Let O(0, 0), A(6, 0), B(6, 6), C(0, 6) let M be the midpoint of OB. find the probability at random:
A) Farther from O than from M.
B) More than twice as far from O as from M.
Answer for a:7/8
answer for b: pi plus 2 over 9
- RealProLv 75 months ago
I am your farther
Do the statements happen to be about randomly picked points in square OABC?
The points closer to one point than another are those on the side of the perpendicular bisector of the line segment between the two points where the closer point is.
The line y = 3 - x defines a small region inside the square where points are closer to O than to M which is 1/8 of the area of the square, hence the rest is 7/8
We want all points P(x,y) such that
|OP| > 2|OM|
Squaring both sides
x^2 + y^2 > 4[ (x-3)^2 + (y-3)^2 ]
3x^2 - 24x + 3y^2 - 24y + 72 < 0
x^2 - 8x + y^2 - 8y + 24 < 0
(x^2 - 8x + 16) + (y^2 - 8y + 16) - 32 + 24 < 0
(x - 4)^2 + (y - 4)^2 < 8
This defines the inside of a circle of radius 2sqrt2 centred at (4,4). Only 2 45° segments of that circle are not in the square, so the area in the square that satisfies the condition is 8pi - 2[ 8pi/4 - 8/2 ] = 4pi + 8
Divide with area of entire square...
(not pi + 2/9)