Probability that cutting a stick at two points forms a triangle?

Question

Math overflow has given many creative answers to this topic at link:https://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-r

I mostly understand but just need one clarification to put me at ease. The solution I am most comfortable with in the context of my homework relates to uniform joint distributions. I refer to the answer given by "Bill the Lizard".

"The three pieces form a triangle if none of the pieces is greater than half the length of the stick. In other words..."

This conclusion leads to:

$(y > 1/2) AND (x < 1/2) AND (y - x) < 1/2$

$(x > 1/2) AND (y < 1/2) AND (x - y) < 1/2$

Now, my confusion. If none of the pieces is greater than half the length of the stick, then why is $y > 1/2 \;\;and\;\; x > 1/2\;\;$ in the above inequalities??

Thank you.

Answer 1

As you see in the image linked to the answer of "Bill the Lizard":

Now, my confusion. If none of the pieces is greater than half the length of the stick, then why is $y>1/2$ and $x>1/2$ in the above inequalities??

In the first image we have $y> 1/2$ since $y$ is greater than the half of the stick. In the second image we have $x>1/2$ since $x$ is greater than the half of the stick.

$(y > 1/2) AND (x < 1/2) AND (y - x) < 1/2$ comes from the first image, as $x$ appears in the left of y.

$(x > 1/2) AND (y < 1/2) AND (x - y) < 1/2$ comes from the second image, as $x$ appears in the right of y. But do not confuse $x$ and $y$ with pieces, they are only points in the stick. The pieces are their difference.