# Geometric Probability of Finding probability that three points within a Square Could form a triangle

I was playing around with different geometrical problems and came across the following problem which I created:

Given a unit square ABCDABCD, what is the probability that after randomly selecting two points P1P_1 and P2P_2 in the square, the lengths AP1AP_1, P1P2P_1P_2, and P2CP_2C could be valid lengths of a triangle’s sides?

I made some rather trivial observations that if a,b,ca,b,c are the sides of such a triangle such that a≤b≤ca \leq b \leq c, then the perimeter PP satisfies √2≤P≤2+√2\sqrt{2} \leq P \leq 2+\sqrt2. Thus:
a+b+c≥√2a+b+c \geq \sqrt 2
a+b≥√2−ca+b \geq \sqrt 2- c
Thus if c≤√22c \leq \frac{\sqrt2}{2}, then clearly a+b≥ca+b \geq c.

I also tried another approach where I considered the circle centered at AA with radius aa and the circle centered at CC with radius cc. Then I tried to find some way to bound the valid area of points from that but could not find anything useful. I would greatly appreciate some assistance on this problem.

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en.wikipedia.org/wiki/… You may use these conditions as an equivalent conditions of the three sides being a valid triangle. You just set A=(0,0)A = (0, 0) and C=(1,1)C = (1, 1), and the xx, yy coordinates of P1,P2P_1, P_2 are just 4 independent uniform (0,1)(0, 1) random variables.
– BGM
2 days ago

Yes I agree that the triangle inequality is very important for this problem. But my main issue was predicting the probability that given the 4 random variables, what probability that they formed a triangle.
– Mathkid
2 days ago

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