# 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

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Begin by drawing a square, say with opposite corners at $(0,0)$ and $(1,1)$. Make a pencil mark inside the square. Then draw lines from the point to the four sides, with each line intersecting its side of the square perpendicularly. Then edit your question to ask more help. – Dilip Sarwate Mar 17 '14 at 22:28
Draw a square, take a point $P$ inside the square. Draw the two lines through $P$ parallel to the sides until they meet the sides. So you are making a "cross." The altitudes are the $4$ lengths. – André Nicolas Mar 17 '14 at 22:29
Consider the condition that makes the three lengths the sides of a triangle. The answer will then be immediate. – André Nicolas Mar 17 '14 at 23:49
But there ll be more probabilities if you extend the sides of the square? Right? – Valkyrie Mar 17 '14 at 23:54
The probability is independent of the side of the square: it is $1$ for all squares. – André Nicolas Mar 17 '14 at 23:56

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 $1$.

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What do mean by the half units? Why those – Valkyrie Mar 18 '14 at 0:07
Also could you attached a diagram – Valkyrie Mar 18 '14 at 0:08
Labeled with s and t, please – Valkyrie Mar 18 '14 at 0:09
The units thing was a joke. The problem said "unit square" but to avoid fractions I wanted to use a $2\times 2$. The answer is clearly independent of the size of the square, everything scales. Attaching a diagram is a lot of work. Not necessary I think. If you draw a line parallel to a side through a point $P$ in the interior of the square, the two pieces that line is cut into by $P$ can be called $1+x$ and $1-x$, since there are two "general" numbers that add up to $2$. We could have called them $y$ and $2-y$, but that is less elegant. – André Nicolas Mar 18 '14 at 0:14
Why can't there be decimals? Is 1 plus x equal to 2? So x could either 1 or -1? – Valkyrie Mar 18 '14 at 0:20