Let's choose two points on the opposite sides of the unit square. What is the probability that the distance between these two points is less than 1.2 
Let's choose two points on the opposite sides of the unit square. What is the probability that the distance between these two points is less than 1.2?
I'm thinking about this problem and I still have no idea how to start it. There are infinite points on the sides of a square and it's a little bit confusing for me. I assume that I should use in some way the Pythagorean theorem, but I still have no idea, how does it work in this case?
 A: Labelling the square's corners as $ABCD$, suppose the opposite sides in question are $AB$ and $DC$, with $P$ on $AB$ such that $AP=a$ and $Q$ on $DC$ such that $QD=b$. Then there is a right-angled triangle with hypotenuse $PQ$ and legs 1 and $|a-b|$. If $PQ\le1.2$, then $|a-b|\le\sqrt{1.2^2-1^2}=\sqrt{0.44}$.
Assuming $a$ and $b$ are uniformly distributed, we can draw a diagram very similar to one I did for a much more popular problem:

The probability is the shaded area. The two unshaded regions form a square with area $(1-\sqrt{0.44})^2=1-2\sqrt{0.44}+0.44$. The probability that $P$ and $Q$ are less than 1.2 apart is then the complement of this, or $2\sqrt{0.44}-0.44=0.88665$.
A: We suppose the points to be uniformly distributed on the respective sides.
You can choose two couples of sides.
 For one couple, denotes by $t$ and $s$ the position of the point on the segment, e.g. $A=(0,t),\;B=(0,s)$,
then you shall have
$$
\left\{ \matrix{
  0 \le t,s \le 1 \hfill \cr 
  0 \le \sqrt {1 + \left( {s - t} \right)^2 }  \le 6/5 \hfill \cr}  \right.\quad 
$$
Now
$$
\eqalign{
  & 0 \le \sqrt {1 + \left( {s - t} \right)^2 }  \le 6/5\quad  \Rightarrow \quad 0 \le 1 + \left( {s - t} \right)^2  \le 36/25\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad 0 \le \left( {s - t} \right)^2  \le 11/25\quad  \Rightarrow \quad  - \sqrt {11} /5 \le \left( {s - t} \right) \le \sqrt {11} /5 \cr} 
$$
And so
$$
\left\{ \matrix{
  0 \le t,s \le 1 \hfill \cr 
   - \sqrt {11} /5 \le \left( {s - t} \right) \le \sqrt {11} /5 \hfill \cr}  \right.
$$
is the portion of the square $1 \times 1$ (on the plane $s,t$) which lies within a stripe of height $\pm  \sqrt {11} /5$ from the diagonal.
That is the same scheme as for the "meeting problem".
Take the area of the two excluded triangles $2\left( {1 - \sqrt {11} /5} \right)^{\,2} /2 = \left( {36 - 10\sqrt {11} } \right)/25$ 
 and deduct from that of the unit square to get 
$$
\left( {10\sqrt {11}  - 11} \right)/25 \approx 0.8866
$$
