I did wrong. Commented May 30, 2017 at 14:31
• @Frpzzd: you did not normalize the weight in your final integral.
– user65203
Commented May 30, 2017 at 14:32
• Oh... so I need to add a $\frac{2}{\pi}$. I see. Thank you! I'll change it. Commented May 30, 2017 at 14:35
• @YvesDaoust Why do we take an expectation? Why do we care about expectation at all? Commented May 30, 2017 at 14:37
• @WunschPunsch: because we are interested in the answer.
– user65203
Commented May 30, 2017 at 15:14

Let the coordinates of the first point be $(x_0,y_0)$ and those of the second point be $(x_1, y_1)$. Then the coordinates of the other two points on the rectangle are $(x_0, y_1)$ and $(x_1, y_0)$.

Let us suppose the first point must fall on the quarter-circle in the first quadrant. This does not affect the probability at all because of the symmetry of the circle. Then we have $y_0=\sqrt{1-x_0^2}$. In order to have the rectangle fully inside of the circle, all of the following must be true: $$x_0^2+y_1^2 \le 1$$ $$x_1^2+y_0^2 \le 1$$ We can substitute into the second requirement $y_0=\sqrt{1-x^2}$ to get $$x_1^2+1-x_0^2 \le 1$$ $$x_1^2-x_0^2 \le 0$$ $$x_0^2 \ge x_1^2$$ $$|x_0| \ge |x_1|$$ Similarly, the first requirement can be replaced by $$|y_0| \ge |y_1|$$ and our requirements are $$|x_0| \ge |x_1|$$ $$|y_0| \ge |y_1|$$ Try and visualize this:

Let $F$ be the event in which the rectangle fits, and $P(F|x_0)$ be the probability that the rectangle falls entirely inside of the circle given the value of $x_0$. All points $(x_1, y_1)$ for which the rectangle would fit in the circle are inside of the green rectangle shown with width $2x_0$ and height $2y_0$. This is now a spatial probability problem. Given $x_0$, we can now say that the probability of the rectangle fitting is the probability that $(x_1, y_1)$ lands in that green rectangle. This probability is given by the area of the rectangle over the area of the circle: $$P(F|x_0)=\frac{4x_0y_0}{\pi}$$ $$P(F|x_0)=\frac{4x_0\sqrt{1-x_0^2}}{\pi}$$ We will have to integrate over this to find the total probability. Our first instinct would be to try $$\int_{-1}^{1} P(F|x_0)dx_0$$ but this counts the points on the circumference of the circle unevenly, and not with equal probability. Instead, we should try $$\int_{0}^{\frac{\pi}{2}} P(F|\cos x_0)dx_0$$ Because this sweeps the circumference of the quarter-circle counting all points equally. Thus the probability should be $$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}} \frac{4\cos x_0\sqrt{1-\cos^2 x_0}}{\pi}dx_0$$ We can simplify this to $$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}} \frac{4\cos x_0|\sin x_0|}{\pi}dx_0$$ This value turns out to be about $0.405$.

• Assume that the probability was $1$ instead of $4x_0y_0/\pi$. Your last integral would yield $\pi/2$ !?
– user65203
Commented May 30, 2017 at 14:31
• @Frpzzd 1) No, I do not have answers 2) Why do we integrate $P(F|\cos{x_0})$ instead of $P(F|x_0)$? Our $x_0 \sim unif(0,1)$ (because it is in the first quadrant). What do you mean saying, that $\int_0^1 P(F|x_0) dx_0$ counts points on circumference unevenly? Commented May 30, 2017 at 14:51
• Here is an example of how it counts the points unevenly. Suppose you took a quarter circle and sliced it with the line $x=\frac{1}{2}$. A longer length of circumference would lie to the right of the line than to the left. However, if we sliced it with the line $y=x$, it divides the quarter circle into two arcs of equal length. When we integrate from $0$ to $1$, it is like we are "wiping" a vertical line over the circle... but this does not cut the circle evenly! Instead, we must "wipe" over the circle with an angle at the origin. Commented May 30, 2017 at 14:55

Due to symmetry, without loss of generality the point on circumference is uniformly selected to by in the first quadrant, that is it makes an angle $\theta$, uniform in $(0,\pi/2)$.

The $x$-component of this point is then $X_0= \cos \theta$ and the $y$-component is $Y_0=\sin \theta$.

As for the second point, write $X$ and $Y$ for its $x$- and $y$- components, respectively. In order for the rectangle to be contained in the circle, we must have $(X,Y) \in [-X_0,X_0]\times [-Y_0,Y_0]$.

Because $(X,Y)$ is uniform in the disk, and is independent of $\theta$, conditioning on $\theta$, the probability of $(X,Y)$ landing in this rectangle is the area of this rectangle divided by $\pi$. That is,

$$\frac{ 2 \cos \theta \times 2 \sin \theta}{\pi} = \frac{ 2 \sin (2\theta) }{\pi}.$$

Integrate with respect to $\theta$ to get the answer:

$$p= \frac{2}{\pi} \int_0^{\pi/2} \frac{ 2\sin (2\theta)}{\pi}d\theta= \frac 2{\pi^2} \int_0^{\pi}\sin(u)du =\frac{4}{\pi^2}.$$

• I believe that you should have integrated from $0$ to $\frac{\pi}{2}$ instead of $0$ to $\pi$. Commented May 30, 2017 at 15:18
• I changed $u=2\theta$, but didn't account for that... Will fix now. Thanks. Commented May 30, 2017 at 15:19