# Probability that the circle will intersect the square.

We are randomly picking up two points inside a square, according to a uniform probability distribution. Draw a line segment which connects those two points. And use the line segment as a diameter for our circle.

What is the probability that the boundary of the circle will intersect the boundary of the square?

I can not come up with the correct and clear idea for the solution. Here is what I tried to do: Let the square sides have length $2a$. Let the points be $(x_1, y_1)$ and $(x_2, y_2)$. Where we have $(0, 0)$ in the leftmost lower corner.

• Are you sure you have it right? You're picking a random point inside a square and using it as the center of a circle. The circle will intersect the square every time. – Jack M Oct 24 '17 at 17:30
• @JackM I'm not sure how it is in English, but where I'm from a disk of radius $r$ and center $c$ is the set of points at distance less than $r$ from $c$. The circle with same center and radius would be the boundary of the disk, or points at distance exactly $r$ from $c$. – N.Bach Oct 24 '17 at 18:08
• I have added the tag "geometric probability". Besides, could you say something about the motivation for this question ? – Jean Marie Oct 24 '17 at 20:04
• I would like to see how much harder it is to solve it in comparison with a general Buffon`s problem. – trafalgarLaww Oct 24 '17 at 20:29

Let $P_1,P_2$ be the picked points and $M$ be the midpoint of $P_1 P_2$.
Our random circle intersects the square iff the distance of $M$ from the boundary of the square is less than the length of $MP$. Thus, assuming that the square is given by $[-1,1]^2$ and $P_1=(x_1,y_1)$, $P_2=(x_2,y_2)$, we want the probability of the event
$$\min\left(1-\left|\frac{x_1+x_2}{2}\right|,1-\left|\frac{y_1+y_2}{2}\right|\right)\leq \frac{1}{2}\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$ with $x_1,x_2,y_1,y_2$ being independent and uniformly distributed random variables over the interval $[-1,1]$. The complementary event is given by $$\left\{\begin{array}{rcl} \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}&\leq& 2-|x_1+x_2|\\\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}&\leq& 2-|y_1+y_2|\end{array}\right.$$ which is equivalent to $$\left\{\begin{array}{rcl} (y_1-y_2)^2&\leq& 4+4x_1x_2-4|x_1+x_2|\\(x_1-x_2)^2&\leq& 4+4y_1y_2-4|y_1+y_2|\end{array}\right.$$ An efficient way for computing this probability is probably to assume that $x_1,x_2$ have already been fixed, compute the area of the subset of the $y_1 y_2$ plane described by the previous inequalities as a function of $x_1$ and $x_2$, then integrate such function over $[-1,1]^2$ with respect to $dx_1\,dx_2$.
• @trafalgarLaww: in order to go from the first system to the second, square both sides and simplify. $|a-b|^2=(a-b)^2$. – Jack D'Aurizio Nov 16 '17 at 16:05