SOLUTION: a point x is selected at a random from a line segment AB with a midpoint O. Find the probability that the line segments AX, XB, and AO can form a triangle?

Question 1156651: a point x is selected at a random from a line segment AB with a midpoint O. Find the probability that the line segments AX, XB, and AO can form a triangle?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Answer: 0.5
This is the same as the fraction 1/2
This is equivalent to 50%

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Explanation:

Set up an xy coordinate system. Plot point A at the origin (0,0).
Plot point B at (1,0). I'm picking this point so that AB is 1 unit long.

Point O is halfway between (0,0) and (1,0), so it is at (0.5, 0)
Note the distance from A to O is 0.5 units.
AO = 0.5

Let point X be some point on segment AB. It has coordinates (k,0) where k is between 0 and 1. It is possible for k to be 0 or 1.
The distance from point A to point X is k units
The distance from point X to point B is 1-k units.
We have these segment lengths
AX = k
XB = 1-k

The triangle we're trying to form has segments that are as long as
AO = 0.5
AX = k
XB = 1-k

Let's say the triangle had side lengths a,b,c such that
a = AO = 0.5
b = AX = k
c = XB = 1-k

By the triangle inequality theorem, the following three inequalities must hold true for a triangle to form
a+b > c
b+c > a
a+c > b
A quick version is that picking any two sides must sum to a value larger than the third side.

Let's see what happens when we substitute in a = 0.5, b = k, c = 1-k

a+b > c turns into 0.5+k > 1-k which solves to k > 0.25
b+c > a turns into k+(1-k) > 0.5 which simplifies to 1 > 0.5 and that is always true regardless of what k is. We can ignore this.
a+c > b turns into 0.5+(1-k) > k which solves to k < 0.75

What we have found is that k > 0.25 and k < 0.75
A more compact way to say this is to write 0.25 < k < 0.75
k is between 0.25 and 0.75; it cannot equal either endpoint.

If 0.25 < k < 0.75, then AO = 0.5, AX = k, XB = 1-k will form lengths that can be made into a triangle. Otherwise, a triangle is not possible.

Note the distance from 0.25 to 0.75 is 0.5 units, which divides over the length of AB to get 0.5/1 = 0.5; this step is fairly trivial.

So overall, that is how the probability is 0.5 which represents a 50% chance. In fraction form, you would write 1/2.

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