A point is selected uniformly at random in the interior of a unit square... From It, the altitude to each side of the square is drawn. For each side, a stick of the altitude's length is obtained. Determine the probability that you can select three of the sticks and arrange them to form a triangle.
Some of the problems I have with this problem are, don't know how to draw the diagram, not understanding the problem.
How could I arrange those sticks. And how to draw the altitudes?
Does the probability vary on the length of squares sides
 A: Without loss of generality we may assume that the square is a $2\times 2$. 
Two what? Two half-units. 
Then the four lengths of the altitudes can be taken to be $1+s$, $1-s$, $1+t$, and $1-t$, where $0\le s\le t\lt 1$. 
We claim that the lengths $1+t$, $1+s$, $1-s$ are the sides of a triangle. 
Since $1+t\ge 1+s\ge 1-s$, it is enough to verify that the sum of the two smaller sides $1+s$ and $1-s$ is greater than the larger side. But this sum is $2$, which is $\gt 1+t$.
The required probability is therefore $1$.  
A: Though it seems correct to have the probability as 1, it is not one because of the fact that there is more than one rule for a triangle. For example, you can have $A + B > C$ and then have $ A>B+C$ So you will need to consider all three of these cases. In order to answer the question, you can view the heights as $x, 1-x, y, 1-y$ and then we will pick 3 of the altitudes. Without loss of generality, let us pick the points $y, 1-y,$ and $x$. We know that $y+1-y> x$ for all x (except exactly one, but we can view this probability as infinitely small.) Then we have $2y+x>1$ and $1+x >2y$ we can graph these two inequalities and then get the area of the intersection of their graphs, this would be the probability. This is because we did not define y and x, so there is no loss of generality. So the probability after graphing this is $1/2$.
A: Let the sides of the square in order be $a,b,c,d$, let the random interior point be $P$ and let the respective altitudes be $t_a$, $t_b$, $t_c$, $t_d$.  Without loss of generality suppose $t_a \leq t_b,t_c,t_d$.
Then we have $t_b=1-t_d\leq 1-t_a=t_c$.
So $t_b\leq t_c$ and similarly $t_d\leq t_c$
Also, since $P$ is an interior point, $t_c<1$ and so $t_c<t_b+t_d=1$.
Therefore the altitudes $t_b,t_c,t_d$ can be arranged to form a triangle, so the probability is 1.
