Finance puzzle
Here is the problem :
Two random points, one red and one blue, are chosen uniformly and independently from the interior of a square. To ten decimal places1, what is the probability that there exists a point on the side of the square closest to the blue point that is equidistant to both the blue point and the red point?
I obtained the following probability :
$P(0\le \frac{x_{r}^{2}+y_{r}^{2}-x_{b}^{2}-y_{b}^{2}}{-2y_{b}+2y_{r}} \le 1)$
with $x_{r}, y_{r}, x_{b}, y_{b}$ being the x-y coordinates of the red and blue point. I considered a 1x1 square, so thoses coordinates are between 0 and 1 (randomly).
Can you help me solve this probability ?
Thank you ! :)
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This would need a more generous bouty, I'm afraid.
how is this a Finance puzzle?
I don't know what your "probability P(0 ≤ ... ≤ 1)" means... But given two points A,B, you can compute the intersection of the "perpendicular bisector" of AB, that is the line (p) that goes through the midpoint with a slope -1/(slope of AB), with the border. You can assume without loss of generality that A is in the "right triangle" 0 < |yA| < xA and the relevant border is {x=1}. (I suggest to consider the square [-1, 1]².)
Then the condition that the perpendicular bisector (p) hits the border is -1 ≤ y(x=1) ≤ 1 when y(x) is the equation of (p) based on the midpoint and the slope, cf earlier. Now, for given A you must find the region of all B for which this condition is satisfied. You will see that geometrically it is the symmetric difference of two quarters of ellipses. Algebraically it corresponds to a quadratic equation that must have a solution which must satisfy a double inequality (from -1 ≤ y ≤ 1)